119689
DIOPHANTUS OF ALEXANDRIA
CAMBRIDGE UNIVERSITY PRESS
KETTTKR LANE, E,C.
C, K. CLAY,
: 100, ^RIIXrCES SXK.EET
: A. ASMKR A3NTI> CO.
: F. A. BROCKIHAUS
: O. JP. JPUOTJSTAM'S SO3STS
MACM IL/tAJST A^TO CO.,
DIOPHANTUS OF ALEXANDRIA
A STUDY IN THE HISTORY
OF
GREEK ALGEBRA
BY
SIR THOMAS L. HEATH, K.C.B.,
SC.D., SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE
SECOND . EDITION
WITH A SUPPLEMENT CONTAINING AN ACCOUNT OF FERMAT'S
THEOREMS AND PROBLEMS CONNECTED WITH DIOPHANTINE
ANALYSIS AND SOME SOLUTIONS OF DIOPHANTINE
PROBLEMS BY EULER
Cambridge :
at the University Press
1910
PREFACE:
HE first edition of this book, which was the first English
Diophantus, appeared in 1885, and has long been out of
print. Inquiries made for it at different times suggested to me
that it was a pity that a treatise so unique and in many respects
so attractive as the Arithmetica should once more have become
practically inaccessible to the English reader. At the same time
I could not but recognise that, after twentyfive years in which so
much has been done for the history of mathematics, the book
needed to be brought up to date, Some matters which in 1885
were still subject of controversy, such as the date of Diophantus,
may be regarded as settled, and some points which then had to
be laboured can now be dismissed more briefly. Practically the
whole of the Introduction, except the chapters on the editions of
Diophantus, his methods of solution, and the porisms and other
assumptions found in his work, has been entirely rewritten and
much shortened, while the chapters on the methods and on the
porisms etc., have been made fuller than before. The new text of
Tannery (Teubner 1893, ^95) has enabled a number of obscure
passages, particularly in Books V and VI, to be cleared up and,
as a basis for a reproduction of the whole work, is much superior to
the text of Bachet. I have taken the opportunity to make my
version of the actual treatise somewhat fuller and somewhat closer
to the language of the original. In other respects also I thought
I could improve upon a youthful work which was my first essay in
the history of Greek mathematics. When writing it I was solely
concerned to make Diophantus himself known to mathematicians,
vi PREFACE
and I did not pay sufficient attention to Fermat's notes on the
various problems. It is well known that it is in these notes that
many of the great propositions discovered by Fermat in the
theory of numbers are enshrined ; but, although the notes are
literally translated in Wertheim's edition, they do not seem to
have appeared in English; moreover they need to be supple
mented by passages from the correspondence of Fermat and from
the Doctrinae analyticae Inventum Novum of Jacques de Billy.
The histories of mathematics furnish only a very inadequate
description of Fermat's work, and it seemed desirable to attempt
to give as full an account of his theorems and problems in
or connected with Diophantine analysis as it is possible to
compile from the scattered material available in Tannery and
Henry's edition of the Oeuvres de Fermat (1891 1896). So much
of this material as could not be conveniently given in the notes
to particular problems of Diophantus I have put together in
the Supplement, which is thus intended to supply a missing
chapter in the history of mathematics. Lastly, in order to make
the book more complete, I thought it right to add some of the
more remarkable solutions of difficult Diophantine problems given
by Euler, for whom such problems had a great fascination ; the last
section of the Supplement is therefore devoted to these solutions.
T. L. H.
October > 1910.
CONTENTS
INTRODUCTION
CHAP. I. Diophantus and his Works
II. The MSS. of and writers on Diophantus
III. Notation and definitions of Diophantus .
IV. Diophantus' methods of solution ....
V. The Porisms and other assumptions in Diophantus
VI. The place of Diophantus
PAGES
113
14 3 i
3253
5493
99no
111127
THE ARITHMETICS
BOOK 1 129143
ll 143156
HI 156168
i> IV 168199
V. 200 225
VI 226246
On Polygonal numbers 247 259
Conspectus of the Arithmetica 260 266
SUPPLEMENT
SECTIONS I V. NOTES, THEOREMS AND PROBLEMS BY FERMAT.
I. On numbers separable into integral squares . . 267277
II. Equation x*~Ay*~i 277 292
III. Theorems and Problems on rational rightangled
triangles 293318
IV. Other problems by Fermat 318 320
V. Fermat's Tripleequations 321328
SECTION VI. SOME SOLUTIONS BY EULER 329380
INDEX: I. GREEK 381382
II. ENGLISH 382387
INTRODUCTION
CHAPTER I
DIOPHANTUS AND HIS WORKS
THE divergences between writers on Diophantus used to begin,
as Cossali said 1 , with the last syllable of his name. There is now,
however, no longer any doubt that the name was Diophantar, not
Diophant&r 2 .
The question of his date is more difficult Abu'lfaraj, the
Arabian historian, in his History of the Dynasties, places Diophantus
under the Emperor Julian (A.D. 3613), but without giving any
authority ; and it may be that the statement is due simply to a
confusion of our Diophantus with a rhetorician of that name,
mentioned in another article of Suidas, who lived in the time of
Julian 8 . On the other hand, Rafael Bombelli in his Algebra,
1 Cossali, Origine, trasporto in Italia, primi prognssi in essa dell* Algebra (Parma,
17979), i. p. 6 1 : "Su la desinenza del nome comincia la diversita tra gli scrittori."
2 Greek authority is overwhelmingly in favour of Diophanto. The following is the
evidence, which is collected in the second volume of Tannery's edition of Diophantus
(henceforward to be quoted as "Dioph.," "Dioph. II. p. 36"" indicating page 36 of
Vol. II., while "Dioph. II. 20" will mean proposition 20 of Book II.): Suidas s.v.
'Tirana (Dioph. n. p. 36), Theon of Alexandria, on Ptolemy's Syntaxis Book I. c. 9
(Dioph. n. p. 35), Anthology, Epigram on Diophantus (Ep. xiv. 126; Dioph. n. p. 60),
Anonymi prolegomena in Introductionem arithmeticam Nicomachi (Dioph. n. p. 73),
Georgii Pachymerae paraphrasis (Dioph. n. p. 122), Scholia of Maximus Planudes
(Dioph. n. pp. 148, 177, 178 etc.), Scholium on lamblichus In Nicomachi arithm. introd.,
ed. Pistelli, p. 127 (Dioph. n. p. 72), a Scholium on Dioph. n. 8 from the MS. "^4"
(Dioph. II. p. 260), which is otherwise amusing (H ^v^ <rov, At^a^re, efy pert, rov
Sarava ft>e/ca TTJS 8v<rKO\ia,s r&v re &\\uv ffov Bewpfi^ruv KO.I By KO.I rov irapbvros Qeujpfj
fiaTos, "Your soul to perdition, Diophantus, for the difficulty of your problems in general
and of this one in particular") ; John of Jerusalem (lothc.) alone ( Vita loawiis Damas
cent xi. : Dioph. n. p. 36), if the reading of the MS. Parisinus 1559 is right, wrote, in
the plural, d>s KvOaybpcu, TJ Ai60<wT<ii, where however At60avrat is clearly a mistake for
3 Atj3<vtos, <ro^WT^s 'Avnoxe^s, r&v
QeoSocriov rov irpe<rflvr4pov
H. D.
2 INTRODUCTION
published in 1572, says dogmatically that Diophantus lived under
Antoninus Pius (138161 A.D.), but there is no confirmation of this
date either.
The positive evidence on the subject can be given very shortly.
An upper limit is indicated by the fact that Diophantus, in his
book on Polygonal Numbers, quotes from Hypsicles a definition
of such a number 1 . Hypsicles was also the writer of the sup
plement to Euclid's Book XIII. on the Regular Solids known as
Book XIV. of the Elements ; hence Diophantus must have written
later than, say, 150 B.C. A lower limit is furnished by the fact that
Diophantus is quoted by Theon of Alexandria 3 ; hence Diophantus
wrote before, say, 350 A.D. There is a wide interval between
150 B.C. and 350 A,D., but fortunately the limits can be brought
closer. We have a letter of Psellus (nth c.) in which Diophantus
and Anatolius are mentioned as writers on the Egyptian method
of reckoning. " Diophantus," says Psellus 3 , " dealt with it more
accurately, but the very learned Anatolius collected the most
essential parts of the doctrine as stated by Diophantus in a
different way (reading eripw) and in the most succinct form,
dedicating (irpooe^v^oG) his work to Diophantus." It would
appear, therefore, that Diophantus and Anatolius were contem
poraries, and it is most likely that the former would be to the
latter in the relation of master to pupil. Now Anatolius wrote
about 2789 A.D., and was Bishop of Laodicea about 280 A.D. We
may therefore safely say that Diophantus flourished about 250 A.D.
or not much later. This agrees well with the fact that he is not
quoted by Nicomachus (about 100 A.D.), Theon of Smyrna (about
130 A,D.) or lamblichus (end of 3rd c.).
1 Dioph. i. p. 4702.
2 Theo Alexandrinus in primum librum Ptolemaci Mathematicae Compositionis (on c.
IX.): see Dioph. II. p. 35, /ca0' & K<d hi6<t>avr6s 0i/<r
K<d foTtixrijs irAvroTe, rb roXXewrXaa w6 JJLCVOV eTSos tic* aiir^jv afrrb TO elflof lomu
* Dioph, II. p. 389 : irepl 8t TTJS alyvirnaKfis ju*0i>Sov Tatirys Atd^cwros
&Kpi(3ffTpoi>t 6 8t \oyic6raros 'ApardXtos T& (rweKTiKdrrara, n&prj TTJS /ear' tKetv
AroXe&ijuej'os ^r^pw (?r^oy or rctfy>y) Aw^^r^ crwoTm/ccirara ff/>o<re0c6viy<re. The MSS.
read frtpw, which is apparently a mistake for &^/>ws or possibly for reti/>v. Tannery con
jectures r$ ra,tp<f 9 but this is very doubtful ; if the article had been there, Aw^<vr^ r$
frratpQ would have been better. On the basis of fr&lpy Tannery builds the further
hypothesis that the Dionysius to whom the Arithmetica is dedicated is none other than
Dionysius who was at the head of the Catechist school at Alexandria 232247 and was
Bishop there 248265 A.D. Tannery conjectures then that Diophantus was a Christian
and a pupil of Dionysius (Tannery, "Sur la religion des derniers mathe'maticiens de
1'antiquiteV* Extrait des Annales de Pkilosophie Chr^tienm, 1896, p. 13 sqq.). It is
however difficult to establish this (Hultsch, art. "Diophantos aus Alexandreia" in Pauly
Wissowa's RealEncyclopddie der classischen. Altertwnswisscnchafteri).
DIOPHANTUS AND HIS WORKS 3
The only personal particulars about Diophantus which are
known are those contained in the epigramproblem relating to him
in the Anthology 1 . The solution gives 84 as the age at which he
died. His boyhood lasted 14 years, his beard grew at 21, he
married at 33; a son was born to him five years later and died, at
the age of 42, when his father was 80 years old. Diophantus' own
death followed four years later 2 . It is clear that the epigram was
written, not long after his death, by an intimate personal friend
with knowledge of and taste for the science which Diophantus
made his lifework 3 .
The works on which the fame of Diophantus rests are :
(1) The Arithmetica (originally in thirteen Books).
(2) A tract On Polygonal Nitmbers.
Six Books of the former and part of the latter survive.
Allusions in the Arithmetica imply the existence of
(3) A collection of propositions under the title of Porisms;
in three propositions (3, 5 and 1 6) of Book V. Diophantus quotes
as known certain propositions in the Theory of Numbers, prefixing
to the statement of them the words " We have it in the Porisms
that ...... " (e%o/iez> ez> rofc Tlopi<rfia<n,v OTL /c.r.e.).
A scholium on a passage of lamblichus where he quotes a
dictum of certain Pythagoreans about the unit being the dividing
line (fjLeffopiov) between number and aliquot parts, says "thus
Diophantus in the Monastica* ...... for he describes as 'parts 1 the
progression without limit in the direction of less than the unit"
Tannery thinks the Mopiaartted may be ancient scholia (now
lost) on Diophantus I. De 3' sqq. 5 ; but in that case why should
Diophantus be supposed to be speaking ? And, as Hultsch
1 Anthology, Ep. xiv. 126; Dioph. n. pp. 60 1.
2 The epigram actually says that his boyhood lasted \ of his life; his beard grew
after tV more ; after J more he married, and his son was born five years later ; the son
lived to half his father's age, and the father died four years after his son. Cantor ( Gesch.
d. Math, I 8 , p. 465) quotes a suggestion of Heinrich Weber that a better solution is
obtained if we assume that the son died at the time when his father's age was double his,
not at an age equal to half the age at which his father died. In that case
$#+3* s # + f* + 5 + (.ff4)+4=,r, or 3^=196 and x=6&.
This would substitute lof for 14, i6 for 21, 25^ for 33, $o for 42, 6i& for 8o r
and 65! for 84 above. I do not see any advantage in this solution. On the contrary,
I think the fractional results are an objection to it, and it is to be observed that the
scholiast has the solution 84, derived from the equation
3 Hultsch, art. Diophantos in PaulyWissowa's RealEncyclopiidie.
4 lamblichus In Nicomachi arithm. introd* p. 127 (ed. Pistelli) ; Dioph. n. p. 72.
5 Dioph. n. p. 72 note.
4 INTRODUCTION
remarks, such scholia would more naturally have been quoted
as o"xp\t,a and not by the separate title Mopiaorifca 1 . It may
have been a separate work by Diophantus giving rules for reckon
ing with fractions ; but I do not feel clear that the reference
may not simply be to the definitions at the beginning of the
Arithmetics
With reference to the title of the Arithmetic^ we may observe
that the meaning of the word dpiQprjTi/cd here is slightly different
from that assigned to it by more ancient writers. The ancients
drew a marked distinction between dpiO^riKri and \oyi,<rrt,Kr),
though both were concerned with numbers. Thus Plato" states
that dpiO/jwjTiKij is concerned with the abstract properties of
numbers (as odd and even, etc.), whereas X07r^ deals with the
same odd and even, but in relation to one another 3 . Geminus also
distinguishes the two terms 3 . According to him dpLffpijTt/ctf deals
with numbers in tltemselves> distinguishing linear, plane and solid
numbers, in fact all the forms of number, starting from the unit,
and dealing with the generation of plane numbers, similar and
dissimilar, and then with numbers of three dimensions, etc.
Xoytor^ on the other hand deals, not with the abstract properties
of numbers in themselves, but with numbers of concrete things
(alcrdyr&v, sensible objects), whence it calls them by the names of
the things measured, eg. it calls some by the names ^\ir^ and
<^aXr779 4 . But in Diophantus the calculations take an abstract
form (except in V. 30, where the question is to find the number
of measures of wine at two given prices respectively), so that the
distinction between \oryi<mKir] and dpidnrjTiicr} is lost.
We find the Arithmetica quoted under slightly different titles,
Thus the anonymous author of prolegomena to Nicomachus*
Introductio Arithmetica speaks of Diophantus' " thirteen Books of
Arithmetic*." A scholium on lamblichus refers to "the last
theorem of the first Book of Diophantus' Elements of Arithmetic
1 Hultsch, lee. cit*
2 Gorgias, 451 B, C : ret jv AXXct KaQAircp $ dpifywjn/cfy ^ Xoyumin^ Jfcet* **^ ^ CL&TQ
yap tan, T6 re apriov ical rb vtparto' faafopet te TWOUTOV, tin Kal irpos atira Kal irpbs
a\\r)\a Trfis ^%et TrX^ous ^TTioxoiret TO irepiTrbv Koi rb Apnov $ Xaywm/nJ.
3 Proclus, Comment, on Euclid I., p. 39, 1440, 7.
4 Cf. Plato, Laws 819 B, c, on the advantage of combining amusement with instruction
in arithmetical calculation, e.g. by distributing apples or garlands M\wv r4 TIVWV
diavojMl Kal <rre<i>di>w) and the use of different bowls of silver, gold, or brass etc. ($t&\as
a/Mi xP&rov Kal XO,\KOV Kal bpytipov Kal roiotrtov nvQv &\\w /tepa?piWes, ol 5 SKas vus
5ia5tfl6vres, 8irep elTrov, els iratSiav frappSTrovres ras r&v AvayieaLw apt^Qif xpfyrw).
5 Dioph. ii. p. 73, 25.
DIOPHANTUS AND HIS WORKS 5
<rT0%eG)<reft>5) V A scholium on one of the epigrams
in Metrodorus' collection similarly speaks of the " Elements of
DiophantusV
None of the MSS. which we possess contain more than the
first six Books of the Arithmetical, the only variation being that
some few divide the six Books into seven 3 , while one or two give
the fragment on Polygonal Numbers with the number VIII. The
idea that Regiomontanus saw, or said he saw, a MS. containing
the thirteen Books complete is due to a misapprehension. There
is no doubt that the missing Books were lost at a very early date.
Tannery 4 suggests that Hypatia's commentary extended only to
the first six Books, and that she left untouched the remaining
seven, which accordingly were first forgotten and then lost; he
compares the case of Apollonius' Conies, the first four Books of
which were preserved by Eutocius, who wrote a commentary on
them, while the rest, which he did not include in his commentary,
were lost so far as the Greek text is concerned. While, however,
three of the last four Books of the Conies have fortunately reached
us through the Arabic, there is no sign that even the Arabians
ever possessed the missing Books of Diophantus. Thus the
second part of an algebraic treatise called the Fakhrl by Abu
Bekr Muh. b. alHasan alKarkhl (d. about 1029) is a collection of
problems in determinate and indeterminate analysis which not
only show that their author had deeply studied Diophantus, but in
many cases are taken direct from t\\t Aritkmetica, with the change,
occasionally, of some of the constants. In the fourth section of
this work, which begins and ends with problems corresponding to
problems in Diophantus Books II. and III. respectively, are 25
problems not found in Diophantus ; but the differences from
Diophantus in essential features (eg. several of the problems lead
to equations giving irrational results, which are always avoided
by Diophantus), as well as other internal evidence, exclude the
hypothesis that we have here a lost Book of Diophantus 6 . Nor is
there any sign that more of the work than we possess was known
1 Dioph. ii. p. 72, 17 ; lamblichus (ed. Pistelli), p. 132, 12.
2 Dioph. ii. p. 62, 25.
8 e.g. Vaticanus gr. 200, Scorialensis 0it5, and the Broscius MS. in the University
Library of Cracow ; the two last divide the first Book into two, the second beginning
immediately after the explanation of the sign for mimts (Dioph. i. p. 14, i).
4 Dioph. II. p. xvii, xviii.
5 See F, Woepcke, Extrait du Fakhri> traitt tfAlgebre par Abou Bekr Mohammed
ben Alhagan Alkarkhi (manuscrit 952, supplement arabe de la biblioth&jue Impfriale), Paris,
1853
<5 INTRODUCTION
to Abu'l Wafa alBuzjam (940998 A.D.), who wrote a "commentary
(tafslr) on the algebra of Diophantus " as well as a " Book of
proofs of the propositions used by Diophantus in his work..."
These facts again point to the conclusion that the lost Books were
lost before the loth c,
Tannery's suggestion that Hypatia's commentary was limited
to the six Books, and the parallel of Eutocius' commentary on
Apollonius 1 Conies, imply that it is the last seven Books, and the
most difficult, which are lost. This view is in strong contrast to
that which had previously found most acceptance among com
petent authorities. The latter view was most clearly put, and
most ably supported, by Nesselmann 1 , though Colebrooke 2 had
already put forward a conjecture to the same effect ; and historians
of mathematics such as Hankel, Moritz Cantor, and Giinther have
accepted Nesselmann's conclusions, which, stated in his own
words, are as follows: (i) that much less of Diophantus is wanting
than would naturally be supposed on the basis of the numerical
proportion of 6 to 13; (2) that the missing portion is not to be
looked for at the end but in the middle of the work, and indeed
mostly between the first and second Books. Nesselmann's general
argument is that, if we carefully read the last four Books, from the
third to the sixth, we find that Diophantus moves in a rigidly
defined and limited circle of methods and artifices, and that any
attempts which he makes to free himself are futile; "as often as
he gives the impression that he wishes to spring over the magic
circle drawn round him, he is invariably thrown back by an
invisible hand on the old domain already known ; we see, similarly,
in halfdarkness, behind the clever artifices which he seeks to use
in order to free himself, the chains which fetter his genius, we hear
their rattling, whenever, in dealing with difficulties only too freely
imposed upon himself, he knows of no other means of extricating
himself except to cut through the knot instead of untying 1 it"
Moreover, the sixth Book forms a natural conclusion to the whole,
in that it consists of exemplifications of methods explained and
used in the preceding Books. The subject is the finding of right
angled triangles in rational numbers such that the sides and area
satisfy given conditions, the geometrical property of the rightangled
triangle being introduced as a fresh condition additional to the
purely arithmetical conditions which have to be satisfied in the
1 Algebra der Griechcn, pp.
3 Algebra of the Hindus^ Note M, p. Ixi.
DIOPHANTUS AND HIS WORKS 7
problems of the earlier Books. But, assuming that Diophantus'
resources are at an end in the sixth Book, Nesselmann has to
suggest possible topics which would have formed approximately
adequate material for the equivalent of seven Books of the
Arithmetica* The first step is to consider what is actually wanting
which we should expect to find, either as foreshadowed by the
author himself or as necessary for the elucidation or completion of
the whole subject. Now the first Book contains problem^ leading
to determinate equations of the first degree ; the remainder of the
work is a collection of problems which, with few exceptions, lead
to indeterminate equations of the second degree, beginning with
simpler cases and advancing step by step to more complicated
questions. There would have been room therefore for problems
involving (i) determinate equations of the second degree and (2)
indeterminate equations of the first There is indeed nothing to
show that (2) formed part of the writer's plan ; but on the other
hand the writer's own words in Def. 1 1 at the beginning of the
work promise a discussion of the solution of the complete or
adfected quadratic, and it is clear that he employed his method of
solution in the later Books, where in some cases he simply states
the solution without working it out, while in others, where the
roots are "irrational/* he gives approximations which indicate
that he was in possession of a scientific method. Pure quadratics
Diophantus regarded as simple equations, taking no account of the
negative root. Indeed it would seem that he adopted as his
ground for the classification of quadratics, not the index of the
highest power of the unknown quantity contained in it, but the
number of terms left in it when reduced to its simplest form. His
words are 1 : " If the same powers of the unknown occur on both
sides, but with different coefficients (JLM) 6fjboir\r)0'fj Se), we must
take like from like until we have one single expression equal to
another. If there are on both sides, or on either side, any terms
with negative coefficients (sv eXXefye<r& TWO, etbrf), the defects must
be added on both sides until the terms on both sides have
none but positive coefficients (Ivwrrap'XpvTa), when we must again
take like from like until there remains one term on each side.
This should be the object aimed at in framing the hypotheses of
propositions, that is to say, to reduce the equations, if possible,
until one term is left equated to one term. But afterwards I will
1 Dioph. I. Def. n, p. 14.
6 INTRODUCTION
to Abu'l Wafa alBuzjam (940998 A.D.), who wrote a "commentary
(tafslr) on the algebra of Diophantus" as well as a "Book of
proofs of the propositions used by Diophantus in his work..,"
These facts again point to the conclusion that the lost Books were
lost before the roth c.
Tannery's suggestion that Hypatia's commentary was limited
to the six Books, and the parallel of Eutocius' commentary on
Apollonius' Conies, imply that it is the last seven Books, and the
most difficult, which are lost. This view is in strong contrast to
that which had previously found most acceptance among com
petent authorities. The latter view was most clearly put, and
most ably supported, by Nesselmann 1 , though Colebrooke 2 had
already put forward a conjecture to the same effect ; and historians
of mathematics such as Hankel, Moritz Cantor, and Giinther have
accepted Nesselmann's conclusions, which, stated in his own
words, are as follows: (i) that much less of Diophantus is wanting
than would naturally be supposed on the basis of the numerical
proportion of 6 to 13; (2) that the missing portion is not to be
looked for at the end but in the middle of the work, and indeed
mostly between the first and second Books. Nesselmann's general
argument is that, if we carefully read the last four Books, from the
third to the sixth, we find that Diophantus moves in a rigidly
defined and limited circle of methods and artifices, and that any
attempts which he makes to free himself are futile ; " as often as
he gives the impression that he wishes to spring over the magic
circle drawn round him, he is invariably thrown back by an
invisible hand on the old domain already known ; we see, similarly,
in halfdarkness, behind the clever artifices which he seeks to use
in order to free himself, the chains which fetter his genius, we hear
their rattling, whenever, in dealing with difficulties only too freely
imposed upon himself, he knows of no other means of extricating
himself except to cut through the knot instead of untying it,"
Moreover, the sixth Book forms a natural conclusion to the whole,
in that it consists of exemplifications of methods explained and
used in the preceding Books. The subject is the finding of right
angled triangles in rational numbers such that the sides and area
satisfy given conditions, the geometrical property of the rightangled
triangle being introduced as a fresh condition additional to the
purely arithmetical conditions which have to be satisfied in the
1 Algtbra der Griechen^ pp. 264273.
3 Algelra of the Hindus, Note M, p. Ixi.
DIOPHANTUS AND HIS WORKS 7
problems of the earlier Books. But, assuming that Diophantus'
resources are at an end in the sixth Book, Nesselmann has to
suggest possible topics which would have formed approximately
adequate material for the equivalent of seven Books of the
Aritkmetica. The first step is to consider what is actually wanting
which we should expect to find, either as foreshadowed by the
author himself or as necessary for the elucidation or completion of
the whole subject. Now the first Book contains problem^ leading
to determinate equations of the first degree ; the remainder of the
work is a collection of problems which, with few exceptions, lead
to indeterminate equations of the second degree, beginning with
simpler cases and advancing step by step to more complicated
questions. There would have been room therefore for problems
involving (i) determinate equations of the second degree and (2)
indeterminate equations of the first. There is indeed nothing to
show that (2) formed part of the writer's plan ; but on the other
hand the writer's own words in Def. 1 1 at the beginning of the
work promise a discussion of the solution of the complete or
adfected quadratic, and it is clear that he employed his method of
solution in the later Books, where in some cases he simply states
the solution without working it out, while in others, where the
roots are " irrational/' he gives approximations which indicate
that he was in possession of a scientific method. Pure quadratics
Diophantus regarded as simple equations, taking no account of the
negative root. Indeed it would seem that he adopted as his
ground for the classification of quadratics, not the index of the
highest power of the unknown quantity contained in it, but the
number of terms left in it when reduced to its simplest form. His
words are 1 : " If the same powers of the unknown occur on both
sides, but with different coefficients (pr) o/j,oTr\y0fj Se), we must
take like from like until we have one single expression equal to
another. If there are on both sides, or on either side, any terms
with negative coefficients (eV eXXefyeo/ TWO, eftty), the defects must
be added on both sides until the terms on both sides have
none but positive coefficients (ewirdp'XpvTa), when we must again
take like from like until there remains one term on each side.
This should be the object aimed at in framing the hypotheses of
propositions, that is to say, to reduce the equations, if possible,
until one term .is left equated to one term. But afterwards I will
1 Dioph. I. Def. u, p. 14.
8 INTRODUCTION
show you also how, when two terms are left equal to one term,
such an equation is solved/' That is to say, reduce the quadratic,
if possible, to one of the forms ax*=.bx> ax*=c> or bx~c\ I will
show later how to solve the equation when three terms are left of
which any two are equal to the third, i.e. the complete quadratic
ax* bx c = o, excluding the case ax* f bx f c  o. The exclusion
of the latter case is natural, since it is of the essence of the work
to find rational and positive solutions. Nesselmann might have
added that Diophantus' requirement that the equation, as finally
stated, shall contain only positive terms, of which two are equated
to the third, suggests that his solution would deal separately with
the three possible cases (just as Euclid makes separate cases of the
equations in his propositions VI. 28, 29), so that the exposition
might occupy some little space. The suitable place for it would
be between the first and second Books. There is no evidence
tending to confirm Nesselmann's further argument that the six
Books may originally have been divided into even more than
seven Books. He argues from the fact that there are often better
natural divisions in the middle of the Books (eg. at II. 19) than
between them as they now stand; thus there is no sign of a
marked division between Books I. and II. and between Books II.
and III., the first five problems of Book II. and the first four of
Book III. recalling similar problems in the preceding Books
respectively. But the latter circumstances are better explained,
as Tannery explains them, by the supposition that the first
problems of Books II. and ill. are interpolated from some ancient
commentary. Next Nesselmann points out that there arc a
number of imperfections in the text, Book V. especially having
been " treated by Mother Time in a very stepmotherly fashion " ;
thus it seems probable that at V. 19 three problems have dropped
out altogether. Still he is far from accounting for seven whole
Books; he has therefore to press into the service the lost
"Porisms" and the tract on Polygonal Numbers.
If the phrase which, as we have said, occurs three times in
Book V., "We have it in the Porisms that.../' indicates that the
" Porisms " were a definite collection of propositions concerning
the properties of certain numbers, their divisibility into a certain
number of squares, and so on, it is possible that it was from the
same collection that Diophantus took the numerous other pro
positions which he assumes, either explicitly enunciating them, or
implicitly taking them for granted. May we not then, says
DIOPHANTUS AND HIS WORKS 9
Nesselmann, reasonably suppose the " Porisms " to have formed
an introduction to the indeterminate and semideterminate analy
sis of the second degree which forms the main subject of the
Arithmetica, and to have been an integral part of the thirteen
Books, intervening, probably, between Books I. and II. ? Schulz, on
the other hand, considered this improbable, and in recent years
Hultsch 1 has definitely rejected the theory that Diophantus filled
one or more Books of his Arithmetic** exclusively with Porisms.
Schulz's argument is, indeed, not conclusive. It is based on the
consideration that ff Diophantus expressly says that his work deals
with arithmetical problems* '; but what Diophantus actually says is
" Knowing you, O Dionysius, to be anxious to learn the solution
(or, perhaps, ( discovery/ evpecrw) of problems in numbers, I have
endeavoured, beginning from the foundations on which the study is
built up, to expound (VTTOCTT^CTYU = to lay down) the nature and
force subsisting in numbers," the last of which words would easily
cover propositions in the theory of numbers, while " propositions/'
not " problems/ 3 is the word used at the end of the Preface, where
he says, "let us now proceed to the propositions (Trpordcrew)
which have been treated in thirteen Books."
On reconsideration of the whole matter, I now agree in the
view of Hultsch that the Porisms were not a separate portion of
the Arithmetica or included in the Arithmetica at all. If they had
been, I think the expression " we have it in the Porisms " would
have been inappropriate. In the first place, the Greek mathe
maticians do not usually give references in such a form as this
to propositions which they cite when they come from the same
work as that in which they are cited ; as a rule the propositions
are quoted without any references at all. The references in this
case would, on the assumption that the Porisms were a portion of
the thirteen Books, more naturally have been to particular pro
positions of particular Books (cf. Eucl. XII. 2, " For it was proved
1 Hultsch, /oc. cit.
2 The whole passage of Schulz is as follows (pref. xxi) : " Es 1st daher nicht unwahr
scheinlich, class diese Porismen erne eigene Schrift unseres Diophantus waren, welche
vorzuglich die Zusammensetzung der Zahlen aus gewissen Bestandtheilen zu ihrem
Gegenstande hatten. Konnte man diese Schrift als einen Bestandtheil des grossen in
dreizehn BUchern abgefassten arithmetischen Werkes ansehen, so ware es sehr erklarbar,
dass gerade dieser Theil, der den blossen Liebhaber weniger anzog, verloren ging. Da
indess Diophantus ausdrlicklich sagt, sein Werk behandele arithmetische Problems^ so hat
wenigstens die letztere Annahme nur einen geringen Grad von Wahrscheinlichkeil."
io INTRODUCTION
. in the first theorem of the loth Book that..."). But a still vaguer
reference would have been enough, even if Diophantus had chosen
to give any at all ; if the propositions quoted had preceded those
in which they are used, some expression like rovro <yap 7iy>o
i,, (t for this has already been proved,'* or SeSet/crat yap
o, " for this has been shown/' would have sufficed, or, if the
propositions occurred later, some expression like w? l^s Se^rjaerai
or Sei%0?;Wra4 v^ 7)p,S>v vvrepov, " as will be proved in due course"
or "later." The expression "we have it in the Porisms" (in the
plural) would have been still more inappropriate if the "Porisms"
had been, as Tannery supposes 1 , not collected together as one or
more Books of the Arithmetica, but scattered about in the work as
corollaries to particular propositions 2 . And, as Hultsch says, it is
hard, on Tannery's supposition, to explain why the three particular
theorems quoted from "the Porisms" were lost, while a fair
number of other additions survived, partly under the title iropia^a
(cf. I. 34, i. 38), partly as " lemmas to what follows/' X^/xa efc TO
6^9 (cf. lemmas before IV. 34, 35, 36, V. 7, 8, VI. 12, 15). On the
other hand, there is nothing improbable in the supposition that
Diophantus was induced by the difficulty of his problems to give
place in a separate work to the "porisms" necessary to their
solution,
The hypothesis that the Porisms formed part of the Arithmet
ica being thus given up, we can hardly hold any longer to
Nesselmann's view of the contents of the lost Books and their
place in the treatise; and I am now much more inclined to the
opinion of Tannery that it is the last and the most difficult Books
which are lost. Tannery's argument seems to me to be very
attractive and to deserve quotation in full, as finally put in the
preface to Vol. II. of his Diophantus 3 . He replies first to the
assumption that Diophantus could not have proceeded to problems
more difficult than those of Book V. " But if the fifth or the sixth
Book of the Arithmetica had been lost, who, pray, among us would
have believed that such problems had ever been attempted by the
Greeks? It would be the greatest error, in any case in which a
1 Dioph. n. p. xix.
2 Thus Tannery holds (loc. cit.) that the solution of the complete quadratic was given
in the form of corollaries to I. 27, 30 ; and he refers the three "porisms" quoted in V. 3,
5, 16 respectively to a second (lost) solution of m. io, to in. 15, and to iv. i, a.
* Dioph. II. p. xx.
DIOPHANTUS AND HIS WORKS n
thing cannot clearly be proved to have been unknown to all the
ancients, to maintain that it could not have been known to some
Greek mathematician. If we do not know to what lengths
Archimedes brought the theory of numbers (to say nothing of
other things), let us admit . our ignorance. But, between the
famous problem of the cattle and the most difficult of Diophantus'
problems, is there not a sufficient gap to require seven Books to
fill it? And, without attributing to the ancients what modern
mathematicians have discovered, may not a number of the things
attributed to the Indians and Arabs have been drawn from
Greek sources ? May not the same be said of a problem solved by
Leonardo of Pisa, which is very similar to those of Diophantus but
is not now to be found in the Arithmetical In fact, it may fairly
be said that, when Chasles made his reasonably probable restitution
of the Porisms of Euclid, he, notwithstanding the fact that he had
Pappus' lemmas to help him, undertook a more difficult task than
he would have undertaken if he had attempted to fill up seven
Diophantine Books with numerical problems which the Greeks
may reasonably be supposed to have solved."
On the assumption that the lost portion came at the end of the
existing six Books, Schulz supposed that it contained new methods
of solution in addition to those used in Books I. to VI., and in
particular extended the method of solution by means of the double
equation (Si7r\fj laory? or SwrXofardr?^), By means of the double
equation Diophantus shows how to find a value of the unknown
which will make two expressions (linear or quadratic) containing it
simultaneously squares. Schulz then thinks that he went on, in
the lost Books, to make three such expressions simultaneously
squares, i.e> advanced to a triple equation. But this explanation
does not in any case take us very far.
Bombelli thought that Diophantus went on to solve deter
minate equations of the third and fourth degree 1 ; this view,
however, though natural at that date, when the solution of cubic
and biquadratic equations filled so large a space in contemporary
investigations and in Bombelli's own studies, has nothing to
support it
Hultsch 2 seems to find the key to the question in the fragment
of the treatise on Polygonal Numbers and the developments to
1 Cossali, I. pp. 75, 76. 2 Hultsch, loc. cit.
12 INTRODUCTION
which it might have been expected to lead. In this he differs
from Tannery, who says that, as Serenus' treatise on the sections
of cones and cylinders was added to the mutilated Conies of
Apollonius consisting of four Books only, in order to make up a
convenient volume, so the tract on Polygonal Numbers was added
to the remains of the Arithmetic^ though forming no part of the
larger work 1 . Thus Tannery would seem to deny the genuineness
of the whole tract on Polygonal Numbers, though in his text he
only signalises the portion beginning with the enunciation of the
problem " Given a number, to find in how many ways it can be
a polygonal number "as a "vain attempt by a commentator " to
solve this problem. Hultsch, on the other hand, thinks we may
conclude that Diophantus really solved the problem. He points
out moreover that the beginning of the tract is like the beginning
of Book I. of the Arithmetica in containing definitions and pre
liminary propositions. Then came the difficult problem quoted,
the discussion of which breaks off in our text after a few pages ;
and to this it would be easy to tack on a great variety of other
problems. Again, says Hultsch, the supplementary propositions
added by Bachet may serve to give an approximate idea of the
difficulty of the problems which were probably treated in Books VII.
and the following. And between these and the bold combination
of a triangular and a square number in the CattleProblem
stretches, as Tannery says, a wide domain which was certainly
not unknown to Diophantus, but was his huntingground for the
most various problems. Whether Diophantus dealt with plane
numbers, and with other figured numbers, such as prisms and
tetrahedra, is uncertain.
The name of Diophantus was used, as were the names of Euclid,
Archimedes and Heron in their turn, for the purpose of palming
off the compilations of much later authors. Tannery prints in
his edition three fragments under the head of "Diophantus
Pseudepigraphus." The first 2 , which is not " from the Arithmetic
of Diophantus " as its heading states, is worth notice as containing
some particulars of one of " two methods of finding the square
root of any square number"; we are told to begin by writing the
number " according to the arrangement of the Indian method," ie,
according to the Indian numerical notation which reached us
through the Arabs. The fragment is taken from a Paris MS.
1 Dioph. ii. p. xviii. 2 Dioph, n. p, 3, 314.
DIOPHANTUS AND HIS WORKS 13
(Supplem, gr, 387), where it follows a work with the title '
T?)9 fieyaXys ical 'Iz/Sttf?)? ^fyifyopias (i.e. ^^o^opia^ written in
1252 and raided about half a century later by Maximus Planudes,
The second fragment 1 is the work edited by C. Henry in 1879 as
Opusculum de multiplicatione et dwisione sexagesimalibus Dwphanto
vel Pappo attribmndum. The third 2 , beginning with kiQ(f>dvrov
emTreSoperpMa, is a compilation made in the Byzantine period out
of late reproductions of the yecofnGrpov/jieva and arepeoiJieTpov^va
of Heron, The second and third fragments, like the first, have
nothing to do with Diophantus,
1 Dioph, IL p. 3, 1515, 17. 3 Dioph. 11. p. 15, 1831, n.
CHAPTER II
THE MSS. OF AND WRITERS ON DIOPHANTUS
FOR full details of the various MSS. and of their mutual
relations, reference should be made to the prefaces to the first and
second volumes of Tannery's edition 1 . Tannery's account needs
only to be supplemented by a description given by Gollob 2 of
another MS. supposed by Tannery to be nonexistent, but actually
rediscovered in the Library of the University of Cracow (Nr 544).
Only the shortest possible summary of the essential facts will be
given here.
After the loss of Egypt the work of Diophantus long remained
almost unknown among the Byzantines ; perhaps one copy only
survived (of the Hypatian recension), which was seen by Michael
Psellus and possibly by the scholiast to lamblichus, but of which
no trace can be found after the capture of Constantinople in 1204.
From this one copy (denoted by the letter a in Tannery's table of
the MSS.) another MS. (a) was copied in the 8th or pth century ;
this again is lost, but is the true archetype of our MSS, The
copyist apparently intended to omit all scholia, but, the distinction
between text and scholia being sometimes difficult to draw, he
included a good deal which should have been left out. For
example, Hypatia, and perhaps scholiasts after her, seem to have
added some alternative solutions and a number of new problems ;
some of these latter, such as II. 17, 17, 18, were admitted into the
text as genuine,
The MSS. fall into two main classes, the antePlanudcs class,
as we may call it, and the Planudean. The most ancient and the
best of all is Matritensis 48 (Tannery's A\ which was written in
the 1 3th century and belongs to the first class; it is evidently a
most faithful copy of the lost archetype (a). Maximus Planudes
wrote a systematic commentary on Books Land IL,and his scholia,
1 Dioph. i. pp. iiiv, n. pp. xxiixxxiv.
2 Eduard Gollob, "Ein wiedergefundener Diophantuscodex " in Zeitschrift fur Math*
u. Physik, XLIV. (1899), hist.litt. Abtheilung, pp. 137140.
THE MSS. OF AND WRITERS ON DIOPHANTUS 15
which are edited by Tannery for the first time, are preserved in the
oldest representative which we possess of the Planudean class,
namely, Marcianus 308 (Tannery's JSJ, itself apparently copied
from an archetype of the I4th century now lost, with the exception
of ten leaves which survive in Ambrosianus Et 157 sup.
Tannery shows the relation of the MSS. in the following
diagram :
(a) Lost copy of the Hypatian recension.
(a) Lost copy, of eighth or ninth c.
(FIRST CLASS) (PLANUDEAN CLASS)
;
[. Matritensis 48 = A*
1 3th c.
\. Vaticanus gr. 1 9 1 = J 7 ,
second half of 1 5th c.
;. Vaticanus gr. 304,
beginning of i6th c.
o. Lost MS. of the i4th c. of which ten leaves
1 are extant in Ambrosianus Et 157 sup.
10. Marcianus 308 = B l ,
beginning of 15 th c.
ir. Guelferbytanus
Gudianus i, i5th c.
14. Ambrosianus
A 91 sup.
(1545)
15. Vaticanus gr. 200
(1545)
4. Parisinus 2379 =C
(after first two
Books),
middle of i6th c.
5. Parisinus 2378 =P,
middle of i6th c.
6. Neapolitanus
III C 17,
middle of i6th c.
7. Urbinas gr. 74,
end of 1 6th c.
8. Oxon. Baroccianus
166 (part of Book L
only)
12. Palatinus gr. 391,
end of 1 6th c.
13. Reginensis 128,
end of 1 6th c.
1 6. Scorialensis TIu
d545)
4. Parisinus 2379 C
I (first two Books)
17. Parisinus 2485=^
middle of 1 6th c*
1 8. Scorialensis
RIIIiS,
middle of ro"th c.
19. Ambrosianus
Q 121 sup. (part of
Book I.),
middle of 1 6th c.
20. TaurinensisCIIIi6
21. Parisinus Ars. 8406
22. Scorialensis flI 1 5,
1 middle of 6th c.
23. Scorialensis RII3,
end of 1 6th c.
24. Oxon. Savilianus,
end of i6th c.
Auria's recension made up out of MSS. 2, 3, 15 above and Xylander's
translation: 25. Parisinus 2380 =/?.
26. Ambrosianus E 5 sup.
27. MS. (Patavinus) of Broscius (Brozek) now at Cracow.
28. Lost MS. of Cardinal du Perron.
1 6 INTRODUCTION
The addition of a few notes as regards the most important and
interesting of the MSS., in the order of their numbers in Tannery's
arrangement, will now sufficiently complete the story.
1. The best and most ancient MS., that of Madrid (Tannery's
A\ was unfortunately spoiled at a late date by corrections made,
especially in the first two Books, from some MS. of the Planudean
class, in such a way that the original reading is sometimes entirely
erased or made quite illegible. In these cases recourse must be
had to the Vatican MS. 191.
2. The MS. Vaticanus graecus 191 was copied from A before
it had suffered the general alteration by means of a MS. of the
other class, though not before various other corrections had been
made in different hands not easily distinguished ; thus V some
times has readings which Tannery found to have arisen from some
correction in A, A appears to have been at Rome for a con
siderable period at the time when V was copied ; for the librarian
who wrote the old table of contents 1 at the beginning of V inserted
in the margin in one place 2 the word dp%dftvo$ 3 which had been
omitted, direct from the original (A).
3. Vat. gr. 304 was copied from F, not from A ; Tannery
inferred this mainly from a collation of the scholia, and he notes
that the word apjapevo? above mentioned is here brought into the
text by the erasure of some letters. This MS. 304, being very
clearly written, was used thenceforward to make copies from. The
next five MSS. do not appear to have had any older source.
4. The MS. Parisinus 2379 (Tannery's C) was that used by
Bachet for his edition. It was written by one loannes Hydruntinus
after 1545, and has the peculiarity that the first two Books were
copied from the MS. Vat gr. 200 (a MS. of the Planudean class),
evidently in order to include the commentary of Planudes, while
the MS. Vat. gr. 304 belonging to the prePlanudes class was
followed in the remaining Books, no doubt because it was con
sidered superior. Thus the class of which C is the chief repre
sentative is a sort of mixed class.
5. 6. Parisinus 2378 = /*, and Neapolitans III C 17, were
copied by Angelus Vergetius. In the latter Vergetius puts the
1 The MS. V was made up of various MSS. before separated. The old table of
contents has AtofAvrov fyiBMTucfi' appoint didfopa.. The Ap/iow/cd include the Intro
duction to Harmony by Cleonides, but without any author's name. This fact sufficiently
explains the error of Ramus in saying, Schola, mathematics Bk I. p. 35, e *Scripserat et
Diophantus harmonica."
2 Dioph. i. p. 2, 6.
THE MSS. OF AND WRITERS ON DIOPHANTUS 17
numbers A, B, F, A, E, Z, H at the top of the pages (as we put
headlines) corresponding to the different Books, implying that he
regarded the tract on Polygonal Numbers as Book VII.
The other MSS. of the first class call for no notice, and we pass
to the Planudean class.
9. Tannery, as he tells us, congratulated himself upon finding
in Ambrosianus Et 157 sup. ten pages of the archetype of the
class, and eagerly sought for new readings. So far, however, as he
was able to carry his collation, he found no difference from the
principal representative of the class (BJ next to be mentioned.
10. The MS. Marcianus 308 (=J9i) of the I5th century formerly
belonged to Cardinal Bessarion, and was seen by Regiomontanus
at Venice in 1464. It contains the recension by Planudes with his
commentary.
11. It seems certain that the Wolfenbiittel MS. Guelferbytanus
Gudianus I (iSth c.) was that which Xylander used for his
translation ; Tannery shows that, if this was not the MS. lent
to Xylander by Andreas Dudicius Sbardellatus, that MS. must
have been lost, and there is no evidence in support of the latter
hypothesis. It is not possible to say whether the Wolfenbiittel
MS. was copied from Marcianus 308 (JB^) or from the com
plete MS. of which Ambrosianus Et 157 sup. preserves the ten
leaves.
12. Palatinus gr. 391 (end of i6th c.) has notes in German in
the margin which show that it was intended to print from it ; it
was written either by Xylander himself or for him. It is this MS.
of which Claudius Salmasius (Claude de Saumaise, 15881653)
told Bachet that it contained nothing more than the six Books,
with the tract on Polygonal Numbers.
13. Reginensis 128 was copied at the end of the i6th century
from the Wolfenbuttel MS.
14. 15. Ambrosianus A 91 sup. and Vaticanus gr. 200 both
come from B l ; as they agree in omitting V. 28 of Diophantus, one
was copied from the other, probably the latter from the former.
They were both copied by the same copyist for Mendoza in 1545.
Vat. gr. 200 has headings which make eight Books ; according to
Tannery the first Book is numbered a', the fourth S* ; before v. 20
(in Bachet's numbering) should this be IV. 20? is the heading
kwfydvTov e", before the fifth Book Aio^dvrov r ol/ , before the sixth
&io<f>dvrov f ov , and before the tract on Polygonal Numbers
Ov if v ; this wrong division occurs in the next three MSS.
H. D. 2
i8 INTRODUCTION
(16, 17, 18 in the diagram), all of which seem to be copied from
Vat. 200.
The MSS. numbered 20, 21, 22, 23 in the diagram are of the
hybrid class derived from Parisinus 2379 (C). Scorialensis flI15
and Scorialensis RII3, the latter copied from the former, have
the first Book divided into two (cf. p. 5 above), and so make
seven Books of the Arithmetica and an eighth Book of the
Polygonal Numbers.
27. The Cracow MS. has the same division into Books as the
MSS. last mentioned. According to Gollob, the collation of this
MS., so far as it was carried in 1899, showed that it agrees in the
main with A (the best MS.), B l (Marcianus 308) and C (Parisinus
2379) ; but, as it contains passages not found in the two latter, it
cannot have been copied from either of them.
25. Parisinus 2380 appears to be the copy of Auria's
Diophantus mentioned by Schulz as having been in the library of
Carl von Montchall and bearing the title " Diophanti libri sex, cum
scholiis graecis Maximi Planudae, atque liber de numeris poly
gonis, collati cum Vaticanis codicibus, et latine versi a Josepho
Auria 1 ."
The first commentator on Diophantus of whom we hear is
Hypatia, the daughter of Theon of Alexandria ; she was murdered
by Christian fanatics in 415 A.D. According to Suidas she wrote
commentaries on Diophantus, on the Astronomical Canon (sc. of
Ptolemy) and on the Conies of Apolloniusl Tannery suggests
that the remarks of Michael Psellus (nth c.) at the beginning of
his letter about Diophantus, Anatolius, and the Egyptian method
of arithmetical reckoning were taken bodily from some MS. of
Diophantus containing an ancient and systematic commentary ;
and he believes this commentary to have been that of Hypatia. I
have already mentioned the attractive hypothesis of Tannery that
Hypatia's commentary extended only to our six Books, and that
this accounts for the loss of the rest
Georgius Pachymeres (1240 to about 1310) wrote in Greek a
paraphrase of at least a portion of Diophantus. Sections 2544 of
1 Schulz, Diophantus, pref. xliii,
2 Suidas s.v. 'TTrarfo: ^ypa^ev Mfjuty/Mt els &t6<f>avrov t <eh> rbv &<rrpovoiUK&v Kw6va y
els rd. KWUC& 'ATToXXwy/ov MfwrifM. So Tannery reads, following the best MSS. ; he
gives ample reasons for rejecting Kuster's conjecture ds &io<j>Avrov rd? farpovojuK&v xcuwa,
viz. (i) that the order of words would have been rdy Aio0&>rov d(rr/>ovoA/c6i> Kw6m,
(i) that there is nothing connecting Diophantus with astronomy, while Suidas mention*,
s.v. Slew, a commentary efr T&> IlroXe/to/oi; irp
THE MSS. OF AND WRITERS ON DIOPHANTUS 19
this survive and are published by Tannery in his edition of
JDiophantus. 1 . The chapters lost at the beginning may have con
tained general observations and introductions to the first two
paragraphs of Book I. ; section 25 begins with the third paragraph
(Def. i), and the rest of the fragment takes us up to the problem
in I. ii.
Soon afterwards Maximus Planudes (about 12601310) wrote
a systematic commentary on Books I., IL This is also included by
Tannery in his edition 2 .
There are a number of other ancient scholia, very few of which
seemed to Tannery to be worth publication 3 .
But in the meantime, and long before the date of Georgius
Pachymeres, the work of Diophantus had become known in Arabia,
where it was evidently the subject of careful study. We are told
in the Fihrist, the main part of which was written in the year
987 A.D., (i) that Diophantus was a Greek of Alexandria who
wrote a book "On the art of algebra 4 ," (2) that Abu'l Wafa
alBuzjam (940998) wrote (a) a commentary (tafsir) on the
algebra of Diophantus and (b) a book of " proofs to the pro
positions used by Diophantus in his book and to that which
he himself (Abu'l Wafa) stated in his commentary V (3) that
Qusta b. Luqa alBa e labakki (died about 912) wrote a "com
mentary on three and a half Books of Diophantus' work on
arithmetical problems 6 ." Qusta b. Luqa, physician, philosopher,
astronomer, mathematician and translator, was the author of works
on Euclid and of an " introduction to geometry " in the form of
question and answer, and translator of the socalled Books XIV,, XV,
of Euclid ; other Arabian authorities credit him with an actual
" translation of the book of Diophantus on Algebra r ." Lastly, we
are told by Ibn abl Usaibi f a of "marginal glosses which Ishaq b.
Yiinis (died about 1077), the physician of Cairo, after Ibn al
Haitham, added to the book of Diophantus on algebraic problems."
The title is somewhat obscure ; probably Ibn alHaitham (about
9651039), who wrote several works on Euclid, wrote a commentary
on the Arithmetica and Ishaq b. Yunis added glosses to this
commentary 8 .
1 Dioph. ii. pp. 78122. 2 Dioph. II. pp. 125255.
3 The few that he gives are in Vol. II. pp. 256260; as regards the collection in
general cf. Hultsch in Berliner philologische Wochen$chrift> 1896, p. 615.
4 Fihristt ed. Suter, p. 22. 6 ibid. p. 39, 6 ibid* p. 43.
7 Suter, Die Mathematiker und Astronomen der Araber, 1900, p. 41.
8 Suter, <#. V. pp. 1078. Cf. Bibliotheca Mathematics iv s , 19034, p. 296.
20 INTRODUCTION
To Regiomontanus belongs the credit of being the first to call
attention to the work of Diophantus as being extant in Greek.
We find two notices by him during his sojourn in Italy, whither he
journeyed after the death of his teacher Georg von Peurbach,
which took place on the 8th April, 1461. In connexion with
lectures on the astronomy of Alfraganus which he gave at Padua
he delivered an Gratia introductoria in omnes scientias mathe
matical 1 . In this he observed : " No one has yet translated from
the Greek into Latin the fine thirteen Books of Diophantus, in
which the very flower of the whole of Arithmetic lies hid, the ars
rei et census which today they call by the Arabic name of
Algebra 2 ." Secondly, he writes to Bianchini, in answer to a letter
dated 5th February, 1464, that he has found at Venice "Diofantus,"
a Greek arithmetician, who has not yet been translated into Latin ;
that in his preface Diophantus defines the various powers up to
the sixth ; but whether he followed out all the combinations of
these Regiomontanus does not know : " for not more than six
Books are found, though in the preface he promises thirteen. If
this book, which is really most wonderful and most difficult, could
be found entire, I should like to translate it into Latin, for the
knowledge of Greek which I have acquired while staying with my
most reverend master [Bessarion] would suffice for this.,./* He
goes on to ask Bianchini to try to discover a complete copy and,
in the meantime, to advise him whether he should begin to translate
the six Books 8 . The exact date of the ratio is not certain.
Regiomontanus made some astronomical observations at Viterbo
in the summer and autumn of 1462, He is said to have spent a
year at Ferrara, and he seems to have gone thence to Venice.
Extant letters of his written at Venice bear dates from 2/th July,
1463, to 6th July, 1464, and it may have been from Venice
that he made his visit to Padua. At all events the Oratio at
Padua must have been near in time to the discovery of the
MS. at Venice.
Notwithstanding that attention was thus called to the work, it
1 Printed in the work Rudimenta astronomica Alfragani^ NUrnberg, 1537.
a As the ars rei et census, the solution of determinate quadratic equations, is not found
in our Diophantus, it would seem that at the time of the Oratio Regiomontanus had only
looked at the MS. cursorily, if at all.
8 The letter to Bianchini is given on p. 135 of Ch. Th. v. Murr's Memorabilia,
Norimbergae, 1786, and partly in Doppelmayer's Historische Nachricht von den Niirn*
bergiscken Mathematicis und Kiimtkm (NUrnberg, 1730), p. 5, notej'*
THE MSS, OF AND WRITERS ON DIOPHANTUS 21
seems to have remained practically a closed book from the date of
Maximus Planudes to about 1570. Luca Paciuolo, towards the
end of the I5th c., Cardano and Tartaglia about the middle of the
i6th, make no mention of it. Only Joachim Camerarius, in a
letter published in I556 1 , mentions that there is a MS. of
Diophantus in the Vatican which he is anxious to see. Rafael
Bombelli was the first to find a MS. in the Vatican and to conceive
the idea of publishing the work. This was towards 1570, for in his
Algebra 2 published in 1572 Bombelli tells us that he had in the
years last past discovered a Greek book on Algebra written by " a
certain Diofantes, an Alexandrine Greek author, who lived in the
time of Antoninus Pius " ; that, thinking highly of the contents of
the work, he and Antonio Maria Pazzi determined to translate it ;
that they actually translated five books out of the seven into
which the MS. was divided ; but that, before the rest was finished,
they were called away from it by other labours. Bombelli did not
carry out his plan of publishing Diophantus in a translation, but
he took all the problems of the first four Books and some of those
of the fifth, and embodied them in his Algebra, interspersing them
with his own problems. He took no pains to distinguish
Diophantus' problems from his own ; but in the case of the former
he adhered pretty closely to the original, so that Bachet admits his
obligations to him, remarking that in many cases he found
1 De. Graeds Latinisqtte numerorum notis et praeterea Saracenis seu Indicis, etc. etc.,
studio Joachimi Camerarii, Papeberg, 1556.
3 Nesselmann tells us that he has not seen this work but takes his information about
it from Cossali, I was fortunate enough to find in the British Museum one of the copies
dated 1579 (really the same as the original edition of 1572 except that the titlepage and
date are new, and a dedicatory letter on pp. 38 is reprinted; there were not two
separate editions). The title is L' } Algebra, opera di Rafael Bombelli da Bologna diuisa in
ire Libri* In Bologna, Per Giovanni Rossi, MDLXXIX, The original of the passage
from the preface is :
"Questi anni passati, essendosi ritrouato una opera greca di questa disciplina nella
libraria di Nostro Signore in Vaticano, composta da un certo Diofante Alessandrino Autor
Greco, il quale fii a tempo di Antonin Pio, e havendomela fatta vedere Messer Antonio
Maria Pazzi Reggiano, publico lettore delle Matematiche in Roma, e giudicatolo con lui
Autore assai intelligente de' numeri (ancorche non tratti de' numeri irrationali, ma solo
in lui si vede vn perfetto ordine di operare) egli, ed io, per arrichire il mondo di cosl fatta
opera, ci dessimo a tradurlo, e cinque libri (delli sette che sono) tradutti ne habbiamo ; lo
restante non hauendo potuto finire per gli trauagli auenuti all' uno, e all' altro ; e in detta
opera habbiamo ritrouato, ch' egli assai volte cita gli Autori Indiani, col che mi ha fatto
conoscere, che questa disciplina appo gF indiani prima fu, che a gli Arabi." The last
words stating that Diophantus often quotes from Indian authors are no doubt due to
Bombelli's taking for part of Diophantus the tract of Maximus Planudes about the Indian,
method of reckoning.
2 2 INTRODUCTION
Bombelli's translation better than Xylander's and consequently
very useful for the purpose of amending the latter 1 .
It may be interesting to mention a few points of notation in
this work of Bombelli. At the beginning of Book II. he explains
that he uses the word "tanto" to denote the unknown quantity,
not " cosa" like his predecessors ; and his symbol for it is i, the
square of the unknown O 2 ) is i, the cube i; and so on. For plus
and minus (pib and mend) he uses the initial letters / and ;//.
Thus corresponding to x + 6 we should find in Bombelli il/. 6,
and for x* + S^r 4, i^p. 5! m. 4. This notation shows, as will be
seen later, some advance upon that of Diophantus in one important
respect.
The next writer upon Diophantus was Wilhelm Holzmann who
published, under the Graecised form of his name, Xyhuider, by
which he is generally known, a work bearing the title : Diophanti
Alexandrini Rerum Arithmeticarum Libri sex, quorum primi duo
adiecta habent Scholia Maximi (tit coniectura esf] Planudis. Item
Liber de Numeris Polygonis seu Multangulis. Opus incomparabile,
lierae Arithmeticae Logisticae perfectionem continens^ panels adhuc
uisum. A Guil. Xylandro Augustano incredibili labore Latino
redditum, et Commentariis explanatum, inque lucem editum, ad
Illwtriss* Principem Ltidovicum Vidrtembergensem. Basileae per
Eusebium Episcopium^ et Nicolai Fr. haeredes. MDLXX V. Xylander
was according to his own statement a "public teacher of Aristotelian
philosophy in the school at Heidelberg V J He was a man of almost
universal culture 3 , and was so thoroughly imbued with the classical
literature, that the extraordinary aptness of his quotations and his
wealth of expression give exceptional charm to his writing whenever
he is free from the shackles of mathematical formulae and techni
calities. The Epistola Nimcupatoria is addressed to the Prince
Ludwig, and Xylander neatly introduces it by the line "Offerimu.s
numeros, numeri sunt principe digni." This preface is very quaint
and interesting. He tells us how he first saw the name of
Diophantus mentioned in Suidas, and then found that mention
1 "Sed snas Diophanteis quaestionibus ita immiscuit, ut has ab illis distingucre non
sit in promptu, neque vero sc fidum satis interpretem praebuit, cum passim vcrba
Diophanti immutet, bisque pleraque addat, pleraque pro arbitrio detrahat. In inultia
nihilominus interpretationem Bombellii, Xilandriana praestare, et ad hanc emcudandam
me adjuvisse ingenue fateor." Ad lectorem.
2 "Publicus philosophiae Aristoteleae in schola Heidelbergensi doctor.**
 8 Even Bacbet, who, as we shall see, was no favourable critic, calls him " Vir omnibus
disciplinis excultus."
THE MSS. OF AND WRITERS ON DIOPHANTUS 23
had been made of his work by Regiomontanus as being extant
in an Italian library and having been seen by him. But, as the
book had not been edited, he tried to think no more of it but,
instead, to absorb himself in the study of such arithmetical books
as he could obtain, and in investigations of his own 1 . Selftaught
except in so far as he could learn from published works such as
those of Christoff Rudolff (of the "Coss"), Michael Stifel, Cardano,
Nunez, he yet progressed so far as to be able to add to, modify
and improve what he found in those works. As a result he fell
into what Heraclitus called oft/ow, lepav votrov, that is, into the
conceit of "being somebody" in the field of Arithmetic and
"Logistic"; others too, themselves learned men, thought him an
arithmetician of exceptional ability. But when he first became
acquainted with the problems of Diophantus (he continues) right
reason brought such a reaction that he might well doubt whether he
ought previously to have regarded himself as an object of pity or of
derision. He considers it therefore worth while to confess publicly his
own ignorance at the same time that he tries to interest others in
the work of Diophantus, which had so opened his eyes. Before this
critical time he was so familiar with methods of dealing with surds
that he had actually ventured to add something to the discoveries
of others relating to them ; the subject of surds was considered to
be of great importance in arithmetical questions, and its difficulty
1 I cannot refrain from quoting the whole of this passage : "Sed cum ederet nemo :
cepi desiderium hoc paulatim in animo consopire, et eorum quos consequi poteram
Arithmeticorum librorum cognitione, et meditationibus nostris sepelire. Veritatis porro
apud me est autoritas, ut ei coniunctum etiam cum dedecore meo testimonium lubentissime
perhibeam. Quod Cossica seu Algebrica (cum his enim reliqua comparata, id sunt quod
umbrae Homerice in Necya ad animam Tiresiae) ea ergo quod non assequebar mod6,
quanquam mutis duntaxat usus preceptoribus caetera a.&ro8l$aKTo$, sed et augere, uariare,
adeoque corrigere in loco didicissem, quae summi et fidelissimi in docendo uiri Christifer
Rodolphus Silesius, Micaelus Stifelius, Cardanus, Nonius, aliique litteris mandauerant:
incidi in o^otv, lep&v vbffov, ut scite appellauit Heraclitus sapientior multis aliis philoso
phis, hoc est, in Arithmetica, et uera Logistica, putaui me esse aliquid : itaque de me
passim etiam a multis, iisque doctis urns iudicatum fuit, me non de grege Arithmeticum
esse. Verum ubi primum in Diophantea incidi : ita me recta ratio circumegit, ut flendusne
mihi ipsi antea, an uer6 ridendus fuissem, haud iniuria dubitauerim. Operae precium est
hoc loco et meam inscitiam inuulgare, et Diophantei opens, quod mihi nebulosam istam
caliginem ab oculis detersit, imm6 eos in coenum barbaricum defossos eleuauit et repur
gaitit, gustum aliquem exhibere. Surdorum ego numerorum tractationem ita tenebam,
ut etiam addere aliorum inuentis aliquid non poenitendum auderem, atque id quidem in
rebus arithmeticis magnum habetur, et dif&cultas istarum rerum multos a mathematibus
deterret. Quanto autem hoc est praeclarius, in iis problematis, quae surdis etiam
numeris uix posse uidentur explicari, rem eo deducere, ut quasi solum arithmeticum
uertere iussi obsurdescant illi plane, et ne mentio quidem eorum in tractatione ingemo*
sissimarum quaestionum admittatur."
54 INTRODUCTION
was even such as to deter many from the study of mathematics.
"But how much more splendid" says Xylander, "in the case of
problems which seem to be hardly capable of solution even with
the help of surds, to bring the matter to the point that, while the
surds, when bidden (so to speak) to plough the arithmetic soil,
become true to their name and deaf to entreaty, they are not so
much as mentioned in these most ingenious solutions ! " He then
describes the enormous difficulties which beset his work owing
to the corruptions in his text. In dealing, however, with the
mistakes and carelessness of copyists he was, as he says, no novice;
for proof of which he appeals to his editions of Plutarch, Stephanus
and Strabo. This passage, which is good reading, but too long
to reproduce here, I give in full in the note 1 . Next Xylander
tells us how he came to get possession' of a manuscript of Dio
phantus. In October of the year 1571 he made a journey to
Wittenberg; while there he had conversations on mathematical
subjects with two professors, Sebastian Theodoric and Wolfgang
Schuler by name, who showed him a few pages of a Greek
1 " Id uer6 mihi accidit durum el uix superabile incommodum, quod mirifice deprauata
omnia inueni, cum neque problernatum expositio interdum integra esset, ac passim rmmeri
(in quibus sita omnia esse in hoc argumento, quis ignorat?) tarn problematum quam
solutionum sine explicationum corruptissimi. Non pudebit me ingenufc fateri, qualem me
heic gesserim. Audacter, et summo cum feruore potius quam alacritate animi opus ip.sum
initio sum aggressus, laborque mihi omnis uoluptati fuit, tantus est meus rerum arithmeti
carum amor, quin etgratiam magnam me apud omnes liberalium scientiarum amatores ac
patronos initurum, et praeclare de rep. litteraria meritumm intelligebam, eamque rem
mihi laudi (quam a bonis profectam nemo prudens aspernatur) gloriaeque fortasse etiam
emolument fore sperabam. Progressus aliquantulum, in salebras incidi : quae tantum
abest ut alacritatem meam retuderint, ut etiam animos mihi addiderint, ncque enim mihi
novum aut insolens est aduersns librariorum incuriam cerlamen, et hoc in re mililaui, (ut
Horatii nostri uerbis utar) non sine gloria, quod me non arroganter diccro, I)i<>,
Plutarchus, Strabo, Stephanusque nostri testantur. Sed cum mox in ipsum pclagus
monstris scatens me cursus abripuit : non despondi equidem animum, ncque mamis cledi,
sed tamen saepius ad oram unde soluissem respexi, quam portum in quern essct euadcndum
cogitando prospicerem, depraehendique non minus uere quam elcganter ea cccinlssc
Alcaeum, quae (si possum) Latin& in hac quasi uotiua mea tabula scribam.
Qui uela uentis uult dare, dum licet,
Cautus futuri praeuideat modum
Cursus. mare ingressus, raarino
Nauiget arbitrio necesse est.
Sane quod de Echeneide pisce fertnr, eum nauim cui se adplicet remorari, poeno crcdibile
fecit mihi mea cymba tot mendorum remoris retardata. Kxpediui tamgn me ita, ut facile
omnes mediocri de his rebus iudicio pracditi, intellecturi sint incredibilem me laborcm ct
aerumnas difficilimas superasse : pudore etiam stimulatum oneris quod ultro mihi impos
uissem, non perferendi. Paucula quaedam non plane explicata, studio et certis rle cnusis
in alium locum reiecimus. Opus quidem ipsum ita absoluimus ut netjue eius nos pudere
debeat, et Arithmeticae Logisticesque studiosi nobis se plurimum debere sint haud dubie
professuri,'*
THE MSS, OF AND WRITERS ON DIOPHANTUS 25
manuscript of Diophantus and informed him that it belonged to
Andreas Dudicius whom Xylander describes as "Andreas Dudicius
Sbardellattis, hoc tempore Imperatoris Romanorum apud Polonos
orator/' On his departure from Wittenberg Xylander wrote out
and took with him the solution of a single problem of Diophantus,
to amuse himself with on his journey. This he showed at Leipzig
to Simon Simonius Lucensis, a professor at that place, who wrote to
Dudicius on his behalf. A few months afterwards Dudicius sent
the MS. to Xylander and encouraged him to persevere in his
undertaking to translate the Arithmetica into Latin. Accordingly
Xylander insists that the glory of the whole achievement belongs
in no less but rather in a greater degree to Dudicius than to
himself. Finally he commends the work to the favour of Prince
Ludwig, extolling the pursuit of arithmetical and algebraical
science and dwelling in enthusiastic anticipation on the influence
which the Prince's patronage would have in helping and advancing
the study of Arithmetic 1 , This Epistola Nuncupatoria bears the
date I4th August, I574 2 . Xylander died on the loth of February
in the year following that of the publication, 1576.
Tannery has shown that the MS. used by Xylander was
Guelferbytanus Gudianus I. Bachet observes that he has not been
able to find out whether Xylander ever published the Greek text,
though parts of his commentary seem to imply that he had, or at
least intended to do so. It is now clear that he intended to bring
out the text, but did not carry out his intention. Tannery observes
that the MS. Palatinus gr. 391 seems to have been written either by
Xylander himself or for him, and there are German notes in the
margin showing that it was intended to print from it.
Xylander's achievement has been, as a rule, quite inadequately
appreciated. Very few writers on Diophantus seem to have studied
the book itself: a fact which may be partly accounted for by its
rarity. Even Nesselmann, whose book appeared in 1842, says that
he has never been able to find a copy. Nesselmann however seems
to have come nearest to a proper appreciation of the value of the
work : he says " Xylander's work remains, in spite of the various
1 ** Hoc non mod6 tibi, Princeps Illustrissime, honorificum erit, atque gloriosum ; sed
te labores nostros approbante, arithmeticae studium ciim alibi, turn in tua Academia et
Gymnasiis, excitabitur, confirmabitur, prouehetur, et ad perfectam ems scientiam. multi tuis
auflpiciis, nostro labore perducti, magnam hac re tuis in remp. beneficiis accessionem
factam esse gratissima commemoratione praedicabunt."
2 " Heidelberga. postrid. Eidus Sextiles CD ID LXXIV."
26 INTRODUCTION
defects which are unavoidable in a first edition of so difficult an
author, especially when based on only one MS. and that full of
errors, a highly meritorious achievement, and does not deserve
the severe strictures which it has sometimes had passed upon it.
It is true that Xylander has in many places not understood his
author, and has misrepresented him in others ; his translation is
often rough and unLatin, this being due to a too conscientious
adherence to the actual wording of the original ; but the result
was none the less brilliant on that account. The mathematical
public was put in possession of Diophantus' work, and the
appearance of the translation had an immediate and enormous
influence on the development and shaping of Algebra 1 /' As a
rule, the accounts of Xylander's work seem to have been based
on what Bachet says about it and about his obligations to it.
When I came to read Bachet myself and saw how disparaging,
as a rule, his remarks upon Xylander were, I could not but suspect
that they were unfair. His repeated and almost violent repudiation
of obligation to Xylander suggested to me the very thing which he
disclaimed, that he was under too great obligation to his predecessor
to acknowledge it duly. I was therefore delighted at my good
fortune in finding in the Library of Trinity College, Cambridge,
a copy of Xylander, and so being able to judge for myself of
the relation of the later to the earlier work. The result was to
confirm entirely what I had suspected as to the unfair attitude
taken up by Bachet towards his predecessor. I found it every
where; even where it is obvious that Xy lander's mistakes or
difficulties are due only to the hopeless state of his solitary MS.
Bachet seems to make no allowance for the fact. The truth is that
Bachet's work could not have been as good as it was but for the
pioneer work of Xylander; and it is the great blot in Bachct's
otherwise excellent edition that he did not see fit to acknowledge
the fact.
I must now pass to Bachet's work itself. It was the first
edition published which contained the Greek text, and appeared
in 1621 bearing the title: Diopfianti Alexandrini Arithmeticorum
libri sex, et de numeris multang^il^s liber unus. Nunc primtim
Graece et Latinb editi, atque absolutissimis Commentariis illustrate
Auctore Claudia Gaspare Bactieto Meziriaco Sebusiano, V.C* Lutetiae
Parisiorum, Sumptibus Hieronymi Drovart'\ via Jacobaea> sub Scuto
1 Nesselmann, p. 27980.
2 For " sumptibus Hieronymi Drovart etc. " some copies have " sumplibus Sebastiani
THE MSS. OF AND WRITERS ON DIOPHANTUS 27
Solari. MDCXXL Bachet's Greek text is based, as he tells us,
upon a MS. which he calls "codex Regius," now in the Bibliotheque
Nationale at Paris (Parisinus 2379) ; this MS. is his sole authority,
except that Jacobus Sirmondus had part of a Vatican MS. (Vat.
gr. 304) transcribed for him. He professes to have produced a
good Greek text, having spent incalculable labour upon its emenda
tion, to have inserted in brackets all additions which he made to it,
and to have given notice of all corrections, except those of an
obvious or trifling nature ; a few passages he has left asterisked, in
cases where correction could not be safely ventured upon. He
is careful to tell us what previous works relating to the subject he
had been able to consult. First he mentions Xylander (he spells
the name as Xz'lander throughout), who had translated the whole of
Diophantus, and commented upon him throughout, "except that
he scarcely touched a considerable part of the fifth book, the whole
of the sixth and the treatise on multangular numbers, and even
the rest of his work was not very successful, as he himself admits
that he did not thoroughly understand a number of points." Then
he speaks of Bombelli (as already mentioned) and of the Zetetica of
Vieta (in which the author treats in his own way a large number
of Diophantus' problems : Bachet thinks that he so treated them
because he despaired of restoring the book completely). Neither
Bombelli nor Vieta (says Bachet) made any attempt to demonstrate
the difficult porisms and abstruse theorems in numbers which
Diophantus assumes as known in many places, or sufficiently
explained the causes of his operations and artifices. All these
omissions on the part of his predecessors he thinks he has supplied
in his notes to the various problems and in the three books of
"Porisms" which he prefixed to the work 1 . As regards his Latin
translation, he says that he gives us Diophantus in Latin from the
version of Xylander most carefully corrected, in which he would
have us know that he has done two things in particular, first,
Cramoisy, via Jacobaea, sub Ciconiis." The copy (from the Library of Trinity College,
Cambridge) which I used in preparing my first edition has the former words ; a copy in
the Library of the Athenaeum Club has the latter.
1 On the nature of some of Bachet's proofs Nicholas Saunderson (formerly Lucasian
Professor) remarks in Elements of Algebra^ 1740, apropos of Dioph. ill. 15 : " M. Bachet
indeed in the i6th and 1 7th props, of his second book of Porisms has given us demonstra
tions, such as they are, of the theorems in the problem : but in the first place he
demonstrates but one single case of those theorems, and in the next place the demonstra
tions he gives are only synthetical, and so abominably perplexed withal, that in each
demonstration he makes use of all the letters in the alphabet except I and O, singly to
represent the quantities he has there occasion for."
28 INTRODUCTION
corrected what was wrong and filled the numerous lacunae,
secondly, explained more clearly what Xylander had given in
obscure or ambiguous language; "I confess however," he says,
"that this made so much change necessary, that it is almost
fairer to attribute the translation to me than to Xilander. But if
anyone prefers to consider it as his, because I have held fast, tooth
and nail, to his words when they do not misrepresent Diophantus,
I have no objection 1 ." Such sentences as these, which are no
rarity in Bachet's book, are certainly not calculated to increase
our respect for the author. According to Montucla 2 , " the historian
of the French Academy tells us " that Bachet worked at this edition
during the course of a quartan fever, and that he himself said that,
disheartened as he was by the difficulty of the work, he would never
have completed it, had it not been for the stubbornness which his
malady generated in him.
As the first edition of the Greek text of Diophantus, this work,
in spite of any imperfections we may find in it, does its author all
honour.
The same edition was reprinted and published with the addition
of Per mat's notes in 1670 : Diophanti Alexandrini Arithmeticorum
libri seX) et de numeris multangulis liber unus. Cum cowwentariis
C. G. Backeti V.C. et obseruationibus D. P. de Fermat Senatoris
TolosanL Accessit Doctrinae Analyticae inmntwn nouum, collectum
ex variis eiusdem D. de Fermat Epistolis. Tolosae, Excudebat
Bernardus Bosc> e Region e Collegii Societatis Jesu. MDCLXX.
This edition was not published by Fermat himself, but by his
son after his death. S. Fermat tells us in the preface that this
publication of Fermat's notes to Diophantus 3 was part of an
attempt to collect together from his letters and elsewhere his
contributions to mathematics. The "Doctrinae Analyticae In
uentum nouum" is a collection made by Jacobus de Billy 4
1 Deinde Latinum damus tibi Diophantum ex Xilawlri vcrsionc accuratissime castigata,
in qua duo potissimum now praestitisse scias velim, nam ct deprauala corrextmus, hiantesque
passim lacunas repleuimus : et quae subobscure, vel ambiguc fucrat interpretalus Xilnnder,
dilucidius exposuiraus; fateor tamen, inde tantam inductam esse mutationcm, vt prope
modum aequius sit veisionem istam nobis quam Xilamlro tribuere. Si quis autem potius
ad eum pcrtincre contendal, qu6d eius verba, quatemis Diophanto frauclx non crant,
mordicus relinuimus, per me licet.*' u i. 323.
s Now published in (Euvres de Fermat by I*. Tannery and C. Henry, Vol. I. (1891),
pp. 289342 (the Latin original), and Vol. in. (1896), pp. 141274 (French translation).
4 Now published in (Euvres de format, in. 323398 (French translation). DC
Billy had already published in i6(5o a book under the title Diofhanltts gwmetra sim
opus contixtum ex arithmetica et geometrta.
THE MSS. OF AND WRITERS ON DIOPHANTUS 29
from various letters which Fermat sent to him at different times.
The notes upon Diophantus' problems, which his son hopes will
prove of value very much more than commensurate with their
bulk, were (he says) collected from the margin of his copy of
Diophantus. From their brevity they were obviously intended
for the benefit of experts 1 , or even perhaps solely for Fermat's
own, he being a man who preferred the pleasure which he had
in the work itself to any reputation which it might bring
him. Fermat never cared to publish his investigations, but was
always perfectly ready, as we see from his letters, to acquaint
his friends and contemporaries with his results. Of the notes
themselves this is not the place to speak in detail. This edition
of Diophantus is rendered valuable only by the additions in it
due to Fermat; for the rest it is a mere reprint of that of 1621.
So far as the Greek text is concerned, it is very much inferior
to the first edition. There is a far greater number of misprints,
omissions of words, confusions of numerals ; and, most serious of
all, the brackets which Bachet inserted in the edition of 1621 to
mark the insertion of words in the text are in this later edition
altogether omitted. These imperfections have been already noticed
by Nesselmann 2 . Thus the reprinted edition of 1670 is untrust
worthy as regards the text.
In 1585 Simon Stevin published a French version of the first
four books of Diophantus 3 . It was based on Xylander and was
a free reproduction, not a translation, Stevin himself observing that
the MS. used by Xylander was so full of mistakes that the text of
1 Lectori Beneuolo, p. iii : "Doctis tantum quibus pauca sufficiunt, harum obserua
tionum auctor scribebat, vel potius ipse sibi scribens, his studiis exerceri malebat quam
gloriari; adeo autem ille ab omni ostentatione alienus erat, vt nee lucubrationes suas
typis mandari curauerit, et suorum quandoque respousorum autographa nullo seruato
exemplar! petentibus vltr6 miserit ; norunt scilicet plerique celeberrimorum huius saeculi
Geometrarum, quam libenter ille et quantS. humanitate, sua iis inuenta patefecerit."
2 "Was dieser Abdruck an ausserer Eleganz gewonnen hat (denn die Bachet'sche
Ausgabe ist mit ausserst unangenehmen, namentlich Griechischen Lettern gedruckt), das
hat sie an innenn Werthe in Bezug auf den Text verloren. Sie ist nicht bloss voller
Druckfehler in einzelnen Worten und Zeichen (z. B. durchgehends IT slatt "^, 900)
sondern auch ganze Zeilen sind ausgelassen oder doppelt gedruckt (z. B. in. 12 eine
Zeile doppelt, IV. 25 eine doppelt und gleich hinterher eine ausgelassen, IV. 52 eine
doppelt, v. ri eine ausgelassen, desgleichen v. 14, 25, 33, vi. 8, 13 und so weiter), die
Zahlen versttimmelt, was aber das Aergste ist, die Bachet'schen kritischen Zeichen sind
fast uberall, die Klammer durchgangig weggefallen, so dass diese Ausgabe als Text des
Diophant vollig unbrauchbar geworden ist," p. 283.
3 Included in IlArithmetigw de Simon Stevin de ruges...A Leyde, De rimprimerie
de Christophle Plantin, cio . ID . LXXXV.
3 o INTRODUCTION
Diophantus could not be given word for word 1 . Albert Girard
added the fifth and sixth books to the four, and this complete
Version appeared in i62S 2 .
In 1810 was published an excellent translation (with additions)
of the fragment upon Polygonal Numbers by Posclger : Diopkantus
von Alexandrien uber die PolygonalZahlen. Uebersetzt mit Zusatzen
von F. Tk. Poselger. Leipzig, 1810.
In 1822 Otto Schulz, professor in Berlin, published a very
meritorious German translation with notes : Diophantus von
Alexandria arithmetische Aufgaben nebst dessen Schrift ilbcr die
PolygonZahlen. Aus dem Griechischen ilbersetzt und mit An
merktingen begleitet von Otto Sckuls, Professor am Bcrlinisch
Colnischen Gymnasium ztim grancn Kloster. Berlin, 1822. In dcr
Schlesingerscken Buck und Musikhandhmg. The work of Poselger
just mentioned was with the consent of its author incorporated in
Schulz's edition along with his own translation and notes upon
the larger treatise, the Arithmetica. According to Nesselmann
Schulz was not a mathematician by profession; he produced,
however, a thoroughly useful edition, with notes chiefly upon
the matter of Diophantus and not on the text (with the exception
of a very few emendations) : notes which, almost invariably correct,
help much to understand the author. Schulz's translation is based
upon the edition of Bachet's text published in 1670.
Another German translation was published by G. Wertheim
in 1 890 : Die A rithmetik und die Schrift ilber Polygonalsahlen des
Diophantus von Alexandria, Ubersetzt und mit Anmerknngcn
begleitet von G. WertJuim (Teubner), Though it appeared before
the issue of Tannery's definitive text, it is an excellent translation,
the translator being thoroughly equipped for his task ; it is valuable
also as containing Fermat's notes, also translated into German, with
a large number of other notes by the translator elucidating both
Diophantus and Fermat,and generalising a number of the problems
which, with very few exceptions, receive only particular solutions
from Diophantus himself. Wertheim has also included 46 epigram
problems from the Greek anthology and the enunciation of the
famous CattleProblem attributed to Archimedes.
1 See Bitliotheca Mathsmatica vn s , 19067, p. 59.
2 UArithmetigut fa Simon Stevin de Bruges, Reiteue, eorriget & wtgmatlee dephisieun
traictez et annotation par Albert Girard Samielois Maihematicien. A Leiclo, de
ITmprimerie des Kheviers ciD.io.cxxv. Reproduced in the edition of Les (Euvres
Mathematiques de Simon Steuin de Bruges* Par Albert Gimrd. Leyde, CI;> , u . CXXXJV.
THE MSS. OF AND WRITERS ON DIOPHANTUS 31
No description is necessary of the latest edition, by Tannery,
in which we at last have a definitive Greek text of Diophantus
with the ancient commentaries, etc., Diophanti Alexandrini opera
omnia cum Graecis commentariis. Edidit et Latine interpretatus
est Paulus Tannery (Teubner). The first volume (1893) contains
the text of Diophantus, the second (1895) the Pseudepigrapha,
Testimonia veterum, Pachymeres' paraphrase, Planudes' com
mentary, various ancient scholia, etc., and 38 arithmetical epigrams
in the original Greek with scholia. Any further edition will neces
sarily be based on Tannery, who has added all that is required in
the shape of introductions, etc.
Lastly we hear of other works on Diophantus which, if they
were ever written, are lost or remain unpublished. First, we find
it asserted by Vossius (as some have understood him) that the
Englishman John Pell wrote an unpublished Commentary upon
Diophantus. John Pell (16111685) was at one time professor
of mathematics at Amsterdam and gave lectures there on Dio
phantus, but what Vossius says about his commentary may
well be only a recommendation to undertake a commentary,
rather than a historical assertion of its completion. Secondly,
Schulz states in his preface that he had lately found a note in
Schmeisser's OrtJwdidaktik der Mathematik that Hofrath Kausler
by command of the Russian Academy prepared an edition of
Diophantus 1 . This seems however to be a misapprehension on the
part of Schulz. Kausler is probably referring, not to a translation
of Diophantus, but to his memoir of 1798 published in Nova Acta
Acad. Petropol XI, p. 125, which might easily be described as an
Ausarbeitung of Diophantus' work.
I find a statement in the New American Cyclopaedia (New York,
D. Appleton and Company), Vol. VI., that " a complete translation
of his (Diophantus') works into English was made by the late
Miss Abigail Lousada, but has not been published."
1 The whole passage of Schmeisser is : "Die mechanische, geistlose Behandlung der
Algebra ist ins besondere von Herrn Hofrath Kausler stark gerligt worden. In der
Vorrede zu seiner Ausgabe des Uftakersc/ien Exempelbuchs beginnt er so ; c Seit mehreren
Jahren arbeitete ich fUr die RussischKaiserliche Akademie der Wissenschaften Diophants
unsterbliches Werk liber die Arithmetik aus, und fand darin einen solchen Schatz von
den feinsten, scharfsinnigsten algebraischen Auflosungen, dass mir die mechanische,
geistlose Methode der neuen Algebra mit jedem Tage mehr ekelte u.s.w.' " (p. 33).
CHAPTER III
NOTATION AND DEFINITIONS OF DIOPHANTUS
As it is my intention, for the sake of brevity and per
spicuity, to make use of the modern algebraical notation in giving
my account of Diophantus' problems and general methods, it is
necessary to describe once for all the machinery which our author
uses for working out the solutions of his problems, or the notation
by which he expresses such relations as would be represented fn
our time by algebraical equations, and, in particular, to illustrate
the extent to which he is able to manipulate unknown quantities.
Apart, however, from the necessity of such a description for the
proper and adequate comprehension of Diophantus, the general
question of the historical development of algebraical notation
possesses great intrinsic interest. Into the general history of this
subject I cannot enter in this essay, my object being the elucidation
of Diophantus ; I shall accordingly in general confine myself to an
account of his notation solely, except in so far as it is interesting
to compare it with the corresponding notation of his editors and
(in certain cases) that of other writers, as, for example, certain of
the early Arabian algebraists.
First, as to the representation of an unknown quantity. The
unknown quantity, which Diophantus defines as containing Tr\yj0og
fjiovdStov dfyicTTov, i.e. an undefined number of units (def. 2), is
denoted throughout by what was printed in the editions before
Tannery's as the Greek letter 9 with an accent, thus 9', or in the
form $ 6 . This symbol in verbal description he calls o dpiOpo?, " the
number," i.e., by implication, the number par excellence of the problem
in question. In the cases where the symbol is used to denote in
flected forms, e.g.) the accusative singular or the dative plural, the
terminations which would have been added to the stem of the full
word dpiOfto? were printed above the symbol 9 in the manner of an
NOTATION AND DEFINITIONS OF DIOPHANTUS 33
exponent, thus 9 NV (for apiff/aov, as T NV for rov), 9, the symbol being
in addition doubled in the plural cases, thus 99', 99 ov/ff , 99*", 99 orff , for
dpiOjjwl tf.r.e. When the symbol is used in practice, the coefficient
is expressed by putting the required Greek numeral immediately
after it, thus 99 l ~ta corresponds to I L#, 9' a to x and so on.
Tannery discusses the question whether in the archetype (a) of
the MSS. this duplication of the sign for the plural and this
addition of the terminations of the various cases really occurred 1 .
He observes that any one accustomed to reading Greek MSS. will
admit that the marks of cases are common in the later MSS. but
are very frequently omitted in the more ancient. Further, the
practice of duplicating a sign to express the plural is more ancient
than that of adding the caseterminations. Tannery concludes that
the caseterminations (like the final syllables of abbreviations used
for other words) were very generally, if not always, wanting in the
archetype (a). If this seems inconsistent with the regularity with
which they appear in our MSS., it has to be remembered that A
and &! do not represent the archetype (a) but the readings of a, the
copyist of which probably took it upon himself to substitute the
full word for the sign or to add the caseterminations. Tannery's
main argument is the frequent occurrence of instances where the
wrong caseending has been added, e.g., the nominative for the
genitive ; the conclusion is also confirmed by instances in which
different cases of the word a/nfyto?, e.g. api,9p,ov, dpiOpov, and even
apiOfM&v written in full are put by mistake for /cat owing to the
resemblance between the common abbreviation for /cai and the
sign for dpidp.6^ and of course in such cases the abbreviation would
not have had the endings. As regards the duplication of the sign
for the plural, Tannery admits that this was the practice of the
Byzantines ; but he considers that the evidence is against sup
posing that Diophantus duplicated the sign ; he does not do so
with any other of his technical abbreviations, those for pavd^
$vvajj,i,$, etc. Accordingly in his text of Diophantus Tannery has
omitted the caseendings and written the single sign for aptSpo?
whether in the singular or in the plural ; in his second volume,
however, containing the scholia, etc., he has retained the duplicated
sign.
On the assumption that the sign was the Greek final sigma, it
was natural that Nesselmann should explain it by the supposition
1 Dioph. II. pp. xxxivxxxix.
H, D. 3
34 INTRODUCTION
that Diophantus, in search of a convenient symbol for his unknown
quantity, would select the only letter of the Greek alphabet which
was not already appropriated as a numeral 1 . But he made the
acute observation 2 that, as the symbol occurred in many places (of
course in Bachet's text) for apidpw used in the ordinary un
technical sense, and was therefore, as it appeared, not exclusively
used to designate the unknown quantity, the technical dpidpo?, it
must after all be more of the nature of an abbreviation than an
algebraical symbol like our x. It is true that this uncertainty in
the use of the sign in the MSS. is put an end to by Tannery, who
uses it for the technical dpidfio^ alone and writes the untechnical
dpiOfjbo? in full ; but, even if Diophantus' practice was as strict as
this, I do not think this argues any difference in the nature of the
abbreviation. There is also a doubt whether the final sigma, 9,
was developed as distinct from the form a so early as the date of
the MSS. of Diophantus, or rather so early as the first copy of his
work, if the author himself really gave the explanation of the sign
as found in our text of his second definition. These considerations
suggested to me that the sign was not the final sigma at all, but
must be explained in some other way. I had to look for con
firmation of this to the precise shape of the sign as found in extant
MSS. The only MS. which I had the opportunity of inspecting
personally was the MS. of the first ten problems of Diophantus in
the Bodleian ; but here I found strong confirmation of my view in
the fact that the sign appeared as'<^, quite different in shape from,
and much larger than, the final sigma at the end of words in the
same MS. (There is in the Oxford MS. the same irregularity as
was pointed out by Nesselmann in the use of the sign sometimes
for the technical, and sometimes for the untechnical, aptdyudv".)
But I found evidence that the sign appeared elsewhere in some
what different forms. Thus Rodct in the Journal Asiatique of
January, 1878, quoted certain passages from Diophantus for the
purpose of comparison with the algebra of Muhammad b. MUSH
alKhuwarazml. Rodet says he copied these passages exactly
from Sachet's MS. ; but, while he generally gives the sign as the
final sigma, he has in one case ii ot for dpifffiot. In this last case
1 Nesselmann, pp. 2901. 2 ibid* pp. 3001.
3 An extreme case is raa ro TO\) devrfyov '< dpil9/*oC &>6s, where the sign (contrary to
what would be expected) means the untechnical ApiBpfo, and the technical is written in
full. Also in the definition 6 5 (MI&V rofotav r&v Ifawn&rw KTV}ff<ifi&os.,,dpt0juJ>s wxXeZrat
the word apt&/A&s is itself denoted by the symbol, showing that the word and the symbol
are absolutely convertible.
NOTATION AND DEFINITIONS OF DIOPHANTUS 35
Bachet himself reads 9? 01 . But the same form L)LJ ot which Rodet
gives is actually found in three places in Bachet's own edition,
(i) In his note to IV. 3 he gives a reading from his own MS. which
he has corrected in his own text and in which the signs ia and
41(97 occur, evidently meaning dpiBpfc a and dpid^ol TJ, though the
sign should have been that for dpiBfjuotnov (= i\x). (2) In the text of
IV. 1 3 there is a sentence (marked by Bachet as interpolated) which
contains the expression tjijr, where the context again shows that
tj4 is for dpid/jiol. (3) At the beginning of V. g there is a difficulty
in the text, and Bachet notes that his MS. has prjre 6 SiTrXaoitov
avrov L where a Vatican MS. reads dpiQpov (Xylander notes that
his MS. had in this place fAijre 6 St7r\aa[(ov avrov dp fMo
It is thus clear that the MS. (Paris. 2379) which Bachet
sometimes has the sign for dpiOpos in a form which is a^d be
sufficiently like i to be taken for it. Tannery states that tfi H pp ro 
of the sign found in the Madrid MS. (A) is q, while B^ has/ see in
form (S) nearly approaching Bachet's reproduction of it. ""
It appeared also that the use of the .sign, or something like
it, was not confined to MSS. of Diophantus ; on reference to
Gardthausen, Griechiscke Pctlaeographie^ I found under _ he head
"hieroglyphischconventionell" an abbreviation 9, 9^ fjcr apiQ^,
ot, which is given as occurring in the Bodleian M?g of Euclid
(D'Orville 301) of the Qth century. Similarly Lehmanfii 1 notes as
a sign for dpiOjios found in that MS. a curved line similar to that
which was used as an abbreviation for /cai. He adds that the
ending is placed above it and the sign is doubled for the plural.
Lehmann's facsimile is like the form given by Gardthausen, but has
the angle a little more rounded. The form iji ot above mentioned
is also given by Lehmann, with the remark that it seems to be
only a modification of the other. Again, from the critical notes to
Heiberg's texts of the Arenarius of Archimedes it is. clear that the
sign for dpidfjuos occurred several times in the MSS. in a form
approximating to that of the final sigma, and that there was the
usual confusion caused by the similarity of the signs for dptBfjuos
and /cai 2 . In Hultsch's edition of Heron, similarly, the critical
notes to the Geodaesia show that one MS. had an abbreviation for
1 Lehmann, Die tachygraphischen Abkurzungen der griechischen Handsckriften^ 1880,
p. 107 : " Von Sigeln, welchen ichauch anderwarts begegnet bin, sind 211 nennen d/wfyids,
das in der Oxforder Euclidhandschrift mit einer der Note Kal ahnlichen Schlangenlinie
bezeichnet wird."
2 Cf. Heiberg, Quatstiones Archimedean, pp. 172, 174, 187, 188, 191, 192; Archimedis
opera omnia, II., pp. 268 sqq.
36 INTRODUCTION
dpi0fj,6<i in various forms with the caseendings superposed ; some
times they resembled the letter , sometimes p, sometimes o and
once f 1 . Lastly, the sign for dpid^ resembling the final sigma
evidently appeared in a MS. of Theon of Smyrna 2 ,
All these facts strongly support the assumption that the sign
was a mere tachygraphic abbreviation and not an algebraical
symbol like our x> though discharging much the same function.
The next question is, what is its origin ? The facts (i) that the
sign has the breathing prefixed in the Bodleian MS., which writes
*^> for apiOpos, and (2) that in one place Xylander's MS. read dp
tor the full word, suggested to me the question whether it could
k^ c a contraction of the first two letters of dpL0fjLo<; ; and, on con
su * en ation, this seemed to me quite possible when I found a
cotttr "^jon f or a p gf ven by Gardthausen, namely <Jp. It is easy to
see ' ;: a simplification of this in different ways would readily
Ste^x u sl '^ ns ^ke tne different forms shown above. This then
was th'f hypothesis which I put forward twentyfive years ago, and
which 1\ still hold to be the easiest and best explanation. Two
alternatives are possible, (i) Diophantus may not have made the
contraction himself. In that case I suppose the sign to be a cur
sive contSlf tion made by scribes ; and I conceive it to have come
about through the intermediate form <p. The loss of the downward
stroke, or * the loop, would produce a close approximation to
the forms vvte} 1 we know. (2) Diophantus may have used a sign
approximately/%4jot exactly, like that which we find in the MSS.
For it is from a papft^of 154 A,D., in writing of the class which
Gardthausen calls the u M^MWtoWf8W l 8f^ffL ll flrL contraction c/p for
the two letters is taken. The great advantage of my hypothesis is
that it makes the sign for dpiQp,6$ exactly parallel to those for the
powers of the unknown, e.g. y A Y for ^vvapi? and K Y for Kv/3o^ t ancl
o
to that for the unit yu,om? which is denoted by M, with the sole
difference that the letters coalesce into one instead of being
written separately.
Tannery's views on the subject are, I think, not very con
sistent, and certainly they do not commend themselves to me. He
seems to suggest that the sign is the ancient letter Koppa, perhaps
slightly modified ; he first says that the sign in Diophantus is
peculiar to him and that, although the word dptG^ is very often
* Heron, ed. Hultsch, pp. 146, 148, 149, 150*
2 Theon of Smyrna, ed. Hiller, p. 56, critical notes.
NOTATION AND DEFINITIONS OF DIOPHANTUS 37
represented in mathematical MSS. by an abbreviation, it has much
oftener the form v or something similar, closely resembling the
ancient Koppa. In the next sentence he seems to say that " on
the contrary the Diophantine abbreviation is an inverted di
gamma " ; yet lower down he says that the copyist of a" (copied
from the archetype a} got the form \\ by simplifying tKe more
complicated Koppa. And, just before the last remark, hennas
stated that in the archetype a the form must have been 5 or
like it, as is shown by the confusion with the s l ign for /caL
is so, it can hardly have been peculiar to Diophantus, seeing/that
the same confusion occurs fairly often in the MSS. of 1 other
authors, as above shown.) I think the last consideration (:the con
fusion with fcai) is very much against the Koppahypothests ; and,
in any case, it seems to rne very unlikely that a sign would be
used by Diophantus for the unknown which was already appro
priated to the number go. And I confess I am unable t# see in
the sign any resemblance to an inverted digamma.
Hultsch 1 regards it as not impossible that Diophantus may
have adopted one of the signs used by the Egyptians for their
unknown quantity hau, which, if turned round from left to right,
would give V; but here again I see no particular ^semblance.
Prof. D'Arcy Thompson 2 has a suggestion that the sign might be
the first letter of <r&>p69, a heap. But, apart from ^ 7 fact that the
final sigma (9) is not that first letter, there is no trace whatever
in Diophantus of such a use of the word <j<w/>6<? ; and, when
Pachy meres 3 speaks of a number being owpeia fcoi/aStwi/, he means
no more than the fr\f}0o$ /jLovdSav which he is explaining : his
words have no connexion with the Egyptian hau.
Notwithstanding that the sign is not the final sigma, I shall
not hesitate to use 9 for it in the sequel, for convenience of
printing. Tannery prints it rather differently as .
We pass to the notation which Diophantus used to express the
different powers of the unknown quantity, corresponding to ^, x* t
and so on. He calls the square of the unknown quantity SvvafMs,
and denotes it by the abbreviation J r . The word Svi>a/u9,
literally " power," is constantly used in Greek mathematics for
1 Art. Diophantus in PaulyWissowa's RealEncydopadie der classischen Aliertums
ivissmschaften.
3 Transactions of the Royal Society of Edinburgh, Vol. xxxvni. (1896), pp. 6079.
3 Dioph. II. p. 78, 4. Cf. lamblichus, ed. Pistelli, p. 7, 7 ; 34, 3 ; 81, 14, where wpefa
is similarly used to elucidate T
38 INTRODUCTION
square^. With Di9phntus, however, it is not any square, but only
the square of tfre unknown ; where he speaks of any particular
square number, it is rerpayfovos dpiffpos. The higher powers of the
unknown^quantity which Diophantus makes use of he calls /n;/3o9,
SwafioSi'ivafus, Swapofcvfto?, fcvfto/cvfio?, corresponding respectively
to x 8 , $f t x? } x*. Beyond the sixth power he does not go, having
no 9ccasion for higher powers in the solutions of his problems. For
these powers he uses the abbreviations K Y , A r A, AK Y , K Y K re
There is>a difference between Diophantus' use of the word
i>i$ and of the complete words for the third and higher powers,
namely that the latter are not always restricted like Svvajj,i$ to powers
of the unknown, but may denote powers of ordinary known num
bers as well This is no doubt owing to the fact that while there
are two v \yords Svvajw and TGrpdyavo? which both signify " square,"
there is "only one word for a third power, namely /cv/3o$. It is
important, however, to observe that the abbreviations K r , A r A,
AK Y , K Y K } are, like Svvafu? and A r > only used to denote powers
of the unknown. The coefficients of the different powers of the
unknown, like that of the unknown itself, are expressed by the
addition of the Greek letters denoting numerals, e.g., AK Y r cor
responds ijp&x*. Thus in Diophantus' system of notation the signs
A Y and theVest represent not merely the exponent of a power like
the 2 in x, but the whole expression x*. There is no obvious
connexion between the symbol A Y and the symbol <? of which it is
the square, as there is between x* and x, and in this lies the great
inconvenience of the notation. But upon this notation no advance
was made by Xylander, or even by Bachet and Fermat They wrote
N (which was short for Numerus) for the 9 of Diophantus, Q (Quad
ratus) for A Y , C (Cubus) for K Y , so that we find, for example,
i Q 4 5^24, corresponding to x* + 5# = 24. Other symbols were
however used even before the publication of Xylander's Diophantus,
e.g. in Bombelli's Algebra. Bombelli denotes the unknown and its
powers by the symbols . 1, L, and so on. But it is certain that
up to this time (1572) the common symbols had been R (Radix
or Res), Z (Zensus, i.e. square), C (Cubus). Apparently the first
important step towards ^ x 3 , etc., was taken by Vieta (1540
1 In Plato we have dtivafju.* used for a square number (Timatus, 31) and also
(Theaetetus, 147 D) for a square root of a number which is not a complete square, i.e. for
a surd ; but the commonest use is in geometry, in the form Sw<fyc, " in square," e.g. "AB
is Swdpei double of BC" means
NOTATION AND DEFINITIONS OF DIOPHANTUS 39
1603), who wrote Aq, Ac, Aqq, etc. (abbreviated for A quadratics
and so on) for the powers of A. This system, besides showing the
con .exion between the different powers, has the infinite advantage
that by means of it we can use in one and the same solution any
number of unknown quantities. This is absolutely impossible with
the notation used by Diophantus and the earlier algebraists.
Diophantus in fact never uses more than one unknown quantity in
the solution of a problem, namely the api0p,6<$ or 9.
Diophantus has no symbol for the operation of multiplication ;
it is rendered unnecessary by the fact that his coefficients are all
definite numbers or fractions, and the results are simply put down
without any preliminary step which would call for the use of a
symbol. On the ground that Diophantus uses only numerical
expressions for coefficients instead of general symbols, it might
occur to a superficial observer that there must be a great w t ant
of generality in his methods, and that his problems, being solved
with reference to particular numbers only, would possess the
attraction of a clever puzzle rather than any more general interest.
The answer to this is that, in the first place, it was absolutely
impossible that Diophantus should have used any other than
numerical coefficients, for the reason that the available symbols of
notation were already employed, the letters of the Greek alphabet
always doing duty as numerals, with the exception of the final 9.
In the second place, it is not the case that the use of none but
numerical coefficients makes his solutions any the less general.
This will be clearly seen when I come to give an account of his
problems and methods.
Next as to Diophantus' expressions for the operations of
addition and subtraction. For the former no symbol at all is
used: it is expressed by mere juxtaposition, thus JK r aA r ~vy<iG
corresponds to ^f i3# 2 f $x. In this expression, however, there
is no absolute term, and the addition of a simple numeral, as
for instance y8, directly after e, the coefficient of ?, would cause
confusion. This fact makes it necessary to have some expression
to distinguish the absolute term from the variable terms. For this
purpose Diophantus uses the word /uoz/aSe?, or units, and denotes
them after his usual manner by the abbreviation M. The number
of units is expressed as a coefficient Thus corresponding to
the expression ^+13^ + 5^ + 2 we should find in Diophantus
K Y a A T T*j<;eMfi. As Sachet uses the sign + for addition, he
40 INTRODUCTION
has no occasion for a distinct symbol to mark an absolute term.
He accordingly writes iC+ i$Q + $N+ 2. It is worth observing,
however, that the Italians do use a symbol in this case, namely N
(Nttmero), the first power of the unknown being with them R
(Radio). Cossali 1 makes an interesting comparison between the
terms used by Diophantus for the successive powers of the unknown
and those employed by the Italians after their instructors, the
Arabians. He observes that Fra Luca (Paciuolo), Tartaglia, and
Cardano begin their scale of powers from the power o, not from the
power I, as does Diophantus, and he compares the scales thus :
Scala Diofantea. Scala Araba.
i. Numero. ..il Noto.
x i. Numero... I' Ignoto. 2. Cosa, Radice, Lalo.
j? 2. Podesta. 3. Ccnso.
jfl 3. Cubo. 4. Cubo.
;t 4 4. PodestaPodesta. 5. Censo di Censo.
x* 5. PodestaCubo. 6. Relate i.
x* 6. CuboCubo. 7. Censo di Cubo, o Cubo di Censo.
x 7 7 8. Relatos .
jc 8 8 9. Censo di Censo di Censo.
x* 9 to. Cubo di Cubo.
and so on. So far, however, as this is meant to be a comparison
between Diophantus and the early Arabian algebraists themselves
(as the title " Scala Araba" would seem to imply), there appears to
be no reason why Cossali should not have placed some term to
express Diophantus' povdSe? in the same line with Numero in the
Arabian scale, and moved the numbers i, 2, 3, etc, one place
upwards in the first scale, or downwards in the second. As
Diophantus does not go beyond the sixth power, the last three
places in the first scale are left blank. An examination of these
two scales will show also that the evolution of the successive
powers differs in the two systems. The Diophantine terirft for
them are based on the addition of exponents, the Arabic on
1 Upon Wallis' comparison of the Diophantine with the Arabian scale Cossali
remarks : "ma egli non ha riflettuto a due altre differenze tra le scale medesime. La
prima si e, che laddove Diofanto denomina con singolarita Numero il nurnero ignoto,
denominando Monade il numero dato di comparazione : gli antichi italiani degli arabi
seguaci denominano questo il Numero ; e Radice, o Lato, o Cosa il numero sconoficiuto.
La seconda e, che Diofanto comincia la scala dal numero ignoto ; e Fra Luca, Tarlaglia 1
Cardano la incominciano dal numero noto. Ecco le due scale di rincontro, onde meglio
risaltino all' occhio le differenze loro V, i. p. T9$.
NOTATION AND DEFINITIONS OF DIOPHANTUS 41
their multiplication 1 . Thus the "cubecube" means in Diophantus
x* 3 while the Italian and Arabian system uses the expression " cube
of cube " and applies it to x? t The first system may (says Cossali)
be described as the method of representing each power by the
product of the two lesser powers which are the nearest to it, the
method of multiplication] the second the method of elevation, i.e. the
method which forms by the process of squaring and cubing all
powers which can be so formed, as the 4th, 6th, 8th, pth, etc.
The intermediate powers which cannot be so formed are called
in Italian Relati. Thus the fifth power is Relato i, x 1 is
Relato 2, x is Censo di Relato i, x is Relato 3, and so on.
Another name for the Relati in use among European algebraists in
the 1 6th and I7th centuries was sursolida, with the variants sttper
solida and surdesolida.
It is interesting to compare with these systems the Egyptian
method described by Psellus* The next power after the fourth
), Le* x 5 , the Egyptians called " the first undescribed "
here apparently meaning that of which no account can
be given), because it is neither a square nor a cube ; alternatively
they called it " the fifth number, 1 ' corresponding to the fifth power
of x. The sixth power they apparently called " cubecube " ; but
the seventh was "the second undescribed" (aXoyo? Seurepo?), as
being the product of the square and the " first undescribed," or,
alternatively, the "seventh number." The eighth power was the
"quadruplesquare" (rerpairKri SvW/w), the ninth the "extended
cube" (tfi5/3o9 e'feTu/ero?).. Thus the "first undescribed" and the
"second undescribed" correspond to "Relato i" and c< Relato 2"
respectively, but the "quadruplesquare" exhibits the additive
principle.
For subtraction Diophantus uses a symbol. His full term for
negation or wanting is A,e^t9, corresponding to virapfys which
denotes the opposite. The symbol used to denote it in the MSS.,
and corresponding to our for minus, is (Def. 9 ical TTJS Xenjreco?
eXXiTre? Kara vevov. A) " an inverted ty with the top
1 This statement of Cossali's needs qualification however. There is at least one Arabian
algebraist, alKarkh! (died probably about 1029), the author of the Fakhrl, who uses the
Diophantine system of powers of the unknown depending on the addition of exponents.
AlKarkhi, namely, expresses all powers of the unknown above the third by means of
mdl, his term for the square, and kab, his term for the cube of the unknown, as follows.
The fourth power is with him mdl mal, the fifth mat kdb, the sixth kdb kdb, the seventh
mal mal kdb, the eighth mdl kab kab, the ninth kdb kdb kdb, and so on. Among the
Italians too there was an exception, Leonardo of Pisa, who proceeded on the additive
principle (Sibliotheca Mathematics vi 3 , 19056, p. 310). 2 Dioph. n. p. 3738.
42 INTRODUCTION
shortened, A" As Diophantus uses no distinct sign for f, it
is clearly necessary, in order to avoid confusion, that all the
negative terms in an expression should be placed together
after all the positive terms. And so in fact he does place them.
Thus corresponding to ^5^+8^1, Diophantus would write
o
K Y a$ril^k Y lMa. With respect to this curious sign, given in
the MSS. as T and described as an inverted truncated M', I believe
that I was the first to suggest that it could not be what it is
represented as being. Even when, as in Sachet's edition, the
sign was printed as ^ I could not believe that Diophantus used
so fantastic a sign for minus as an inverted truncated ty". In
the first place, an inverted W seems too farfetched ; to one who
was looking for a symbol to express minus many others more
natural and less fantastic than ^ must have suggested themselves.
Secondly, given that Diophantus used an inverted M/ 1 , why should he
truncate it ? Surely that must have been unnecessary ; we could
hardly have expected it unless, without it, confusion was likely
to arise ; but ^ could not well have been confused with anything.
This very truncation itself appears to throw doubt on the description
of the symbol as we find it in the MS. I concluded that the con
ception of this symbol as an inverted truncated "SP was a mistake,
and that the description of it as such is not Diophantus' description,
but an explanation by a scribe of a symbol which he did not
understand 1 . I believe that the true explanation is the following.
Diophantus here took the same course as in the case of the other
symbols which we have discussed (those for apiO^os, Svva/jux, etc.).
As in those cases he took for his abbreviation the first letter of the
word with such an addition as would make confusion with numbers
impossible (namely the second letter of the word, which in each of
the cases happens to come later in the alphabet than the corre
sponding first letter), so, in seeking an abbreviation for Xet^Jw
and cognate inflected forms developed from XATT, he began by
taking the initial letter of the word. The uncial 2 form is A.
Clearly A by itself would not serve his purpose, since it denotes
a number. Therefore an addition is necessary. The second letter
is E, but AE is equally a number. The second letter of the stem
1 I am not even sure that the description can be made to mean all that it is intended
to mean. &Ar& scarcely seems to be sufficiently precise. Might it not be applied to
4> with any part cut off, and not only the top ?
3 I adhere to the uncial form above for clearness' .sake. If Diophantus used the
" Majuskelcursive " form, the explanation will equally apply* the difference of form being
for our purpose negligible.
NOTATION AND DEFINITIONS OF DIOPHANTUS 43
\LTT is I, but Al is open to objection when so written. Hence
Diophantus placed the I inside the A, thus, A. Of the possibility
of this I entertain no doubt, because there are undoubted cases
of combination, even in uncial writing, of two letters into one sign.
I would refer in particular to X, which is an uncial abbreviation for
TAAANTON. Now this sign, /K, is an inverted and truncated "9
(written in the uncial form, t) ; and we can, on this assumption,
easily account for the explanation of the sign for minus which is
given in the text.
The above suggestion, made by me twentyfive years ago,
seems to be distinctly supported by what Tannery says of the form
in which the sign appears in the MSS. 1 Thus he remarks (i) that
the sign in the MSS. is often made to lean to the right so that it
resembles the letter Lambda, (2) that Planudes certainly wrote fc as
if he meant to write the first letter of \etyei, and (3) that the
letter A appears twice in A where it seems to mean XOOTO?. Yet
in his edition of Diophantus Tannery did not adopt my explanation
or even mention it, but explained the sign as being in reality
adapted from the old letter Sampi (^), the objection to which
suggestion is the same as that to which the identification of 9 with
Koppa is open, namely that "^ represented the number 900, as ?
represented 90. Tannery however afterwards 2 saw reason to
abandon his suggestion that the symbol was originally an archaic
form of the Greek Sampi rather than " un monogramme se
rattachant a la racine de Xer^?." The occasion for this change
of view was furnished by the appearance of the same sign in the
critical notes to Schone's edition of the Metrica of Heron 3 , which
led Tannery to reexamine the evidence of the MSS. of Diophantus
as to the sign and as to the exact word or words which it re
presented in different places, as well as to search for any similar
expressions denoting subtraction which might occur in the works
of other Greek mathematicians. In the MSS. of Diophantus,
when the sign is resolved by writing a full word instead of it,
it is generally resolved into Xe/tyret, the dative of XeZ^rt? ; in such
cases the only grammatical possibility is to construct it with the
genitive case of the quantity subtracted, the meaning then being
"with the wanting, or deduction, of ...". But the best MS. (A)
1 Dioph. ii. p. xli.
2 Bibliotheca Mathematica V 3 , 19045, pp. 58.
:J fferonis Alexandrini opera^ Vol. III., 1903, pp, 156, 8, 10. The MS. reading is
tf 5', the meaning of which is 74  T V
44 INTRODUCTION
has in some places the nominative Xe^ Awhile in others it has the
symbol instead of parts of the verb Xe/7rez>, namely Xwrcwy or
Xe&S/ra? and once even XLTTWOI ; hence we may conclude that in the
cases where A and B l have Xen^et followed by the accusative (which
is impossible grammatically) the sign was wrongly resolved, and
the full word should have been a participle or other part of the
verb \ehrew governing the accusative. The question therefore
arises whether Diophantus himself used the dative Xe^JreA at all
or whether it was introduced into the MSS. later. Certain it is
that the use is foreign to Classical Greek ; but, even if it began
with Diophantus, it did not finally hold the field before the time of
Planudes. No evidence for it can be found in Greek mathe
maticians before Diophantus. Ptolemy has in two places Xet^az/
and \elirovoav respectively, followed by the accusative, and in
one case TO airo T^S FA Xet<0ez/ VTTO rofi aVo rfjs ZF (where the
meaning is ZP  FA S ). Consequently we cannot suppose that the
sign where it occurs in the Mctrica of Heron represents the dative
Xen^a; it must rather stand for a participle, active or passive.
Tannery suggests that the full expression in that passage was
/jiov(i$Q>v oS \i<j>devTo$ Teooapa/caiSejcdrov, the participle being
passive and the construction being the genitive absolute; but I
think a perhaps better alternative would be fMovdSwv oS \w^ra,cr)v
TeooapafccuSe/caTov, where the active participle would govern
the accusative case of the term subtracted* From all this we
may infer that the sign had no exclusive reference to the sub
stantive Xen/r9, still less to the dative case of that substantive, but
was a conventional abbreviation associated with the root of the
verb Xe/7mz>. In these circumstances I think I may now fairly
claim Tannery as, substantially, a convert to my view of the
nature of the sign.
For division it often happens that no symbol is necessary,
i.e. in the cases where the divisor divides the dividend without
a remainder. In other cases the quotient has to be expressed
as a fraction, whether the divisor is a specific number or contains
the variable. The case of division comes then under that of
fractions.
Fractions are represented in different ways according as they are
submultiples (fractions with unity as numerator) or not In the
case of submultiples the Greeks did not write the numerator, but
only the denominator, distinguishing the submultiple from the
cardinal number itself by affixing a certain sign. In more recent
NOTATION AND DEFINITIONS OF DIOPHANTUS 45
MSS. a double accent was used for this purpose: thus 7" = ^.
Diophantus follows this plan in the hypothesis and analysis of his
problems, though in the solutions he seems to have written the
numerator a and assimilated the notation to that used for other
fractions. The sign, however, added to the cardinal number to
express the submultiple takes somewhat different forms in A :
sometimes it is a simple accent, sometimes more elaborate, as ^
above the letter and to the right, or actually forming a continuation
of the numeral sign, e.g. 0^ = . Tannery adopts as the genuine
mark in Diophantus the affix * in place of the accent : thus y x = \.
For \ he writes L 1 as being most suitable for the time of Diophantus,
though A has ^ ', sometimes without the dot.
Of the other class of fractions (numerator not unity) f stands by
itself, having a peculiar sign of its own j curiously enough it occurs
only four times in Diophantus. A has a sign for it which was
confused with that for dpt,0p,os in one place ; Tannery judges from
the Greek mathematical papyrus of Achrnlm 1 that its original form
was <y ; he himself writes in his text the common form 6^. In the
rare cases where the first hand in the oldest MS. (A) has fractions
as such with numerator and denominator written in full, the
denominator is written above the numerator. Tannery therefore
adopts, in his text, this way of writing fractions, separating the
121
numerator and denominator by a horizontal line: thus prca ^.
It is however better to omit the horizontal line (cf. p in Kenyon
Papyri II. No. cclxv. 40 ; also the fractions in Schone's edition
of Heron's Metrica). Once we find in the same MS. (A) in the first
hand the form ie 5 = ^.. In this latter method of writing fractions
the denominator is written as we write exponents ; and this is the
method adopted by Planudes and by Bachet in his edition.
Another alternative is to write the numerator first, and then the
denominator after it in the same line, marking the denominator with
the submultiple sign in some form ; thus 78" would mean f ; this is
the most convenient method for purposes of printing. Or the de
nominator may be written as an abbreviation for the ordinal number,
and the casetermination may be added higher up ; eg. v K^ V 50
twentythirds. But the denominators are nearly always omitted
1 Published by Baillet in Mtmoires publik par les Membres de la Mission arckfologique
franfaise au Catre, T. Ix, Fascicule i, pp. r88. Paris, 1897.
46 INTRODUCTION
altogether in the first hand of A ; in the first two Books S l and the
second hand of A give the denominator in the place in which we write
an exponent, following the method of Planudes ; in the last four
Books both MSS. almost invariably omit the denominator. In
some cases the omission is not unnatural, i.e. where the denominator
has once been given, and it is almost superfluous to repeat it
in other fractions immediately following which have the same
denominator ; in other cases it was probably omitted because the
superposed denominator was taken by the copyist to be an inter
linear scholium. A few examples of fractions from Diophantus
may be added :
A$ I7 fyfi 2 ,6 a.oa a
* = i (v. 10) ; fiwr = % (iv. 28) ; /6 , = L (v. 9) ;
. 39) :
V '
152
Diophantus however often expresses fractions by putting eV
or fiopiov between the numerator and denominator, i.e. he
r __
says one number divided by another. Cf.
KS . ,/QpAtS = 1507984/262144 (IV. '28), where of course M ~
(tens of thousands); ft., 6% Iv popta p/cft . ,a/ee = 25600/1221025
(v. 22). As we said, the most orthodox way of writing a sub
multiple was to omit the numerator (unity) and use the denominator
with a distinguishing sign attached, e.g. r^ or r' = jV. But m his
solutions Diophantus often uses the form applicable to fractions
l
other than submultiples ; eg. he writes a for  (iv. 28).
512
Numbers partly integral and partly fractional, where the
fraction is a submultiplc or the sum of submultiples, are written
much as we write them, the fraction simply following the integer ;
e.g. d 7 X = ! ; in the Lemma to V, 8 we have /9 L ' r' = 2 1 J or 2j,
where is decomposed into submultiples as in Heron. Cf. also
(m. ii)^V x =37oi^.
Before leaving the subject of numerical notation, it may be
convenient to refer to the method of writing large numbers.
r
Myriads (tens of thousands) are expressed by M, myriads to the
NOTATION AND DEFINITIONS OF DIOPHANTUS 47
second power by MM 7 or, in words, Sevrfya pvpid$. The de
nominator 187474560 in V. 8 would thus be written popiov Sevrepas
a /col pvptdScov 7rpcil>Tcov flty/A^ xal M ,S$, and the fraction
131299224/1629586560 would be written Sei/re/oa p,vpia<$ a
M flojeS jjiopiov Sevrepwv p,vpt,ct$a)v
' MT^ 1 
But there is another kind of fraction, besides the purely
numerical one, which is continually occurring in the Arithmetic^
such fractions namely as involve the unknown quantity in some
form or other in their denominators. The simplest case is that in
which the denominator is merely a power of the unknown, 9.
Concerning fractions of this kind Diophantus says (Def. 3): "As
fractions named after numbers have similar names to those of the
numbers themselves (thus a third is named from three, a fourth
from four), so the fractions homonymous with the numbers just
defined are called after them ; thus from apiBpos we name
the fraction dpLdfioarov \i.e. \l% from #], TO Swaftoorov from
SiW/u<?, TO Kvftocrrov from Kv/3o$, TO Svva/AO&vva/jiooTov from
8vva]jt,oBvvafi,i,<s, TO SwafJiotcv^oo'Tov from 8v^a/irO/cv^8o9, and TO
Kvfto/cvpoaTov from Kvpo/cv/3o$. And every such fraction shall
have, above the sign for the homonymous number, a line to
indicate the species." Thus we find, for example, IV. 3, 9 X 77 cor
responding to S/x and, IV. 15, 9 X Xe for 35/r. Cf. A Y *<rv for 250^.
Where the denominator is a compound expression involving the
unknown and its powers, Diophantus uses the expedient which he
often adopts with numerical fractions when the numerators and
denominators are large numbers, namely the insertion of ev popi
or popiov between the expressions for the numerator and de
nominator. Thus in VI. 12 we have
Y , , r* r
A M /3<b/c ev LiOpLco A A a M ~^j A A f
= (6o# 2 4 2520)^ + 900 
and in VI. 14
For &ro?, equal, connecting the two sides of an equation, the
sign in the archetype seems to have been t? ; but copyists intro
1 Hultsch, loc. cit.
48 INTRODUCTION
duced a sign which was sometimes confused with the sign t for
a/H0/ios ; this was no doubt the same abbreviation LJ as that shown
(with terminations of cases added above) in the list given at
the end of Codex Parisinus 2360 (Archimedes) of contractions
found in the "very ancient" MS. from which it was copied and
which was at one time the property of Georgius Valla 1 .
Diophantus evidently put down his equations in the ordinary
course of writing, i.e. they were written straight on, as are the
steps in the propositions of Euclid, and not put in separate lines for
each step in the process of simplification. In the scholia of
Maxim,us Planudes however we find conspectuses of the problems
with steps in separate lines which, except for the slightly more
cumbrous notation, make the work scarcely more difficult to follow
than it is in our notation' 2 . Though in the MSS. we have the
abbreviation I* to denote equality, Bachet makes no use of any
symbol for the purpose in his Latin translation. He uses
throughout the full Latin word. It is interesting however to observe
that in the notes to his earlier translation (1575) Xylander had
already used a symbol to denote equality, namely , two short
vertical parallel lines. Thus we find, for example (p, 76),
which we should express by tf + 12 = #* + 6x+ 9.
Now that we have described in detail Diophantus' method of
expressing algebraical quantities and relations, it is clear that it is
essentially different in its character from the modern notation.
While in modern times signs and symbols have been developed
1 Heiberg, Quaestiones Archimedean p. 1 15.
2 One instance will suffice. On the left Planudes has abbreviations for the words
showing the nature of the steps or the operations they involve, e.g. ZK&. = &#ri? (setting
out), TeTp. = TTpa.y(ji)Vurfj,6'i (squaring), trtivd. = (rtiv0e<ns (adding), d0. = cl0atyris (.subtrac
tion)! pep, ~ /te/wr/^y (division), %ir. = tiirap%i$ (resulting fact).
Dioph. I. 28.
Planudes. Modern equivalent.
[Given numbers] 20, 208
r > y
rerp. A r as$Kfjf>p A y fj.
pep.
Put for the numbers #+ 10, 10  jr.
wSquaring, we have V 2 i20Jf+ 100,
JP 2 + IOO20JT.
Adding, 2 Jf 2 4 2 00 = 208.
Subtracting, 2J^=8.
Dividing, ^=4.
r
Result : [the numbers are] 12, 8.
NOTATION AND DEFINITIONS OF DIOPHANTUS 49
which have no intrinsic relationship to the things which they
represent, but depend for their use upon convention, the case
is quite different in Diophantus, where algebraic notation takes
the form of mere abbreviation of words which are considered as
pronounced or implied.
In order to show in what place, in respect of systems of
algebraic notation, Diophantus stands, Nesselmann observes that
we can, as regards the form of exposition of algebraic operations
and equations, distinguish three historical stages of development,
well marked and easily discernible, (i) The first stage Nessel
mann represents by the name Rhetorical Algebra or "reckoning by
complete words." The characteristic of this stage is the absolute
want of all symbols, the whole of the calculation being carried on
by means of complete words, and forming in fact continuous prose.
As representatives of this first stage Nesselmann mentions lambli
chus (of whose algebraical work he quotes a specimen in his fifth
chapter) "and all Arabian and Persian algebraists who are at
present known." In their works we find no vestige of algebraic
symbols; the same may be said of the oldest Italian algebraists
and their followers, and among them Regiomontanus. (2) The
second stage Nesselmann proposes to call the Syncopated Algebra.
This stage is essentially rhetorical^ and therein like the first in
its treatment of questions ; but we now find for oftenrecurring
operations and quantities certain abbreviational symbols. To
this stage belong Diophantus and, after him, all the later
Europeans until about the middle of the seventeenth century
(with the exception of Vieta, who was the first to establish,
under the name of Logisfica speciosa, as distinct from Logistica
ntimerosa, a regular system of reckoning with letters denoting
magnitudes and not numbers only). (3) To the third stage
Nesselmann gives the name Symbolic Algebra, which uses a com
plete system of notation by signs having no visible connexion
with the words or things which they represent, a complete language
of symbols, which supplants entirely the rhetorical system, it being
possible to work out a solution without using a single word of the
ordinary written language, with the exception (for clearness' sake)
of a connecting word or two here and there, and so on 1 . Neither
1 It may be convenient to note here the beginnings of some of oar ordinary algebraical
symbols. The signs + and  first appeared in print in Johann Widman's arithmetic
(1489), where however they are scarcely used as regular symbols of operation ; next they
are found in the Rechenbuch of Henricus Grammateus (Schreiber), written in 1518 but
perhaps not published till 1521, and then regularly in Stifel's Arithmetica integra (1544)
H. D. 4
S o INTRODUCTION
is it the Europeans from the middle of the seventeenth century
onwards who were the first to use symbolic forms of Algebra.
In this they were anticipated by the Indians.
Nesselmann illustrates these three stages by three examples,
quoting word for word the solution of a quadratic equation
by Muhammad b. Musa as an example of the first stage, and
the solution of a problem from Diophantus as representing the
second.
First Stage. Example from Muhammad b. Musa (ed, Rosen,
p. 5). "A square and ten of its roots are equal to nine and thirty
dirhems, that is, if you add ten roots to one square, the sum is equal
to nine and thirty. The solution is as follows. Take half the number
of roots ) that is in this case five; then multiply this by itself, and
the result is five and twenty. Add this to the nine and thirty,
which gives sixtyfour; take the square root, or eight, and subtract
from it half the number of roots > namely five, and there remain
three: this is the root of the square which was required, and the
square itself is nine 1 . 1 '
Here we observe that not even are symbols used for numbers,
so that this example is even more " rhetorical" than the work of
lamblichus who does use the Greek symbols for his numbers.
as well as in his edition of Rudolffs Coss (1553). Vieta (15401603) has, in addition,
= for <. Robert Recorde (15101558) had already in his Algebra (The Whetstone of
WitU) 1557) used = (but with much longer lines) to denote equality ("bicause noe.s.
thynges, can be inoare equalle"). Harriot (15601621) denoted multiplication by a dot,
and also by mere juxtaposition of letters; Stifel (14871567) had however already
expressed the product of two magnitudes by the juxtaposition of the two letters represent
ing them. Oughtred (15741660) used the sign x for multiplication. Harriot also
introduced the signs > and < for greater and less respectively. 5 for division is found
in Rahn's Algebra (1659). Descartes introduced in his Geometry (1637) our method of
writing powers, as a?, at etc. (except a*, for which he wrote aa) ; but this notation was
practically anticipated by Pierre Herigone (Court mathtmatique> 1634), who wrote ai t a$,
04, etc., and the idea is even to be found in the Rechenbuch of Grammateus above
mentioned, where the successive powers of the unknown are denoted by pri, se, ter, etc.
The use of x for the unknown quantity began with Descartes, who first used z, thenjy, and
then x for this purpose, showing that he intentionally chose his unknowns from the last
letters of the alphabet. V for the square root is traceable to Rudolff, with whom it had
only two strokes, the first (down) stroke being short, and the other relatively long.
1 Thus Muhammad b. Miisa states in words the following solution.
therefore #45 = 8,
#=3.
NOTATION AND DEFINITIONS OF DIOPHANTUS 51
Second Stage. As an example of Diophantus I give a trans
lation word for word of II. 8. So as to make the symbols correspond
exactly I use 5 (Square) for A Y (StW/u?), N (Number) for 9, U
o
(Units) for M (/touaSe?).
"To divide the proposed square into two squares. Let it be
proposed then to divide 16 into two squares. And let the first be
supposed to be 1 5; therefore the second will be 16 U i S. Thus
16 7 1 5 must be equal to a square. I form the square from any
number of N*s minus as many U's as there are in the side of
16 7's. Suppose this to be 2N 4Z7. Thus the square itself will
be 4^ i6U i6N. These are equal to i6U iS. Add to each
the negative term (?? Xeijw, the deficiency) and take likes from
likes. Thus 5 S are equal to i6N, and the N is 16 fifths. One
[square] will be ^ 6 , and the other ^ and the sum of the two
makes up ^, or i6U> and each of the two is a square."
Of the third stage any exemplification is unnecessary.
To the form of Diophantus' notation is due the fact that he
is unable to introduce into his solutions more than one unknown
quantity. This limitation has made his procedure often very dif
ferent from our modern work. In the first place we can begin
with any number of unknown quantities denoted by different
symbols, and eliminate all of them but one by gradual steps in the
course of the work; Diophantus on the other hand has to perform
all his eliminations beforehand, as a preliminary to the actual
work, by expressing every quantity which occurs in the problem
;n terms of only one unknown. This is the case in the great
majority of questions of the first Book, which involve the solu
tion of determinate simultaneous equations of the first degree
with two, three, or four variables ; all these Diophantus expresses 11
in terms of one unknown, and then proceeds to find it from a
simple equation. Secondly, however, this limitation affects much of
Diophantus' work injuriously; for, when he handles problems which
are by nature indeterminate and would lead with our notation to an
indeterminate equation containing two or three unknowns, he is
compelled by limitation of notation to assume for one or other of
these some particular number arbitrarily chosen, the effect of the
assumption being to make the problem a determinate one. How
ever, it is but fair to say that Diophantus, in assigning an arbitrary
value to a quantity, is careful to tell us so, saying, "for such and
such a quantity we put any number whatever, say such and such a
42
52 INTRODUCTION
number." Thus it can hardly be said that there is (as a rule) any
loss of generality. We may say, then, that in general Diophantus is
obliged to express all his unknowns in terms, or as functions, of
one variable. He compels our admiration by the clever devices
by which he contrives so to express them in terms of his single
unknown, 9, as to satisfy by that very expression of them all
conditions of the problem except one, which then enables us to
complete the solution by determining the value of 9. Another
consequence of Diophantus' want of other symbols besides 9 to
express more variables than one is that, when (as often happens)
it is necessary in the course of a problem to work out a subsidiary
problem in order to obtain the coefficients etc. in the functions of 9
which express the numbers to be found, the unknown quantity
which it is the object of the new subsidiary problem to find is also
in its turn denoted by the same symbol 9 ; hence we often have
in the same problem the same variable 9 used with two different
meanings. This is an obvious inconvenience and might lead to
confusion in the mind of a careless reader. Again we find two
cases, II. 28 and 29, where for the proper workingout of the
problem two unknowns are imperatively necessary. We should of
course use x and^; but Diophantus calls the first 9 as usual; the
second, for want of a term, he agrees to call "one zmit" i.e. i.
Then, later, having completed the part of the solution necessary to
find 9, he substitutes its value, and uses 9 over again to denote
what he had originally called "i" the second variable and so
finds it. This is the most curious case of all, and the way in which
Diophantus, after having worked with this " I " along with other
numerals, is yet able to put his finger upon the particular place
>where it has passed to, so as to substitute 9 for it, is very remark
able. This could only be possible in particular cases such as those
which I have mentioned; but, even here, it seems scarcely possible
now to work out the problem by using x and I for the variables
as originally taken by Diophantus without falling into confusion.
Perhaps, however, in working out the problems before writing them
down as we have them Diophantus may have given the " I " which
stood for a variable some mark by which he could recognise it
and distinguish it from other numbers.
Diophantus will have in his solutions no numbers whatever
except "rational" numbers; and in pursuance of this restriction he
excludes not only surds and imaginary quantities, but also negative
quantities. Of a negative quantity per se, Le. without some positive
NOTATION AND DEFINITIONS OF DIOPHANTUS 53
quantity to subtract it from, Diophantus had apparently no con
ception, Such equations then as lead to surd, imaginary, or
negative roots he regards as useless for his purpose: the solution
is in these cases ofiiWw, impossible. So we find him (v, 2)
describing the equation 4=4^+20 as aWo?, absurd, because it
would give #=4. Diophantus makes it his object throughout
to obtain solutions in rational numbers, and we find him frequently
giving, as a preliminary, the conditions which must be satisfied in
order to secure a result rational in his sense of the word, In the "
great majority of cases, when Diophantus arrives in the course of
a solution at an equation which would give an irrational result, he
retraces his steps and finds out how his equation has arisen, and
how he may, by altering the previous work, substitute for it
another which shall give a rational result, This gives rise, in
general, to a subsidiary problem the solution of which ensures
a rational result for the problem itself. Though, however, Dio
phantus has no notation for a surd, and does not admit surd
results, it is scarcely true to say that he makes no use of quadratic
equations which lead to such results. Thus, for example, in v, 30
he solves such an equation so far as to be able to see to what
integers the solution would approximate most nearly,
CHAPTER IV
DIOPHANTUS' METHODS OF SOLUTION
BEFORE I give an account in detail of the different methods
which Diophantus employs for the solution of his problems, so far
as they can be classified, it is worth while to quote some remarks
which Hankel has made in his account of Diophantus 1 . Hankel,
writing with his usual brilliancy, says in the place referred to, "The
reader will now be desirous to become acquainted with the classes
of indeterminate problems which Diophantus treats of, and with
his methods of solution. As regards the first point, we must observe
that included in the 130 (or so) indeterminate problems, of which
Diophantus treats in his great work, there are over 50 different
classes of problems, strung together on no recognisable principle
of grouping, except that the solution of the earlier problems facili
tates that of the later. The first Book is confined to determinate
algebraic equations; Books II. to V. contain for the most part
indeterminate problems, in which expressions involving in the first
or second degree two or more variables are to be made squares or
cubes. Lastly, Book VI. is concerned with rightangled triangles
regarded purely arithmetically, in which some linear or quadratic
function of the sides is to be made a square or a cube. That is all
that we can pronounce about this varied series of problems without
exhibiting singly each of the fifty classes. Almost more different
in kind than the problems are their solutions, and we are completely
unable to give an even tolerably exhaustive review of the different
turns which his procedure takes. Of more general comprehensive
methods there is in our author no trace discoverable: every
question requires a quite special method, which often will not
serve even for the most closely allied problems. It is on that
1 Zur Geschichte der Mtihtuwtik in Alterthum uncl Mtttelditr, Leipzig, 1874,
pp. 1645,
DIOPHANTUS' METHODS OF SOLUTION 55
account difficult for a modern mathematician even after studying
too Diophantine solutions to solve the xoist problem; and if we
have made the attempt, and after some vain endeavours read
Diophantus' own solution, we shall be astonished to see how
suddenly he leaves the broad highroad, dashes into a sidepath
and with a quick turn reaches the goal, often enough a goal with
reaching which we should not be content; we expected to have
to climb a toilsome path, but to be rewarded at the end by an
extensive view; instead of which our guide leads by narrow,
strange, but smooth ways to a small eminence; he has finished!
He lacks the calm and concentrated energy for a deep plunge
into a single important problem; and in this way the reader also
hurries with inward unrest from problem to problem, as in a
game of riddles, without being able to enjoy the individual one.
Diophantus dazzles more than he delights. He is in a wonderful
measure shrewd, clever, quicksighted, indefatigable, but does not
penetrate thoroughly or deeply into the root of the matter. As
his problems seem framed in obedience to no obvious scientific
necessity, but often only for the sake of the solution, the solution
itself also lacks completeness and deeper signification. He is a
brilliant performer in the art of indeterminate analysis invented by
him, but the science has nevertheless been indebted, at least directly,
to this brilliant genius for few methods, because he was deficient
in the speculative thought which sees in the True more than the
Correct That is the general impression which I have derived from
a thorough and repeated study of Diophantus' arithmetic."
It might be inferred from these remarks of Hankel that
Diophantus' object was less to teach methods than to obtain a
multitude of mere results. On the other hand Nesselmann
observes 1 that Diophantus, while using (as he must) specific
numbers for numbers which are " given " or have to be arbitrarily
assumed, always makes it clear how by varying our initial as
sumptions we can obtain any number of particular solutions of
the problem, showing "that his whole attention is directed to
the explanation of the method, to which end numerical examples
only serve as means"; this is proved by his frequently stopping
short, when the method has been made sufficiently clear, and
the remainder of the work is mere straightforward calculation.
The truth seems to be that there is as much in the shape of general
* Algebra der Griechen, pp. 3089*
56 INTRODUCTION
methods to be found in Diophantus as his notation and the nature
of the subject admitted of. On this point I can quote no better
authority than Euler, who says 1 : " Diophantus himself, it is true,
gives only the most special solutions of all the questions which he
treats, and he is generally content with indicating numbers which
furnish one single solution. But it must not be supposed that his
method was restricted to these very special solutions. In his
time the use of letters to denote undetermined numbers was not
yet established, and consequently the more general solutions which
we are now enabled to give by means of such notation could not
be expected from him. Nevertheless, the actual methods which he
uses for solving any of his problems are as general as those which
are in use today; nay, we are obliged to admit that there is
hardly any method yet invented in this kind of analysis of which
there are not sufficiently distinct traces to be discovered in Dio
phantus."
In his 8th chapter, entitled "Diophantus' treatment of equations V*
Nesseltnann gives an account of Diophantus' solutions of (i) Deter
minate, (2) Indeterminate equations, classified according to their
kind. In chapter 9, entitled "Diophantus' methods of solution 3 ,"
he classifies these " methods " as follows 4 : (i) " The adroit assump
tion of unknowns," (2) "Method of reckoning backwards and
auxiliary questions," (3) "Use of the symbol for the unknown in
different significations" (4) "Method of Limits," (5) "Solution by
mere reflection," (6) "Solution in general expressions," (7) "Arbi
trary determinations and assumptions," (8) "Use of the right
angled triangle."
At the end of chapter 8 Nesselmann observes that it is not
his solutions of equations that we have to wonder at, but the art,
amounting to virtuosity, which enabled Diophantus to avoid such
equations as he could not technically solve. We look (says Nessel
mann) with astonishment at his operations, when he reduces the
most difficult problems by some surprising turn to a quite simple
vi Commentarii Academiae Petropolitanae, 17567, Vol. VL (1761), p. i$$zz
mentationes arithmetical collectae (ed. Fuss), 1849, ** P r 93*
3 " Diophant's Behandlung der Gleichungen."
3 " Diophant's Auflosungsmethoden."
4 W "Die geschickte Armahme der Unbekannten," (2) " Methode der Zuriick
rechnung und Nebenaufgabe," (3) "Gebrauch des Symbols fur die Unbekaimte in
verschiedenen Bedeutungen," ( 4 ) "Methode der Grenzen," (5) " Auflosung durch blosse
Reflexion," (6) " Auflb'sung in allgeraeinen Ausdrucken, " (7) " WillkUhrliche Bestim
mungen und Annahmen," (8) "Gebrauch des rechtwinkligen Dreiecks."
DIOPHANTUS' METHODS OF SOLUTION 57
equation. Then, when in the 9th chapter Nesselmann passes to the
"methods," he prefaces it by saying: "To give a complete picture
of Diophantus' methods in all their variety would mean nothing else
than copying his book outright. The individual characteristics of
almost every problem give him occasion to try upon it a peculiar
procedure or found upon it an artifice which cannot be applied to
any other problem.... Mean while, though it may be impossible to
exhibit all his methods in any short space, yet I will try to describe
some operations which occur more often or are particularly re
markable for their elegance, and (where possible) to bring out
the underlying scientific principle by a general exposition and by
a suitable grouping of similar cases under common aspects or
characters." Now the possibility of giving a satisfactory account of
the methods of Diophantus must depend largely upon the meaning
we attach to the word "method." Nesselmann's arrangement seems
to me to be faulty inasmuch as (i) he has treated Diophantus'
solutions of equations, which certainly proceeded on fixed rules,
and therefore by "method? separately from what he calls "methods
of solution," thereby making it appear as though he did not
look upon the "treatment of equations" as "methods"; (2) the
classification of the "Methods of solution" seems unsatisfactory.
Some of the latter can hardly be said to be methods of solution at
all; thus the third, "Use of the symbol for the unknown in different
significations," might be more justly described as a "hindrance to
the solution"; it is an inconvenience to which Diophantus was
subjected owing to the want of notation. Indeed, on the as
sumption of the eight "methods," as Nesselmann describes them,
it is really not surprising that no complete account of them
could be given without copying the whole book. To take the
first, "the adroit assumption of unknowns." Supposing that a
number of essentially different problems are proposed, the differences
make a different choice of an unknown in each case absolutely
necessary. That being so, how could a rule be given for all cases?
The best that can be done is to give a number of typical instances.
Precisely the same remark applies to "methods" (2), (5), (6), (7).
The case of (4), " Method of Limits," is different ; here we have
a " method " in the true sense of the term, *>. in the sense of an
instrument for solution. And accordingly in this case the method
can be exhibited, as I hope to show later on; (8) also deserves
to some extent the name of a " method."
58 INTRODUCTION
In one particular case, Diophantus formally states a method or
rule ; this is his rule for solving what he calls a " doubleequation/'
and will be found in II. n, where such an equation appears for the
first time. Apart from this, we do not find in Diophantus' work
statements of method put generally as bookwork to be applied to
examples. Thus we do not find the separate rules and limitations
for the solution of different kinds of equations systematically
arranged, but we have to seek them out laboriously from the
whole of his work, gathering scattered indications here and there,
and to formulate them in the best way that we can.
I shall now attempt to give a short account of those methods
running through Diophantus which admit of general statement
For the reasons" which I have stated, my arrangement will be
different from that of Nesselmann ; I shall omit some of the heads
in his classification of "methods of solution"; and, in accordance
with his remark that these "methods" can only be adequately
described by a transcription of the entire work, I shall leave them
to be gathered from a perusal of my reproduction of Diophantus'
book.
I shall begin my account with
I. DIOPHANTUS' TREATMENT OF EQUATIONS.
This subject falls naturally into two divisions: (A) Determinate
equations of different degrees, (B) Indeterminate equations.
(A) Determinate equations.
Diophantus was able without difficulty to solve determinate
equations of the first and second degrees; of a cubic equation we
find in his Arithmetica only one example, and that is a very
special case. The solution of simple equations we may pass over;
we have then to consider Diophantus' methods of solution in the
case of (i) Pure equations, (2) Adfected, or mixed, quadratics.
(i) Pure determinate equations.
By pure equations I mean those equations which contain only
one power of the unknown, whatever the degree. The solution is
effected in the same way whatever the exponent of the term in the
DIOPHANTUS' METHODS OF SOLUTION 59.
unknown; and Diophantus treats pure equations of any degree
as if they were simple equations of the first degree.
He gives a general rule for this case without regard to the
degree 1 : "If a problem leads to an equation in which any terms
are equal to the same terms but have different coefficients, we must
take like from like on both sides, until we get one term equal to
one term. But, if there are on one side or on both sides any negative
terms, the deficient terms must be added on both sides until all the
terms on both sides are positive. Then we must take like from like
until one term is left on each side." After these operations have
been performed, the equation is reduced to the form Art = B and
is considered solved. The cases which occur in Diophantus are
cases in which the value of x is found to be a rational number,
integral or fractional. Diophantus only recognises one value of x
which satisfies this equation ; thus, if m is even, he gives only the
positive value, excluding a negative value as "impossible." In the
same way, when an equation can be reduced in degree by dividing
throughout by any power of x y the possible values, #=o, thus
arising are not taken into account. Thus an equation of the form
& = ax, which is of common occurrence in the earlier part of the
book, is taken to be merely equivalent to the simple equation x=a.
It maybe observed that the greater proportion of the problems
in Book I. are such that more than one unknown quantity is sought.
Now, when there are two unknowns and two conditions, both
unknowns can easily be expressed in terms of one symbol. But,
when there are three or four quantities to be found, this reduction
is much more difficult, and Diophantus shows great adroitness in
effecting it: the ultimate result being that it is only necessary
to solve a simple equation with one unknown quantity.
(2) Mixed quadratic equations.
After the remarks in Def. 1 1 upon the reduction of equations
until we have one term equal to another term, Diophantus
adds 2 : "But we will show you afterwards how, in the case also
when two terms are left equal to a single term, such an equation
can be solved." That is to say, he promises to explain the
solution of a mixed quadratic equation. In the Aritkmetica,
as we possess the book, this promise is not fulfilled. The first
1 Def. ii.
2 tirrcpw 64 (rot Setjopev /cat TTWJ 8tio elduv taw &l K&TaXeLQdfrTW rd TQIOVTOV \tferat.
60 INTRODUCTION
indications we have on the subject are a number of cases in which
the equation is given, and the solution written down, or stated to
be rational, without any work being shown. Thus
(IV. 22) x* = 4#  4, therefore x = 2 ;
(IV. 31) 325^ = 3^+18, therefore # = ^ or &;
(VI. 6) 84# 2 + 7* = 7, whence ^ = \ \
(vi. 7) 84# 2  7* = 7, hence .* = ;
(vi. 9) 630^  73^= 6, therefore ^ = ^ ;
and (VI. 8) 630;^ + 73.r=6, and x is rational.
These examples, though proving that Diophantus had somehow
arrived at the result, are not in themselves sufficient to show that
he was necessarily acquainted with a regular method for the
solution of quadratics ; these solutions might (though their variety
makes it somewhat unlikely) have been obtained by mere trial.
That, however, Diophantus' solutions of mixed quadratics were not
merely empirical is shown by instances in V. 30. In this problem
he shows that he could approximate to the root in cases where it is
not " rational" As this is an important point, I give the substance
of the passage in question: "This is not generally possible unless
we contrive to make x > (jr 2  60) and < \(x 2  60). Let then
&  60 be > 5#, but # 2  60 < 8;r. Since then x z  60 > 5#, let 60 be
added to both sides, so that x* > $x f 60, or x? = ^x + some number
>6o; therefore # must not be less than II. M In like manner
Diophantus concludes that "j* = 8;t;+ some number less than 60;
therefore x must be found to be not greater than 12."
Now, solving for the positive roots of these two equations, we
have
*>i(5 + v^65) and x<4 + */?6 y
or x> 106394... and x< 127177....
It is clear that x may be < n or > 12, and therefore Dio
phantus' limits are not strictly accurate. As however it was
doubtless his object to find integral limits, the limits II and 12
are those which are obviously adapted for his purpose, and are
a fortiori safe.
In the above equations the other roots obtained by prefixing
the negative sign to the radical are negative and therefore would
be of no use to Diophantus. In other cases of the kind occurring
DIOPHANTUS' METHODS OF SOLUTION 61
in Book V. the equations have both roots positive, and we have to
consider why Diophantus took no account of the smaller roots in
those cases.
We will take first the equations in V. 10 where the inequalities
to be satisfied are
170 + 17 ........................ (i).
19 ........................ (2).
Now, if a, $ be the roots of the equation
& px + q= o (p, q both positive),
and if a > /3, then
(a) in order that x* px f <? may be > o
we must have x > a or < /3,
and (b) in order that x* px f q may be < o
we must have x < a and > J3.
(i) The roots of the equation
17=0
36 + V 1007 ._, ^ . 6773... ,426...
are ^=   ; that is, ' /J and   ;
17 17 17
and, in order that ijx 2 72^1 17 may be < o, we must have
 i_ ,. 4'26...
x < ^ but >   .
17 17
(2) The roots of the equation
36 + ^935 ., . . 66577... j 5*422...
are ^ = ^^ ; that is, ^ and  ;
19 19 19
and, in order that ig^r 2  72^ I 19 may be > o, we must have
66*577... S'422...
/z or < ^.tr  .
19 19
Diophantus says that ;tr must not be greater than ff or less than
ff. These are again doubtless intended to be a fortiori limits;
but f should have been f, and the more correct way of stating the
case would be to say that, if x is not greater than ff and not less
than f, the given conditions are a fortiori satisfied.
Now consider what alternative (if any) could be obtained, on
Diophantus' principles, if we used the lesser positive roots of the
62 INTRODUCTION
equations. If, like Diophantus, we were to take a fortiori limits,
we should have to say
which is of course an impossibility. Therefore the smaller roots
are here useless from his point of view.
This is, however, not so in the case of another pair of in
equalities, used later in V. 30 for finding an auxiliary #, namely
# 2 f 6o>
x* + 60 <
The roots of the equation
x* 22x + 60 = o
are u V6i; that is, i8'8i... and 3'i8...;
and the roots of the equation x* 24^ + 60 = o
are 12 ^84; that is, 21*16... and 283....
In order therefore to satisfy the above inequalities we must have
x> i8'8i ... or < 318,..,
and x< 21*16 ... but > 2*83.
Diophantus, taking a fortiori integral limits furnished by the
greater roots, says that x must not be less than 19 but must be
less than 21. But he could also have obtained from the smaller
roots an integral value of x satisfying the necessary conditions,
namely the value x == 3 ; and this would have had the advantage
of giving a smaller value for the auxiliary x than that actually
taken, namely 2O 1 . Accordingly the question has been raised 2
whether we have not here, perhaps, a valid reason for believing
that Diophantus only knew of the existence of roots obtained by
using the positive sign with the radical, and was unaware of the
solution obtained by using the negative sign. But in truth we
can derive no certain knowledge on this point from Diophantus'
treatment of the particular equations in question. Thus, e.g^ if he
chose to use the first of the two equations
17,
19,
for the purpose of obtaining an upper limit only, and the second
1 This is remarked by Loria (Le scienze esatte deW antica Greeia, v. p. 128).
But in fact, whether we take 20 or 3 as the value of the auxiliary unknown, we get
the same value for the original DC of the problem. For the original x has to be found
from x*  60 = (x  nif where m is the auxiliary x ; and we obtain ,* iij whether we
put ^6o=(jr~2o) 2 or ^ 3 6o = (^3) 3 .
2 Loria, op cit. p. 129.
DIOPHANTUS' METHODS OF SOLUTION 63
for the purpose of obtaining a lower limit only, he could only use
the values obtained by using the positive sign. Similarly, if, with
the equations
x* + 60 > 22%,
he chose to use the first in order to find a lower limit only, and
the second in order to find an upper limit only> it was not open to
him to use the values corresponding to the negative sign 1 .
For my part, I find it difficult or impossible to believe that
Diophantus was unaware of the existence of two real roots in
such cases. The numerical solution of quadratic equations by the
Greeks immediately followed, if it did not precede, their geometrical
solution. We find the geometrical equivalent of the solution of
a quadratic assumed as early as the fifth century B.C., namely by
Hippocrates of Chios in his Quadrature of 'tunes*, the algebraic form
of the particular equation being & + Vf ax = a*. The complete
geometrical solution was given by Euclid in VI. 2729: and the
construction of VI. 28 corresponds in fact to the negative sign
before the radical in the case of the particular equation there
solved, while a quite obvious and slight variation of the con
struction would give the solution corresponding to the positive sign.
In VI. 29 the solution corresponds to the positive sign before the
radical; in the case of the equation there dealt with the other sign
would not give a "real" solution 8 . It is true that we do not find
the negative sign taken in Heron any more than in Diophantus,
though we find Heron 4 stating an approximate solution of the
equation
^(14*) = 6720/144,
without showing how he arrived at it; x is, he says, approximately
equal to 8J. It is clear however that Heron already possessed
a scientific method of solution. Again, the author of the socalled
Geometry of Heron 5 practically states the solution of the equation
. ^ /. iJ (1 54 X 212 + 841) 20
m the form x= v v ^  * '  ^ ,
ii
1 Enestrom in Bibliotheca Mathematica IX 8 , 19089, pp. 712.
2 Simplicius, Comment, in Aristot. Phys.^ ed. Diels, p. 64, t8; Rudio, Der Bericht
des Simplicius uber die Qitadraturen des Antiphon und des Hippokrates , 1907, p. 58, 8n.
3 The Thirteen Books of Euclid? s Elements, Cambridge, 1908, Vol. n. pp. 257267.
4 Heron, Metrica, ed. Schone, pp. 148^51. The text has 8 as the approximate solu
tion, but the correction is easy, as the inference immediately drawn is that 14^ = 5^.
5 Heron, ed. Hultsch, p. 133, 1023. See M. Cantor, Geschichte der Math. I 3 , p. 405.
64 INTRODUCTION
showing pretty clearly the rule followed after the equation had
been written in the form
^ = 212 x 154.
We cannot credit Diophantus with less than a similar uniform
method ; and, if he did not trouble to give two roots where both
were real, this seems quite explicable when it is remembered that he
did not write a textbook of algebra, and that his object through
out is to obtain a single solution of his problems, not to multiply
solutions or to show how many can be found in each case.
In solving such an equation as
ax* bx + c = o,
it is our modern practice to divide out by a in order to make the
first term a square. It does not appear that Diophantus divided
out by a\ rather he multiplied by a so as to bring the equation
into the form
a*%? abx + ac = o ;
then, solving, he found
and wrote the solution in the form
a
wherein his method corresponds to that of the PseudoHeron above
referred to.
From the rule given in Def. 1 1 for removing by means of addition
any negative terms on either side of an equation and taking equals
from equals (operations called by the Arabians aljabr and almuka
bala) it is clear that, as a preliminary to solution, Diophantus so
arranged his equation that all the terms were positive. Thus,
from his point of view, there are three cases of mixed quadratic
equations.
Case I. Form mx? +/#= q ; the root is
m
according to Diophantus. An instance is afforded by VI. 6. Dio
phantus namely arrives at the equation 6;ir 2 + 3^ = 7, which, if it is
to be of any service to his solution, should give a rational value
of x \ whereupon he says " the square of half the coefficient 1 of x
1 For " coefficient" Diophantus simply uses TA?70o$, multitude or number; thus
"number of dpi^oi " = coeff. of x. The absolute term is described as the "units."
DIOPHANTUS' METHODS OF SOLUTION 65
together with the product of the absolute term and the coefficient
of x* must be a square number ; but it is not," z>. \p* + mq, or in
this case () 2 f 42, must be a square in order that the root may be
rational, which in this case it is not.
Case 2. Form wix* px 4 q. Diophantus takes
m
An example is IV. 39, where 2;r 3 >6>f 18, Diophantus says:
"To solve this take the square of half the coefficient of x, i.e. 9, and
the product of the absolute term and the coefficient of^r 2 , i.e. 36.
Adding, we have 45, the square root 1 of which is not 2 < 7. Add
half the coefficient of x, [and divide by the coefficient of #*] ; whence
x is not < 5." Here the form of the root is given completely; and
the whole operation of rinding it is revealed, Cf. IV. 31, where
Diophantus remarks that the equation 5^ = 3^118 "is not rational.
But 5, the coefficient of x*, is a square plus I, and it is necessary
that this coefficient multiplied by the 18 units and then added to
the square of half the coefficient of x> namely 3, that is to say 2^,
shall make a square."
Case 3. Form mx* + q ~px. Diophantus' root is
m
Cf. in V. 10 the equation already mentioned, 17^ f 17 <
Diophantus says: " Multiply half the coefficient of x into itself and
we have 1296; subtract the product of the coefficient of x* and the
absolute term, or 289. The remainder is 1007, the square root of
which is not 8 > 31. Add half the coefficient of x, and the result is
not >67. Divide by the coefficient of ^, and x is not > 67/1 7."
Here again we have the complete solution given. Cf. VI. 22, where,
having arrived at the equation 172^=336^ + 24, Diophantus
remarks that "this is not always possible, unless half the coefficient
of x multiplied into itself, minus the product of the coefficient of x*
and the units, makes a square/'
For the purpose of comparison with Diophantus' solutions of
quadratic equations we may refer to a few of his solutions of
1 The "square root" is with Diophantus rXeiyxi, or "side."
3 7, though not accurate, is clearly the nearest integral limit which will serve the
purpose.
8 As before, the nearest integral limit.
H. D. S
66 INTRODUCTION
(3) Simultaneous equations involving quadratics.
Under this heading come the pairs of equations
I use Greek letters to distinguish the numbers which the
problem requires us to find from the one unknown which Dio
phantus uses and which I shall call x.
In the first two of the above problems, he chooses his x thus.
Let, he says,
Then it follows, by addition and subtraction, that
= a + x, r\ a x.
Consequently, in I, 27,
jfr = (* + *) (a jt) = a*x* = B,
and x is found from this "pure" quadratic equation.
If we eliminate f from the original equations, we have
rf  say + .5 = 0,
which we should solve by completing the square (a ^) 2 , whence
(aiiy = #B,
which is Diophantus' ultimate equation with a 77 for x.
Thus Diophantus' method corresponds here again to the ordi
nary method of solving a mixed quadratic, by which we make it
into a pure quadratic with a different x.
In I. 30 Diophantus puts f f 97 = 2^, and the solution proceeds
in the same way as in I. 27.
In I. 28 the resulting equation in x is
( a
(4) Cubic equation.
There is no ground for supposing that Diophantus was acquainted
with the algebraical solution of a cubic equation. It is true that there
is one cubic equation to be found in the Arithmetica, but it is only
a very particular case. In VI. 17 the problem leads to the equation
DIOPHANTUS' METHODS OF SOLUTION 67
and Diophantus says simply " whence x Is found to be 4." All
that can be said of this is that, if we write the equation in true
Diophantine fashion, so that all the terms are positive,
This equation being clearly equivalent to
Diophantus no doubt detected the presence of the common factor
on both sides of the equation. The result of dividing by it
is x = 4, which is Diophantus' solution. Of the other two roots
x= V( i) no account is taken, for the reason stated above.
It is not possible to judge from this example how far Dio
phantus was acquainted with the solution of equations of a degree
higher than the second.
I pass now to the second general division of equations.
(B) Indeterminate equations.
As I have already stated, Diophantus does not, in his
Arithmetica as we have it, treat of indeterminate equations of the
first degree. Those examples in Book I. which would lead to such
equations are, by the arbitrary assumption of a specific value for
one of the required numbers, converted into determinate equations.
Nor is it likely that indeterminate equations of the first degree
were treated of in the lost Books. For, as Nesselmann observes,
while with indeterminate quadratic equations the object is to obtain
a rational result, the whole point in solving indeterminate simple
equations is to obtain a result in integral numbers. But Diophantus
does not exclude fractional solutions, and he has therefore only to
see that his results are positive, which is of course easy. Inde
terminate equations of the first degree would therefore, from
Diophantus* point of view, have no particular significance. We
take therefore, as our first division, indeterminate equations of
the second degree.
(a) Indeterminate equations of the second degree.
The form in which these equations occur in Diophantus is
invariably this: one or two (but never more) functions of the
unknown quantity of the form A^ + Bx + C or simpler forms
are to be made rational square numbers by finding a suitable
value for x. Thus the most general case is that of solving one or
two equations of the form Ax* 4 Bx + C=y.
52
68 INTRODUCTION
(l) Single equation.
The single equation takes special forms when one or more of
the coefficients vanish or satisfy certain conditions. It will be well
to give in order the different forms as they can be identified in
Diophantus, premising that for u =j 2 " Diophantus simply uses the
formula taov rerpaywvq), " is equal to a square," or jroiel Terpdy&vov,
" makes a square."
I. Equations which can always be solved rationally. This is
the case when A or C or both vanish.
Form Bx =j/ 2 . Diophantus puts for y any arbitrary square
number, say m\ Then x = m*jB.
Ex. III. 5: 2^r=j/ 2 ,/ 2 is assumed to be 16, and x= 8,
Form Bx+ C=y*. Diophantus puts for y* any square m*, and
x(m z C)IB. He admits fractional values of x y only taking
care that they are " rational," z>. rational and positive.
Ex. III. 6: 6x+i =y=i2i, say, and;tr=2O.
Form Ax* + Bx = j/ 2 . Diophantus substitutes for y any multiple
AI/M W$
is x: whence Ax+B~ x, the factor x disappearing and
n n* rr
the root x o being neglected as usual. Thus x 2 _ // 8 
Exx. 11.21: 4x* + 3x =y z  ($xf, say, and x = f .
II. 33: i&e a +7^=j?/ 2 = (5^) 2 J say, and #$.
2. Equations which can only be rationally solved if certain
conditions are fulfilled.
The cases occurring in Diophantus are the following.
Form Ax? + C=y\ This can be rationally solved according to
Diophantus
(a) When A is positive and a square, say a z .
Thus tfV f C=jy 2 . In this case jp 3 is put = (ax inf ;
therefore a*x +C=(ax m)*,
, Cm*
and ^ ^^^
(m and the doubtful sign being always assumed so as to give x
a positive value).
(/9) When C is positive and a square number, say c\
Thus Ax* + c* y\ H ere Diophantus puts y = (mx c) ;
DIOPHANTUS' METHODS OF SOLUTION 69
therefore Ax? 4 ^ = (mx <r) 2 ,
, , 2mc
and *=Z^'
(7) When one solution is known, any number of other
solutions can be found. This is enunciated in the Lemma to VI. 15
thus, though only for the case in which Cis negative: "Given two
numbers, if, when one is multiplied by some square and the other
is subtracted from the product, the result is a square, then another
square also can be found, greater than the aforesaid square which
has the same property." It is curious that Diophantus does not
give a general enunciation of this proposition, inasmuch as not
only is it applicable to the cases Ax* C=jP 3 but also to the
general form Ax* 4 Bx + C= j/ 2 .
Diophantus' method of finding other greater values of ^satisfy
ing the equation Ax* C=y z when one such value is known is as
follows.
Suppose that X Q is the value already known and that q is the
corresponding value of y.
Put X*=XQ + % in the original expression, and equate it to
(^ k%)\ where k is some integer.
Since A (x, + ) 2  C = (q  k%)\
it follows (because by hypothesis Ax? C q^) that
, ,.2 (Ax Q 4 kg)
whence ff &~A '
, 2 (Ax* + kg)
and ^=^0+ & A *
fC ^~ JLJL
In the second Lemma to VI. 12 Diophantus does prove that the
equation Ax 2 + C=y* has an infinite number of solutions when
A + C is a square, z>. in the particular case where the value #= I
satisfies the equation. But he does not always bear this in mind;
for in in. 10 the equation 52^+ 12 j 2 is regarded as impossible of
solution although 52 + 12 =64, a square, and a rational solution is
therefore possible. Again in III. 12 the equation 266^ 10 =^ 2 is
regarded as impossible though x = i satisfies it
The method used by Diophantus in the second Lemma to
VI. 12 is like that of the Lemma to VI. 15.
Suppose that A + C = q*.
Put i 4 f for x in the original expression Ax* + C, and equate it
to (q k%f, where k is some integer.
70 INTRODUCTION
Thus A(i + ?+C=(q~kf,
and it follows that 2% (A +?) = ? (& A),
so that 2(A+fy)
so mat ~~ >
,
and ^ I +
It is of course necessary to choose / 3 such that k* > A.
' It is clear that, if x = o satisfies the equation, C is a square, and
therefore this case (7) includes the previous case (/3).
It is to be observed that in VI. 14 Diophantus says that a rational
solution of the equation
Ax*  <? = j 2
is impossible unless A is the sum of two squares.
[In fact, if .#=// satisfies the equation, and Ax* c* = k\
we have Af =
Lastly, we have to consider
Form
This equation can be reduced by means of a change of variable
to the preceding form wanting the second term. For, if we put
D
x = z  ^ , the transformation gives
Diophantus, however, treats this form of the equation quite
separately from the other and less fully. According to him the
rational solution is only possible in the following cases.
(a) When A is positive and a square, or the equation is
(Exx. IL 20, 22 etc.)
^
Diophantus then puts y = (<# mf, whence
m* C
 ~ .
2am + s
(ft) When C is positive and a square, or the equation is
Diophantus puts/ 2 = (c  mxf, whence
IV. 8, 9 etc.)
DIOPHANTUS' METHODS OF SOLUTION 71
(7) When \B*  A C is positive and a square number. Dio
phantus never expressly enunciates the possibility of this case;
but it occurs, as it were unawares, in IV. 31. In that problem
is to be made a square. To solve this Diophantus assumes
which leads to the quadratic 3^+ 18 5# 2 = o; but "the equation
is not rational." Accordingly the assumption 4^ will not do;
"and we must find a square [to replace 4] such that 18 times
(this square + r) + (f) 2 may be a square/' Diophantus then solves
the auxiliary equation i8(m*+ i) + f =7 2 , finding m=iS. He
then assumes yc + 1 8  x? = ( 1 8) 2 ^ 2 , which gives 3 2 $x* 3^  1 8 = o,
"and x=jh> that is &." 1
1 With this solution should be compared the much simpler solution of this case given
by Euler (Algebra, tr. Hewlett, 1840, Part II. Arts. 5053), depending on the separation
of the quadratic expression into factors. (Curiously enough Diophantus does not separate
quadratic expressions into their factors except in one case, vi. 19, where however his
purpose is quite different : he has made the sum of three sides of a rightangled triangle
4# 2 + 6#'+ 2, which has to be a cube, and, in order to simplify, he divides throughout by
x+ i, which leaves 4^+2 to be made a cube.)
Since J2? 2  AC is a square, the roots of the quadratic Ax* + Bx + C=o are real, and
the expression has two real linear factors. Take the particular case now in question,
where Diophantus actually arrives at 3*+ 18.^ as the result of multiplying 6 x and
0;, but makes no use of the factors.
We have $x+ i8x*=(6x) (3 +*).
Assume then (6x)
and we have / 2 (6  x) =^(3 +x),
where/, q may be any numbers subject to the condition that 2/ 2 >^ a . If ^=9, f= 16,
we have Diophantus' solution x= .
In general, if Ax* + Bx+C= (f+gx) (k + kx\
we can put
whence
and
This case, says Euler, leads to a fourth case in which Ax*+ Bx+ C=y* can be solved,
though neither A nor C is a square, and though * ^A C is not a square either. The
fourth case is that in which Ax^ + Bx + C is the sum of two parts, one of which is a square
and the other is the product of two factors linear in x. For suppose
where
72 INTRODUCTION
It is worth observing that from this example of Diophantus we
can deduce the reduction of this* general case to the form
Ax* + C=p
wanting the middle term.
Assume, with Diophantus, that Ax* + Bx + C = m*x z : therefore
by solution we have
and x is rational provided that \B*  A C+ Cm* is a square. This
condition can be fulfilled if \B* AC is a square, by the pre
We can then put Z?+ XY= (z+ X\,
whence K=a^+^AT,
q f*
that is, jc(ff+*p
Ex. i. Equation 2fc 2  1 =>/ a .
Put
Therefore x r = 2^+r
S T
and
As jt 3 is alone found in our equation, we can take either the positive or negative sign
and we may put
Ex. 2 . Equation
Here we put 2^ + 2 = 412(^+1) (x i).
Equating this to j 2 + ~ (x + i )i ,
we have 2(jf i) = 4~+^
or . (  2^) =
and
It is to be observed that this method enables us to solve the equation
A*?*=t?
whenever it can be solved rationally, i.e. whenever A is the sum of two squares (</ a + fl 2 ,
say). For then
^ <=**** + (*) (ex+c).
In cases not covered by any of the above rules our only plan is to try to discover one
solution empirically. If one solution is thus found, we can find any number of others ; if
we cannot discover such a solution by trial (even after reducing the equation to the
simplest form A V s 4 C=y a ), recourse must be had to the method of continued fractions
elaborated byLagrange (cf. Otuvres, n. pp. 377535 and pp. 655726; additions to
Killer's Algtbrd).
DIOPHANTUS' METHODS OF SOLUTION 73
ceding case. If \B*~AC is not a square, we have to solve
(putting, for brevity, D for J5 2 AC} the equation
D + Cm 2 =y\
and the reduction is effected.
(2) Doubleeqttation.
By the name "doubleequation" Diophantus denotes the pro
blem of finding one value of the unknown quantity which will make
two different functions of it simultaneously rational square numbers.
The Greek term for the "doubleequation" occurs variously as SwXot
(TOT779, StTrX^ 6<70T77$ or StTrX?} 6oa>er9. We have then to solve the
equations
in rational numbers. The necessary preliminary condition is that
each of the two expressions can severally be made squares. This
is always possible when the first term (in x*) is wanting. This is
the simplest case, and we shall accordingly take it first.
I. Doubleequation of the first degree.
Diophantus has one general method of solving the equations
ax + a =
taking slightly different forms according to the nature of the
coefficients.
(a) First method of solution of
This method depends upon the identity
If the difference between the two expressions in x can be separated
into two factors p, y, the expressions themselves are equated
to [%(pq)}* respectively. Diophantus himself (II. n) states his
rule thus.
"Observing the difference [between the .two expressions], seek
two numbers such that their product is equal to this difference;
then equate either the square of half the difference of the two
factors to the lesser of the expressions or the square of half the
sum to the greater."
74 INTRODUCTION
We will take the general case and investigate to what particular
classes of cases the method is applicable, from Diophantus' point
of view, remembering that his cases are such that the final quadratic
equation in x always reduces to a simple equation.
Take the equations
tax + a =
Subtracting, we have
(a  )*+ (a  b) = u z  w 2 .
We have then to separate (a /3)x + (a b) into two factors;
let these be/, {(a )# + (<&
We write accordingly
TU 2
Thus ;/ 2
4
therefore {(a  /3) x +  <5 I/ 2 } 2 = 4/ 2 (a^r + a),
or (a  yS) 2 ^ 2 + 24r {(a  j$)( a b +p*) 2p* a} + ( 
that is, (a  j8) 2 ^ 3 + 2^ {(a ft) (a b) /> (a + )}
+ (  *) 2  2/ ( + b) +/ = o.
Now, in order that this equation may reduce to a simple
equation, either
(i) The coefficient of x* must vanish, so that
a = &
or (2) The absolute term must vanish, that is,
or {/ 2  (a f )} 2
so that ab must be a square number.
Therefore either a and i are both squares, in which case we
may substitute ^ and d* for them respectively, / being then equal
to c d, or the ratio of a to b is the ratio of a square to a square.
With respect to (i) we observe that on one condition it is not
necessary that a ft should vanish, i.#. provided we can, before
solving the equations, make the coefficients of x the same in both
expressions by multiplying either equation or both equations by
some square number, an operation which does not affect the
problem, since a square multiplied by a square is still a square.
DIOPHANTUS' METHODS OF SOLUTION 75
In other words, it is only necessary that the ratio of a to ft should
be the ratio of a square to a square 3 .
Thus, if cL/{3 = m^/n* or an* = /3m*, the equations can be solved
by multiplying them respectively by n* and m*\ we can in fact
solve the equations
like the equations
cur 4 a z/ 2
in an infinite number of ways.
Again, the equations under (2)
$x + & = <
can be solved in two different ways according as we write them in
this form or in the form
='M
= 2/ 3 j '
obtained by multiplying them respectively by d*, *, in order that
the absolute terms may be equal.
I shall now give those of the possible cases which we find solved
in Diophantus' own work. These are equations
(i) of the form
1 Diophantus actually states this condition in the solution of iv. 32 where, on arriving
at the equations
he says : " And this is not rational because the coefficients of x have not to one another
the ratio which a square number has to a square number."
Similarly in the second solution of in. 15 he states the same condition along with an
alternative condition, namely that a has to b the ratio of a square to a square, which is
the second condition arrived at under (2) above. On obtaining the equations
Diophantus observes "But, since the coefficients in one expression are respectively greater
than those in the other, neither have they (in either case) the ratio which a square number
has to a square number, the hypothesis which we took is useless."
Cf. also iv. 39 where he says that the equations
2 )
w 2 J
are possible of solution because there is a square number of units in each expression.
76 INTRODUCTION
a case which includes the more common one where the coefficients
of x in both are equal \
(2) of the form
solved in two different ways according as they are written in this
form or in the alternative form
General solution of Form (i) or
am*x + a =
an*x + b =
Multiply by n 2 , w? respectively, and we have to solve the
equations
The difference is an* bm*\ suppose this separated into two
factors /, q>
Let u'w'=p y
u f 4 w =*q\
therefore ^  J (p 4 qf, w f * = i (/  qf>
and m^n^ H an*
or <x.m*n*x 4 ^?^ 2
Either equation gives the same value of x y and
since pq an i///& .
Any factors p, q may be chosen provided that the resulting
value of x is positive.
Ex. from Diophantus :
65 6x = z/ 2 ) , v
* L (iv. 32.)
therefore 260  24^ = u'* }
65
The difference = 195 = 15. 13, say;
therefore (15 13)* = 65 24^; that is, 24^=64, and
DIOPHANTUS' METHODS OF SOLUTION 77
General solution (first method) of Form (2), or
ax 4 c* = t
In order to solve by this method, we multiply by d* 3
respectively and write
u being supposed to be the greater.
The difference = (ad*  ft)x. Let the factors of this be px, q.
Therefore u* = i (px + q)\
Thus x is found from the equation
a
This equation gives
p*x* 4 2
or, since /^r = (ad*
In order that this may reduce to a simple equation, as Dio
phantus requires, the absolute term must vanish,
or
and
Thus our method in this case furnishes us with only one solution
of the doubleequation, q being restricted to the value 2cd, and the
solution is
2 (ad*
Ex. from Diophantus. This method is only used in one par
ticular case (IV. 39), where c* = d* as the equations originally stand,
the equations being
The difference is 2x> and q is necessarily taken to be 2//4> or 45
the factors are therefore \x, 4.
Therefore 8^ + 4 = i (i^ + 4) 2 ,
and #=112.
General solution (second method) of Form (2) or
cur 4 ^ 2 = 2
78 INTRODUCTION
The difference = (a )*+(<* d z ).
Let the factors of this be /, {(a  /3) x + c*  <
Then, as before proved (p. 74), / must be equal to (c
Therefore the factors are
n /?
^j^
and we have finally
^if
4 \c
Therefore (  ) x 2 f ^x \~ ~7  a' r = o,
\c a) ( c a }
which equation gives two possible values for x. Thus in this case
we can find by our method two values of x, since one of the factors
p may be either (c + d) or (c d).
Ex. from Diophantus. To solve the equations
The difference is here $x+ 5, and Diophantus chooses as the
factors 5, x+ I. This case therefore corresponds to the value
c + d of p. The solution is given by
(i^ + 3) 2 = 10^ + 9, whence # = 28.
The other value, J t of/ is in this case excluded, because it
would lead to a negative value of x.
The possibility of deriving any number of solutions of a double
equation when one solution is known does not seem to have been
noticed by Diophantus, though he uses the principle in certain
cases of the single equation (see above, pp. 69, 70). Fermat was the
first, apparently, to discover that this might always be done, if one
value a of x were known, by substituting x f a for x in the equa
tions. By this means it is possible to find a positive solution, even
if a is negative, by successive applications of the principle.
But, nevertheless, Diophantus had certain peculiar artifices by
which he could arrive at a second value. One of these artifices
(which is made necessary in one case by the unsuitableness of the
value of x found by the ordinary method) gives a different way of
solving a doubleequation from that which has been explained, and
is used only in one special case (IV. 39).
DIOPHANTUS' METHODS OF SOLUTION 79
() Second method of solving a doubleequation of the first
degree.
Consider only the special case
Jix + n* = u\
Take these expressions, and 2 , and write them in order of
magnitude, denoting them for convenience by A, B, C.
_ u f
Therefore
AB f
__ = z
Suppose now that lix + n* = (y + iif ;
therefore hx = y* + 2wy,
and AB = j (y* f 2ny\
or
thus it is only necessary to make this expression a square,
Assume therefore that
+./ f 2n + i^ + # = (&  rif ;
and any number of values for y y and therefore for x, can be found,
by varying/.
Ex. from Diophantus (the only one), IV. 39.
In this case there is the additional condition of a limit to the
value of x. The doubleequation
has to be solved in such a manner that x < 2.
Here z_>= i> an ^ B is taken 1 to be (y + 2) 2 .
Therefore ^  B =
therefore ^4 = (y* + 4jj>) +y + 4^ + 4
which must be made a square.
1 Of course Diophantus uses the same variable x where I have for clearness used y.
Then, to express what I have called m later, he says: "I form a square from 3 minus
some number of #'s, and x becomes some number multiplied by 6 and then added to 12,
divided by the difference by which the square of the number exceeds 3."
8o INTRODUCTION
If we multiply by f , we must make
3 / + 1 2y  9 = a square,
where y must be < 2,
Diophantus assumes
, 6m + 12
whence y =   ,
m z 3
and the value of #2 is then taken such as to make j> < 2.
It is in a note on this problem that Bachet shows that the
doubleequation
ax f a = u* \
can be rationally solved by a similar method provided that the
coefficients satisfy either of two conditions, although none of the
coefficients are squares and neither of the ratios a : and a : b is
equal to the ratio of a square to a square. Bachet's conditions are :
(1) That, when the difference between the two expressions
is multiplied or divided by a suitablychosen number, and the
expression thus obtained is subtracted from the smaller of the
original expressions, the result is a square number, or
(2) That, when the difference between the two expressions
is multiplied or divided by a suitablychosen number, and the
smaller of the two original expressions is subtracted from the
expression obtained by the said multiplication or division, the
result is a number bearing to the multiplier or divisor the ratio
of a square to a square 1 .
1 Bachet of course does not solve equations in general expressions (his notation does
not admit of this), but illustrates his conditions by equations in which the coefficients are
specific numbers, I will give one of his illustrations of each condition, and then set the
conditions out more generally.
Case (i) . Equations 3* + 1 3 = t
x+ 7=7
difference ijc+ 6
The suitablychosen number (to divide by in this case) is i \
\ (difference) x + 3,
and (lesser expression)  J (difference) =#+ 7  (#+3) = 4, that is, a square.
We have then to find two squares such that
their difference = 2 (difference between lesser and 4).
Assume that the lesser (y+ 2) 2 , 2 being the square root of 4.
Therefore (greater square) = 3 (lesser)  8
DIOPHANTUS' METHODS OF SOLUTION 81
2. Doubleequation of the second degree,
or the general form
These equations are much less thoroughly treated in Diophan
tus than those of the first degree. Only such special instances
To make 3^+12^ + 4 a square we put
where p must lie between certain limits which have next to be found. The equation gives
__ 1 2 + 4/
y f^
In order that y may be positive, / 2 must be > 3 ; and in order that the second of the
original expressions, assumed equal to (^ + a) 2 , may be greater than 7 (it is in fact # + 7),
we must have 0' + 2)>2J (an a fortiori limit, since 2f > N /7),
Therefore tf + 1 2 >  (J?  3) ,
or i6/H57>3^.
Suppose that 3^ = 1 6p + 5 3$, which gives / = 7! .
Therefore p < 7f , but f > 3.
Put/ =3 in the above equation ; therefore
3^+ 12^+4= (2 3^) 3 ,
and y~*
Therefore x= (y + 2 )  7 = 29.
Gz.s\? (2). Equations
= 2^!
^j ;
difference
The suitablenumber (again to divide by) in this case is a ;
\ (difference) = 2^+11,
and i (difference) (lesser expression) =8 = 2.4,
where 2 is the divisor used and 4 is the ratio of a square to a square.
Hence two squares have to be found such that
(their difference) = 2 (sum of lesser and 8).
If the lesser is^ 5 , the greater is 3^+ 16=(4 py)\ say.
Bachet gives, as limit
and puts ^=3. This gives ^=4, so that x=6^.
Let us now state Sachet's conditions generally.
Suppose the equations to be
The difference is (a  p) x+ (a  b).
This has to be multiplied by 5 which is the "suitable" factor in this case, and, if
ap
we subtract the product from .#+, we obtain
i P i zx al>~ap
:(*, or  f .
OL p a, p
H. D, 6
82 INTRODUCTION
occur as can be easily solved by the methods which we have
described for equations of the first degree.
The following types are found.
(l) p^ 3 + cur + # ==
/> 2 # 2 + $x + b SB
The difference is (at /3)# + (a b\ and, following the usual
course, we may, e.g. y resolve this into the factors
(1) The first of Bachet's conditions is that
^j = a square =//? 2 , say.
(2) The second condition is that
afSab = J? __
a/3 ~ 2 *aj8'
aBab . .
or 3 = a ratio of a square to a square.
P
It is to be observed that the first of these conditions can be obtained by considering
the equation
obtained on page 74 above.
Diophantus only considers the cases in which this equation reduces to a simple
equation; but the solution of it as a mixed quadratic gives a rational value of x provided
that
{(a ]8) (ab) /(a + j8)}3 (a ){(* <$) 2  tf(a + t) +/*} is a square,
that is, if
p*{(a + p)*(ap)*\+2^{(a + l>)(a,p)*(a?p 2 )(a~l>)} is a square,
which reduces to ajS/ 2 + (a  ) (ab  aft) = a square .................... v ... (A).
This can be solved (cf. p. 68 above), if
 ~ is a square. (Bachet's first condition.)
ap
Again take Bachet's second condition
aB  ad r*
Q = a square = ^ say,
and substitute ^/j 2 for apat> in the equation (A) above.
T 2
Therefore ajS/ 2  (a  13) ^ = a square,
or aj8/ 2 (a  ]8) jS = a square.
This is satisfied by /'=r; therefore (p. 69) any number of other solutions can
be found.
The second condition can also be obtained directly by eliminating x from the equations
ax + a =
for the result is ri + a = 2 ,
P P
which can be rationally solved if
aB  a/>
a square.
a = 2 1
=; 2 ] J
P
aB  a/>
 =a
DIOPHANTUS' METHODS OF SOLUTION 83
<
as usual, we put
i /a 8
 tL
o
or

4 V0
In order that x may be rational a condition is necessary ; thus
x is rational if
This is the case in the only instance of the type where a is not
equal to b, namely (in. 13)
the difference is 16^+4, and the resolution of this into the factors
4, 4* + i solves the problem.
In the other cases of the type a = b\ the difference is then
(a f$)x, which is resolved into the factors
and we put pW + out: + a =  ^?^ 4 2/urY ,
or
, a + )8 /a  y
whence  ~ JP = (   a.
2 V 4p
Exx. from Diophantus :
x* + x  i =
and
^r 2 h I = tt^J * v
(2) The second type found in Diophantus 1 is
x* 4 cur + # = u*
where one equation has no term in ^, and p = i, a = b.
1 It is perhaps worth noting that the method of the "doubleequation " has a distinct
advantage in this type of case. The alternative is to solve by the method of Euler (who
does not use the " double equation "), i.e. to put the linear expression equal to/ 2 and then,
substituting the value of x (in terms of/) in the quadratic expression, to solve the
62
84 INTRODUCTION
The difference & + (a #) x is resolved into the factors
and we put fix + a = i (a  /9) 2 ,
which gives #.
resulting equation in p. But the difficulties would generally be great. Take the case of
vi. 6 where
> have to be made squares.
i4*+ij H
If 14*+ i =/>, *=(/ i)/i 4 ;
I P* _ j )2
therefore jc 2 + i = rp + i has to be made a square,
or p*  2/ 2 4 1 97 = a square.
This does not admit of solution unless we could somehow discover empirically one
value of p which would satisfy the requirement, and this would be very difficult.
Let us take an easier case for solution by this method,
= s l
=w 2  '
which is solved by Euler (Algebra^ Part II. Art. 222), and let us compare the working of
the two methods in this case.
I. JSzt/er's method. Assuming x+ 1 =j0 2 and substituting^ 2  1 for x in the quadratic
expression, we have
p* . 2/ 2 + 2 = a square.
This can only be solved generally if we can discover one possible value of p by trial ;
this however is not difficult in the particular case, for/= i is an obvious solution.
To find others we put i H^ instead of / in the expression to be made a square ; this
gives
i+ 4 a + iff + q* = a square.
This can be solved in several ways.
1. Suppose i+4? 2 l4 3 +? 4 =(i+? 2 ) 2 ;
thus 4^ a + 4^ = 2^ 3 , whence ^=, /= and x=.
2. Suppose i+4^ + 4 $r*i/=(i? 2 ) 2 ;
thus 4^ 2 +4 3 = 2^ s , and =, p~~ and x~  .
3. Suppose
and we find, in either case, that 0=  2, so that/=  i, ^=o.
4 . Suppose i+^+^l^fi + a? 2 ) 2 ;
and we have 4? 3 H/*=4 4 , whence ?~, ^= and #=( ~J i=^,
j i) \5/ y
This value of x satisfies the conditions, for
The above five suppositions therefore give only two serviceable solutions
To find another solution, we take one of the values of q already found, say ^= , and
DIOPHANTUS 5 METHODS OF SOLUTION 85
Exx. from Diophantus :
3C ~~* 1 2 U I
(V. I.)
3? H I = U* \
14*+ 1 = w*y
(vi. 6.)
x*~6i44*+ 1048576 =
^464 =
}
(VI. 22.)
substitute r   for ^. This gives ^=r+f=r+, and we substitute this value for / in the
expression / 4  2/ 2 f 2.
We have then ^y^ 2 f2r 3 + ^ 4 tobe made a square, or
25  24? 8H+ 32^*4 lor^a square.
i. We take 5 +^4^ for the root, so that the absolute term and the term in r*
may disappear. We can make the term in r disappear also by putting io/=  24 or
/= W T e then have
(a) The upper sign gives
8 + 32^=40 +/ 2 +8/r,
and r=(/ s + 4 8)/( 3 28/)=~;
thus P=, and X=A*I=^.
20 r 400
(^) The lower sign gives
and r=
t ^i i ^61 . r
thus /=  , and JT=^ as before.
^ 20 400
We have therefore x + 1 = [ } , and x* + 1 =
2. Another solution is found by assuming the root to be 5 +fr+gr* and determining
/and g so that the absolute term and the terms in r, r 2 may vanish ; the result is
whence /=  , ^ =  , *+ ' =
and ^ +I
II. Method of " doubleequation. "
The difference =^JT.
(i) If we take as factors #, x i and, as usual, equate the square of half their
difference, or  , to x+ 1, we have
86 INTRODUCTION
The absolute terms in the last case are made equal by multiply
ing the second equation by (i28) 2 or 16384.
(3) One separate case must be mentioned which cannot be
solved, from Diophantus' standpoint, by the foregoing method,
but which sometimes occurs and is solved by a special artifice.
The form of doubleequation is
cur 2 h <# = 2 ) (l),
$x* + bx=w*} (2).
Diophantus assumes u* = wtx 1 ,
whence, by ( I ), x = a/(m* a),
(2) If we take  #, 2jr2, as factors, half the sum of which is JT i, so that the
absolute terms may disappear in the resulting equation, we have
and
.,
16 2
# .
9
(3) To find another value by means of the first, namely x=  ^, we substitute y
for x in the original expressions. We then have to solve
Multiply the latter by so as to make the absolute terms the same, and we must have
Subtract from the first expression, and the difference is y^ y^y I y~ J ; then,
equating the square of half the difference of the factors to the smaller expression,
we have
4 10
so that 961=4003/4 1 oo.
Therefore
= 861 and JT= 2=^ x i = &\* ,**+ r = f 8  9 Y
40O S ~" 4^400' ~\2O/ ' \4O/
(4) If we start from the known value ~ and put y+ for x in the equations, we
obtain Euler's fourth value of x, namely > ?
3 2965284
Thus all the four values obtained by Euler are more easily obtained by the method of
the "doubleequation."
DIOPHANTUS' METHODS OF SOLUTION 87
and, by substitution in (2), we derive that
Q ( a Y ba ^,
P ~~^  H ;  must be a square.
\m* a/ ra 2 a i
 a)
(OT8 _ a) a  a square.
We have therefore to solve the equation
abm* + a(a$ OL&) =/>,
and this form can or cannot be solved by the methods already
given according to the nature of the coefficients 1 . Thus it can
be solved if (a/3 a&)/a is a square or if a/6 is a square.
Exx. from Diophantus :
63T* ^X =3
(b) Indeterminate equations of a degree higher than the second.
(i) Single equations.
These are properly divided by Nesselmann into two classes ;
the first comprises those problems in which it is required to make
a function of x, of a degree higher than the second, a square ; the
second comprises those in which a rational value of x has to be
found which will make any function of x, not a square, but a higher
power of some number. The first class of problems requires the
solution in rational numbers of
Ax n + Bx n ~ l + ...+Kx+L=y\
the second the solution of
for Diophantus does not go beyond making a certain function of
x a cube. In no instance, however, of the first class does the index
n exceed 6, while in the second class n does not (except in a
special case or two) exceed 3.
1 Diophantus apparently did not observe that the above form of doubleequation can
be reduced to one of the first degree by dividing by or* and substituting y for i/x, when it
becomes
Adapting Bachet's second condition, we see that the equations can be rationally solved
if (/3  ad)ja is a square, which is of course the same as one of the conditions under which
the above equation abnP+a (apati) can be solved.
88 INTRODUCTION
First Class. Equation
The forms found in Diophantus are as follows :
i. Equation Ax* + Bx* + Cx + d 2 =y\
Here, as the absolute term is a square, we might put for y
the expression mx f ^, and determine m so that the coefficient
of x in the resulting equation vanishes. In that case
2md C, and m = C\2d\
and we obtain, in Diophantus' manner, a simple equation for x,
giving
Or we might put for y the expression m*x* + nx+d, and deter
mine m, n so that the coefficients of x, x* in the resulting equation
both vanish, in which case we should again have a simple equa
tion for x. Diophantus, in the only example of this form of
equation which occurs (VL 18), makes the first supposition. The
equation is
and Diophantus assumes y = \x + i, whence x ^ m
2. Equation Ax* f Bx* + Cx 2 + Dx + =y\
In order that this equation may be solved by Diophantus'
method, either A or E must be a square. If A is a square and
n
equal to a*, we may assume y = ax* \  x + n, determining n so
that the term in x* vanishes. If is a square (= e*) } we may write
jj' = #z# 2 + x + e, determining m so that the term in x* in the
resulting equation may vanish. We shall then, in either case,
obtain a simple equation in x.
The examples of this form in Diophantus are of the kind
 Bx* + Cx*
where we can assume y=ax* + kxe, determining k so that in
the resulting equation the coefficient of x* or of x may vanish ;
when we again have a simple equation.
Ex. from Diophantus (IV. 28) :
Diophantus assumes y = 3^ 2  6x+ I, and the equation reduces to
3 2# 3 36^2 = o, whence x = $.
DIOPHANTUS' METHODS OF SOLUTION 89
Diophantus is guided in his choice of signs in the expression
ax* + kx + e by the necessity for obtaining a " rational" result.
Far more difficult to solve are those equations in which, the
lefthand expression being biquadratic, the odd powers of x are
wanting, i.e. the equations Ax* + Cx* + E =f and Ax* + E=y\
even when A or E is a square, or both are so. These cases
Diophantus treats more imperfectly.
3. Equation Ax* + Cx 2 f E =y 5 ,
Only very special cases of this form occur. The type is
a?x*  c*x* + & =/,
which is written
Here y is assumed to be ax or ejx> and in either case we have
a rational value for x.
Exx. from Diophantus :
4*
This is assumed to be equal to
where y* is assumed to be equal to 25/4^*.
4. Equation Ax 4 + =y.
The case occurring in Diophantus is x 4 + 97 = j/ 3 (V, 29). Dio
phantus tries one assumption, y = # 2  10, and finds that this gives
4^ ^, which leads to no rational result. Instead, however, of
investigating in what cases this equation can be solved, he simply
drops the equation x* + 97 =JK S and seeks, by altering his original
assumptions, to obtain, in place of it, another equation of the same
type which can be solved in rational numbers. In this case, by
altering his assumptions, he is able to replace the refractory equa
tion by a new one, ^ 4 h337=y 2 , and at the same time to find a
suitable substitution for y, namely y = x* 25, which brings out
a rational result, namely x^. This is a good example of his
characteristic artifice of " Backreckoning 1 /' as Nesselmann calls it.
5. Equation of sixth degree in the special form
x*  Ax*
" Methode der Zuriickrechnung und Nebenaufgabe."
90 INTRODUCTION
It is only necessary to putj> =x s + c, and we have
* 8=
which gives a rational solution if Bj(A f 2c) is a square.
6. If, however, this last condition does not hold, as in the case
occurring IV. 18, ^ 16^ + ^ 464= j/ 2 , Diophantus employs his
usual artifice of "backreckoning," which enables him to replace
the equation by another, ^  128^ + ^ + 4096 =y\ where the
condition is satisfied, and, by assuming 7=^ + 64, x is found to
be^.
Second Class. Equation of the form
Except for such simple cases as Ax*=y*, Ax*> =y, where it is only
necessary to assume y = mx, the only cases occurring in Diophantus
are of the forms
I. Equation A& + Bx + C=y*.
There are only two examples of this form. First, in VI. i the
expression x* 4^ h 4 is to be made a cube, being already a square.
Diophantus naturally assumes x 2 = a cube number, say 8, and
x~ 10.
Secondly, in VI. 17 a peculiar case occurs. A cube is to be
found which exceeds a square by 2. Diophantus assumes (x i) 3
for the cube and (x+if for the square, and thus obtains the
equation
&  3^ 2 + 3^  i = x* + 2* + 3 ,
or ^ 8 + ^ = 4^r 2 + 4,
previously mentioned (pp. 667), which is satisfied by ^ = 4.
The question arises whether it was accidentally or not that this
cubic took so simple a form. Were ori, jtrf i assumed with
knowledge and intention? Since 27 and 25 are, as Fermat
observes 1 , the only integral numbers which satisfy the conditions,
it would seem that Diophantus so chose his assumptions as to lead
back to a known result, while apparently making them arbitrarily
with no particular reference to the end desired. Had this not
1 Note on vi. 17, Oeuvres, i. pp. 3334, n. p. 434. The fact was proved by Euler
(Algebra, Part n. Arts. 188, 193), See note on VI. 17 infra for the proof.
DIOPHANTUS' METHODS OF SOLUTION 91
been so, we should probably have found him, here as elsewhere
in the work, first leading us on a false tack and then showing us
how we can correct our assumptions. The fact that he here
makes the right assumptions to begin with makes us suspect that
the solution is not based on a general principle but is empirical
merely.
2. Equation Ax* f Ex*
If A or D is a cube number, this equation is easy of solution.
"D
For, first, if A = a?, we have only to write y = ax H  a , and we
obtain a simple equation in x.
c
Secondly, if D = d*, we put y = ^ x + d.
o^
If the equation is a*x*+Bx* + Cx+d*y*, we can use either
assumption, or we may puty = ax+d, obtaining a simple equation
as before.
Apparently Diophantus used the last assumption only; for
in IV. 27 he rejects as impossible the equation
8^3^2+8^ i =j/ 3 ,
because y = 2x i gives a negative value x ^ whereas either
of the other assumptions gives a rational value 1 .
(2) Double equations.
There are a few examples in which, of two functions of x> one
is to be made a square, and the other a cube, by one and the same
rational value of x. The cases are for the most part very simple ;
e.g. in VI. 19 we have to solve
4x + 2 =
2X+ !
thus f = 2# 2 , and z = 2.
A rather more complicated case is VI, 21, where we have the
doubleequation
Diophantus assumes y = mx, whence x = 2/(m* 2), and we have
2 \ 3 / 2 \ 2 2 .
2m 4 
1 There is a special case in which C and D vanish, Ax?+Bx*=y*. Here y is put
equal to mx, and x=l(m*A). Cf. iv. 6, 28 (2).
92 INTRODUCTION
To make 2m* a cube, we need only make 2m a cube or put
m = 4. This gives x = f .
The general case
&tr 2 + ex =
would, of course, be much more difficult; for, putting y^mx, we
have
x=c\(!#b\
and we have to solve
or
of which equation the above corresponding equation is a very
particular case.
Summary of the preceding investigation.
1. Diophantus solves completely equations of the first degree,
but takes pains to secure beforehand that the solution shall be
positive. He shows remarkable address in reducing a number of
simultaneous equations of the first degree to a single equation in
one variable.
2. For determinate equations of the second degree he has
a general method or rule of solution. He takes, however, in the
Arithmetica^ no account of more than one root, even where both
roots are positive rational numbers. But, his object being simply
to obtain some solution in rational numbers, we need not be
surprised at his ignoring one of two roots, even though he knew
of its existence.
3. No equations of a degree higher than the second are solved
in the book except a particular case of a cubic.
4. Indeterminate equations of the first degree are not treated
of in the work. Of indeterminate equations of the sfccond degree,
as Ax* + Bx+ Cy*, only those cases are fully dealt with in which
A or C vanishes, while the methods employed only enable us to
solve equations of the more general forms
Ax*+ C=f and Ax* + Bx+C=f
when A, or C, or \B* AC is positive and a square number, or (in
the case of Ax* C**y*) when one solution is already known.
DIOPHANTUS' METHODS OF SOLUTION 93
5. For doubleequations of the second degree Diophantus has
a definite method when the coefficient of x* in both expressions
vanishes ; the applicability of this method is, however, subject to
conditions, and it has to be supplemented in one or two cases by
another artifice. Of more complicated cases we find only a few
examples under conditions favourable for solution by his method.
6. Diophantus' treatment of indeterminate equations of degrees
higher than the second depends upon the particular conditions of
the problems, and his methods lack generality.
7. More wonderful than his actual treatment of equations are
the clever artifices by which he contrives to avoid such equations
as he cannot theoretically solve, eg. by his device of "back
reckoning," instances of which would have been out of place in
this chapter and can only be studied in the problems themselves.
I shall not attempt to class as "methods" certain headings
in Nesselmann's classification of the problems, such as (a) " Solution
by mere reflection," (b) " Solution in general expressions," of which
there are few instances definitely so described by Diophantus, or
(c) " Arbitrary determinations and assumptions." The most that
can be done by way of describing these " methods " is to quote
a few characteristic instances. This is what Nesselmann has
done, and he regrets at the end of his chapter on " Methods of
Solution " that it must of necessity be so incomplete. To under
stand and appreciate the various artifices of Diophantus it is in
fact necessary to read the problems themselves in their entirety.
With regard to the " Use of the rightangled triangle," all that
can be said of a general character is that only " rational " right
angled triangles (those namely in which the three sides can all be
represented by rational numbers) are used in Diophantus, and
accordingly the introduction of the " rightangled triangle " is
merely a convenient way of indicating the problem of finding
two square numbers, the sum of which is also a square number,
The general form used by Diophantus (except in one case, VI. 19,
q.v.) for the sides of a rightangled triangle is (
2ab 9 which expressions clearly satisfy the condition
The expression of the sides of a rightangled triangle in this form
Diophantus calls "forming a rightangled triangle from the
numbers a and b!' His rightangled triangles are of course
formed from particular numbers. "Forming a rightangled
94 INTRODUCTION
triangle from 7, 2 " means taking a rightangled triangle with sides
(7 2 + 2 2 ), (7 2 ~ 2 2 )> 2 . 7 . 2, or 53, 45, 28.
II. METHOD OF LIMITS.
As Diophantus often has to find a series of numbers in
ascending or descending order of magnitude, and as he does not
admit negative solutions, it is often necessary for him to reject
a solution which he has found by a straightforward method
because it does not satisfy the necessary condition ; he is then
very frequently obliged to find solutions which lie ivit/iin certain
limits in place of those rejected.
1. A very simple case is the following : Required to find
a value of x such that some power of it, x n t shall lie between two
given numbers. Let the given numbers be a, 6. Then Diophantus'
method is to multiply both a and b by 2 n , 3*, and so on, successively,
until some #th power is seen which lies between the two products.
Thus suppose that c n lies between ap n and bf* l \ then we can put
x = cjp, in which case the condition is satisfied, for (cjffi lies
between a and b.
Exx. In IV. 31 (2) Diophantus has to find a square between
i\ and 2. He multiplies both by a square, 64; this gives 80 and
128, and 100 is clearly a square which lies between them; there
fore (*gf or $ satisfies the prescribed condition.
Here, of course, Diophantus might have multiplied by any
other square, as 16. In that case the limits would have become
20 and 32 ; between these lies the square 25, which gives the same
square ff as that before found.
In VI. 21 a sixth power ("cubecube") is sought which shall
lie between 8 and 16. The sixth powers of the first four natural
numbers are I, 64, 729, 4096. Multiply 8 and 16 by 2* or 64, and
we have as limits 512 and 1024, between which 729 lies. There
fore fffi is a sixth power satisfying the given condition. To
multiply by 729 in this case would not give us a solution.
2. Sometimes a value of x has to be found which will give
some function of x a value intermediate between the values of two
other functions of x.
Ex. i. In IV. 25 it is necessary to find a value of x such that
8/(;r 2 + x) shall lie between x and x f i.
The first condition gives 8
DIOPHANTUS' METHODS OF SOLUTION 95
Diophantus accordingly assumes that
which is greater than ;t: 3 j;r 2 .
Thus # = f satisfies one condition. It is also seen to satisfy
Q
the second condition, or  <x+ I. Diophantus, however, says
nothing about the second condition being satisfied ; his method is,
therefore, here imperfect.
Ex. 2. In V. 30 a value of x has to be found which shall make
6o but
that is, x*  60 > $x\
&  60 < &r j '
Hence, says Diophantus, x is not less than u and not greater
than 12. We have already spoken (p. 60 sqq.) of his treatment
of such cases.
Next, the problem in question requires that # 2 60 shall be
a square. Assume then that
and we have x = (m*
Since, says Diophantus, .ar is greater than n and less than 12,
it follows that
m* + 60 > 22m but < 24.m ;
and m must therefore lie between 19 and 21 (cf. p. 62 above).
He puts m = 2O, and so finds #
III. METHOD OF APPROXIMATION TO LIMITS.
We come, lastly, to a very distinctive method called by
Diophantus vrapuroTrjs or Tra/noor^To? 76)777. The object of this
is to solve such problems as that of finding two, or three, square
numbers the sum of which is a given number, while each of them
approximates as closely as possible to one and the same number.
This method can be best exhibited by giving Diophantus' two
instances, in the first of which two such squares, and in the second
three> are required. In cases like this the principles cannot be
so well indicated with general symbols as with concrete numbers,
which have the advantage that their properties are immediately
96 INTRODUCTION
obvious, and the separate expression of conditions is rendered
unnecessary.
Ex. I. Divide 13 into two squares each of which >6 (V. 9).
Take half of 13, or 6J, and find what small fraction i/# 2 added
to it will make it a square : thus
6J 4 , or 26 H  , must be a square.
x y
Diophantus assumes
+ 2 , or
whence/ = 10 and i// = jfa, or i/^ 2 = ^ ; and
6 +4<r = a square, (f)*.
[The assumption of (i^+i) 2 is not arbitrary, for assume
I = G^+i) 2 , andjy is then 2//(26 /*) ; since i\y should be
a small proper fraction, 5 is the most suitable and the smallest
possible value for/, inasmuch as 26 ~/ 2 < 2$ or /* + 2p + i > 27.]
It is now necessary, says Diophantus, to divide 13 into two
squares' the sides of which are both as near as possible to f^.
Now the sides of the two squares into which 13 is naturally
decomposed are 3 and 2, and
3 is > M b y &>
2 is < ft by i
But, if 3  ^ , 2 + M were ta ken as the sides of two squares, the
sum of the squares would be
which is> 13.
Accordingly Diophantus puts 3 9^, 2 + I ix for the sides of
the required squares, where therefore x is not exactly ^ but
near it.
Thus (3  9#) 2 + (2 + i l^r) 2 = 1 3,
and Diophantus obtains #=^.3.
The sides of the required squares are ff , f$f
[It is of course a necessary condition that the original number,
here 13, shall be a number capable of being expressed as the sum
of two squares.]
DIOPHANTUS 5 METHODS OF SOLUTION 97
Ex. 2. Divide 10 into three squares such that each of them
is > 3 (v. 1 1).
[The original number, here 10, must of course be expressible
as the sum of three squares.]
Take onethird of 10, or 3^, and find what small fraction of the
form i/ar 2 added to 3^ will make a square; Le. we have to make
30 + ^ a square, or 3Oj/ 2 + I a square, where 3/# ijy.
Diophantus assumes
whence y = 2 and therefore i/ar* = ^ ; and 3i +^ ff = W> a square.
[As before, if we assume yyf = (j>y + i) 2 ,j> = 2//(3O/ 2 ); and,
since ijy must be a small proper fraction, 30 / 3 should be<2/,
or/ 2 f2/+i>3i. Accordingly Diophantus chooses 5 for/ as
being the smallest possible integral value.]
We have now, says Diophantus, to make each of the sides
of our required squares as near as may be to ^.
Now 10= 9 + 1 = 3 2 + (f) 2 + (f) 2 >
and 3, f,  are the sides of three squares the sum of which is 10.
Bringing (3, f , ) and ^ to a common denominator, we have
<!*,**>*) and ft.
And 3 >b,
If now we took 3JJ f f+j f + J for sides of squares, the
sum of the squares would be 3 (ffi or ^, which is > 10.
Accordingly Diophantus assumes as the sides of the three
required squares
where * must therefore be not exactly ^ but near it.
Solving (335^) 2 + (f+37^) 2 4(H3^) 2 =io
or 10 ii6>f 3SSS ;tr2 = IO J
we find ^
the required sides are therefore
and the required squares
J 7 45041 J 651 a ! 2 1 65 8944
51J55S1 > 505521 ' 505521 *
H, D.
98 INTRODUCTION
Other instances of the application of the method will be found
in v. 10, 12, 13, 14, where, however, the squares are not required to
be nearly equal, but each of them is subject to limits which may
be the same or different; e.g. sometimes each square is merely
required to be less than a given number (10, say), sometimes the
squares lie respectively between different pairs of numbers, some
times they are respectively greater than different numbers, while
they are always subject to the condition that their sum is a given
number.
As it only lies within the scope of this Introduction to explain
what we actually find in Diophantus, I cannot do more than give
a reference to such investigations as those of Poselger in his
"Beitrage zur unbestimmten Analysis" published in the Abkand
lungen der Koniglichm Akademie der Wissenschaften zu Berlin aits
dem fa/ire 1832, Berlin, 1834. One section of this paper Poselger
entitles "Annaherungsmethoden nach Diophantus/' and obtains
in it, on Diophantus' principles, a method of approximation to the
value of a surd which will furnish the same results as the method
of continued fractions, with the difference that the " Diophantine
method" is actually quicker than the method of continued frac
tions, so that it may serve to expedite the latter.
CHAPTER V
THE PORISMS AND OTHER ASSUMPTIONS IN DIOPHANTUS
I HAVE already mentioned (in Chapter I.) the three explicit refer
ences made by Diophantus to " The Porisms " and the possibility
that, if these formed a separate work, it may have been from that
work that Diophantus took a number of other propositions relating
to properties of numbers which he enunciates or tacitly takes for
granted in the Arithmetic**.
I begin with the three propositions for which he expressly
refers to "The Porisms."
Porism i. In V. 3 he says, "We have it in the Porisms that,
' If each of two numbers and their product when severally added to
the same given number produce squares, the squares with which
they are so connected are squares of two consecutive numbers V"
That is to say, if x + # = ^ 2 , y + a = n\ and if xy + a is also a
square, then m~n = i.
The theorem is not correctly enunciated, for it would appear
that m~n=i is not the only condition under which the three
expressions may be simultaneously squares.
For suppose
By means of the first two equations we have
xy 4 a = m z n* a (m z 4 # 2  i) f a\
In order that
mW  a O 2 + &  i) + a*
may be a square certain conditions must be satisfied. One suffi
cient condition is
or m  n = i,
which is Diophantus' condition.
1 Literally "(the numbers) arise from two consecutive squares" (yeybvaffiv &irb 5j5o
rerpay<tjvtav rwv /card rb ^s).
72
too INTRODUCTION
But we may also regard
m*n*  a (m* + n*  i) H a* =/ 2
as an indeterminate equation in m of which we know one solution,
namely m n i.
Other solutions are then found by substituting s + (n i) for
m, whence we obtain the equation
(n* a)z* + 2{n*(n i)a(n i)}s
+ (n*  a)(n i) 2  a (n*  i) + a* =/ s ,
or (n~a)z* + 2(n*a)(n !)* + {;/( i)#} 2 =/> ;
which is easy to solve in Diophantus' manner, since the absolute
term is a square.
But in the problem V. 3 three numbers are required, such that
each of them, and the product of each pair, when severally added
to a given number, produce squares. Thus if the third number be
#, three additional conditions have to be satisfied, namely
z + a = # 2 , zx f a = z/ 2 , zy + a = w\
The two last conditions are satisfied, if m + i = n, by putting
# = 2 (x +7) i = 4m z + 4m +140,
when xz h a = {in (2m + i)  2a}*
and yz + a=* {(m + i)(2m + i) 2a\*\
and perhaps this means of satisfying the conditions may have
affected the formulation of the Porism 1 .
The problem V. 4 immediately following assumes the truth of
the same Porism with a substituted for + a.
Porism 2. In V. 5 Diophantus says, "Again we have it in the
Porisms that, ' Given any two consecutive squares, we can find in
addition a third number, namely the number greater by 2 than the
double of the sum of the two squares, which makes the greatest of
three numbers such that the product of any pair of them added to
either the sum of that pair or the remaining number gives a square/"
That is, the three numbers
1 Euler has a paper describing and illustrating a general method of finding such
"porisms" the effect of which is to secure that, when some conditions are satisfied, the
rest are simultaneously satisfied ("De problematibus indetermmatis quae videntur plus
quam determinata " in JVovi Commentarii Acad. Petrofol. 175657, Vol. VI. (1761),
p. 85 s^Q. Commentationes arithmeticae collcctae> I. pp. 245259). This particular
porism of Diophantus appears as a particular case in 13 of the paper.
THE PORISMS AND OTHER ASSUMPTIONS 101
have the property that the product of any two plus either the sum
of those two or the remaining number gives a square. In fact, if
X, Y, Z denote the numbers respectively,
YZ
Porism 3 occurs in V. 16. Unfortunately the text is defective
and Tannery has had to supply three words 1 ; but there can be no
doubt that the correct statement of the Porism here in question is
" The difference of any two cubes is also the sum of two cubes,"
z>. can be transformed into the sum of two cubes, or two cubes can
be found the sum of which is equal to the difference between any
two given cubes. Diophantus contents himself with the enuncia
tion of the proposition and does not show how to prove it or how
he effected the transformation in practice. The subject of the
transformation of sums and differences of cubes was investigated
by Vieta, Bachet and Fermat.
Vieta (Zetetica^ IV. 1820) has three problems on the subject.
(i) Given two cubes, to find in rational numbers two other
cubes such that their sum is equal to the difference of the given
cubes 2 .
As a solution of a 3 &=*x*+y*, he finds
8tio idpuv TJ ftTrepoxft Ktifiwv <5tfo <rvvQen&
2 The solution given by Vieta is obtainable thus. The given cubes being a 3 , ft, where
a> b, we assume x  b, a  kx as the sides of the required cubes.
Thus (*)3+(0&c) 3 = 3  3 ,
whence ^ (i  #) + 3^ (aX*  b) + $x (P  a*k) = o.
This reduces to a simple equation if we assume
Jtf%=o, or k=Pla\
in which case
and the sides of the cubes are therefore
b (id? 
a* + P
Vieta's second problem is similarly solved by taking a+x t kxb as the sides of the
required cubes, and the third problem by taking x  , kx  a as the sides of the required
cubes respectively.
loa * INTRODUCTION
(2) Given two cubes, to find in rational numbers two
others such that their difference is equal to the sum of the given
cubes.
Solving <z 3 4 b*  x* f> we find that
(3) Given two cubes, to find in rational numbers two cubes
such that their difference is equal to the difference of the given
cubes.
For the equation a*  6 s = x* y\ Vieta finds
as a solution 1 .
In the solution of (i) x is clearly negative if 2&* > a 3 ; therefore,
in order that the result may be " rational,'' a 3 must be > 28*. But
for a " rational " result in (3) we must, on the contrary, have fl 3 < 2# 3 .
Fermat was apparently the first to notice that, in consequence, the
processes in (i) and (3) exactly supplement each other, so that by
employing them successively we can effect the transformation
required in (i) even when a 3 is not > 2# 3 .
The process (2) is always possible; therefore, by a suitable
combination of the three processes, the transformation of a sum
of two cubes into a difference of two cubes, or of a difference of
two cubes into a sum or a difference of two other cubes is always
1 Vieta's formulae for these transformations give any number of very special solutions
(in integers and fractions) of the indeterminate equation j$+y3 + 1! a =z fl j including solutions
in which one of the first three cubes is negative. These special solutions are based on
the assumption that the values of two of the unknowns are given to begin with, Euler
observed, however, that the method does not give all the possible values of the other two
even in that case. Given the cubes g 8 and 4 s , the method furnishes the solution
3 3 + 4 3 +( ) =( ) > ^t not the simpler solution 33+43 + 53=63. Euler ac
cordingly attacked the problem of solving the equation < * 8 + t y 3 +s 3 s=' 3 more generally.
He began with assuming only one, instead of two, of the cubes to be given, and, on that
assumption, found a solution much more general than that of Vieta. Next he gave a
more general solution still, on the assumption that none of the cubes are given to begin
with. Lastly he proceeded to the problem To fend all the sets of three integral cubes the
sum of which is a cttbe and showed how to obtain a very large number of such sets
including sets in which one of the cubes is negative (Now Commentarii Acad. Petropol
i75657> Vol. VI. (1761), p. 155 ^=Commentatione$ arithmeticae, i. pp. 193207).
The problem of solving ^+j/ 3 =^ 3 +2^ in integers in any number of ways had occupied
Fre"nicle, who gave a number of solutions (Oeuvres de Fermat, III. pp. 420, 535) ; but the
method by which he discovered them does not appear.
THE PORISMS AND OTHER ASSUMPTIONS 103
practicable 1 . Fermat showed also how, by a repeated use of the
several processes as required, we can transform a sum of two cubes
into a sum of two other cubes, the latter sum into the sum of two
others and so on ad infinitum*.
Besides the "Porisms" there are many other propositions
assumed or implied by Diophantus which are not definitely called
1 Fermat (note on iv. a) illustrates by the following case :
Given two cztbes 125 and 64, to transform their difference into the sum of two other
cubes.
Here 0=5, 6 =4, and so 23 3 ><z 3 ; therefore we must first apply the third process
by which we obtain
,._, _
5 4 U 3 ) U
"^ S ( 7T ) > 2 ( n ) ' we can ' k v ^ e ^ rst P roc ess, turn the difference between the
cubes f ~ J and (~ J into the sum of two cubes.
" In fact," says Fermat, "if the three processes are used in turn and continued
ad infinitum> we shall get a succession ad infinitum of two cubes satisfying the same
condition ; for from the two cubes last found, the sum of which is equal to the difference
of the two given cubes, we can, by the second process, find two more cubes the difference
of which is equal to the sum of the two cubes last found, that is, to the difference
between the two original cubes; from the new difference between two cubes we can
obtain a new sum of two cubes, and so on ad infinitum"
As a last illustration, to show how a difference between cubes can be transformed into
the difference between two other cubes even where the condition for process (3) is not
satisfied, Fermat takes the case of 8 ~ i, i.e. the case where
0=2, =i and
First use process (i) and we have
Then use process (2), and
2 Suppose it required to solve the fourth problem of transforming the sum of two cubes
into the sum of two other cubes.
Let it be required so to transform 2 3 4 1 3 or 9.
First transform the sum into a difference of two cubes by process (2). This gives
The latter two cubes satisfy the condition for process (3) and, applying that process,
we get
f 2?V  I^Y ^
V 7/ \7 /
\ 9 39i
The cubes last found satisfy the condition for process (i), and accordingly the difference
between the said last cubes, and therefore the sum of the original cubes, is at last trans
formed into the sum of two other cubes.
io 4 INTRODUCTION
porisms, though some of them are of the same character as the
three porisms above described.
Of these we may distinguish two classes.
I. The first class of theorems or facts assumed without ex
planation by Diophantus are more or less of the nature of identical
formulae. Some are quite simple, <?.. the facts that the expressions
{ \(a + b) } 2  ab and # 2 (a + I ) 2 f a* + (a 4 I ) 2 are respectively
squares, that the expression a (a? a) f a f (/z 3 a) is always a
cube, and the like.
Others are more difficult and betoken a certain facility in work
ing with quasialgebraical expressions. Examples of this kind are
the following :
(a) If X = a*x 4 2a> Y = (a 4 1 ) 2 x + 2 (a + I ), or, in other words,
if xX+i =(^+i) s , xY+i ={(a+ i)#hi} 2 , then XY+i is a
square [IV. 20], As a matter of fact,
JCY+ i ={a(a+ i)x + (2a+ i)} 2 .
(#) 8 times a triangular number plus i gives a square [iv. 38],
In fact, 8/ ( * +l) +i=(2*+ i) 2 .
2
(7) If Xa = m\ Ya = (m+i)*, and Z=2 (X+ F) i,
then the expressions YZ a, ZX a, X Y a are all squares.
(The upper signs refer to the assumption in v. 3, the lower to that
in v. 4.)
In fact, YZ a = {(m + i)(2m + i) + 2a}*,
(S)
then the six expressions
are all squares [v. 6].
In fact,
2. The second class is much more important, consisting of a
number of propositions in the Theory of Numbers which we find
THE PORISMS AND OTHER ASSUMPTIONS 105
first stated or assumed in the Arithmetic**. It was, in general, in
explanation or extension of these that Fermat's most famous notes
were written. How far Diophantus possessed scientific proofs of
the propositions which he assumes, as distinct from a merely
empirical knowledge of them, must remain to a great extent
matter of speculation.
(a) Theorems in DiopJtantus respecting the composition of num
bers as the sum of two squares.
(1) Any square number can be resolved into two squares in
any number of ways, II. 8.
(2) Any number which is the sum of two squares can be
resolved into two other squares in any number of ways, II. 9,
N.B. It is implied throughout that the squares may be frac
tional as well as integral.
(3) If there are two whole numbers each of which is the sum of
two squares ', their product can be resolved into the sum of two squares
in two ways, III. 19.
The object of III, 19 is to find four rational rightangled triangles
having the same hypotenuse. The method is this. Form two
rightangled triangles from (a, b) and (c, d) respectively, i.e. let
the sides of the triangles be respectively
and c* + d\ c*d\ 2cd.
Multiplying all the sides in each by the hypotenuse of the other,
we have two triangles with the same hypotenuse, namely
, 2a
and (a* + b*)(c* + &\ (a* + &)(#  d^ 2cd (a* + S 3 ).
Two other triangles having the same hypotenuse are obtained
by using the theorem enunciated. In fact,
(tf + P)(t + d*) = (ac bdj + (ad + bcj
and the triangles are formed from acbd> ad+bc, being the
triangles
 d*\ 2 (ac + bd)(ad bc\
2 (ac  bd)(ad+ be).
106 INTRODUCTION
In the case taken by Diophantus
a* + b* = 2 2 + I 3 = 5 ,
and the four triangles are respectively
(65, 52, 39), (65> 60, 25), (65, 63, 16), (65, 56, 33>
(If certain relations 1 hold between a, &, c, d, this method fails.
Diophantus has provided against them by taking two triangles " in
the smallest numbers" (VTTO eXa^/crnoi/ apiQp&v), namely 3,4, 5 and
5, 12, 13.)
Upon this problem III. 19 Fermat has a long and important
note which begins as follows 2 :
"[i] A prime number of the form 4h I is the hypotenuse of
a rightangled triangle in one way only, its square is so in two
ways, its cube in three, its biquadrate in four ways, and so on ad
infinitum.
"[2] The same prime number 4^+ I and its square are the
sum of two squares in one way only, its cube and its biquadrate
in two ways, its fifth and sixth powers in three ways, and so on ad
infinitum.
"[3] If & prime number which is the sum of two squares be
multiplied into another prime number which is also the sum of
two squares, the product will be the sum of two squares in two
ways ; if the first prime be multiplied into the square of the second
1 (i) We must not have a\bc\d or a\b~d\c> for in either case one of the perpendiculars
of one of the resulting triangles vanishes, making that triangle illusory. Nor (a) must
c\d be equal to (a + b)l(ab) or to (a~Z>)/(a + &), for in the first case acbd^ad+bc,
and in the second case ac+bd=adbc, so that one of the sums of squares equal to
(a 2 + ^) (fi+d?) is the sum of two equal squares, and consequently the triangle formed
from the sides of these equal squares is illusory, one of its perpendicular sides vanishing.
8 G. Vacca (in Bibliotheca. Mathematics 113. 1901, pp. 3589) points out that Fermat
seems to have been anticipated, in the matter of these theorems, by Albert Girard, who
has the following note on Diophantus v. 9 (Qeuvres mathhnatiques de Simon Stevin par
Albert Girard, 1634, p. 156, col i):
" ALB. GiR. Determinaison tfun nombre qui se peut diviser en deux guarrez entiers.
I. Tout nombre quarr.
II. Tout nombre premier qui excede un nombre quaternaire de Tumie*.
III. Le produict de ceux qui sont tels.
IV. Et le double d'un chacun d'iceux.
Laquelle determinaison n'estant faicte n'y de 1'Autheur n'y des interpretes, servira Jant
en la presente et suivante comme en plusieurs autres. "
Now Girard died on 9 December, 1632 j and the Theorems of Fermat above
quoted are apparently mentioned by him for the first time in his letter to Mersenne of
25 December, 1640 (Oeuvres de Fermat^ n. p. 213). Was the passage of Girard known
to Fermat ?
THE PORISMS AND OTHER ASSUMPTIONS 107
prime, the product will be the sum of two squares in three ways ;
if the first prime be multiplied into the cube of the second, the
product will be the sum of two squares in four ways, and so on
ad infinitum 1 "
It is not probable that Diophantus was aware that prime num
bers of the form 4^ + I and numbers arising from the multiplication
of such numbers are the only classes of numbers which are always
the sum of two squares ; this was first proved by Euler 2 . But it
is remarkable that Diophantus should have selected the first two
prime numbers of the form 4^+1, namely 5 and 13, which are
both sums of two squares, as the hypotenuses of his first two right
angled triangles and then made their product, 65, the hypotenuse
of other rightangled triangles, that product having precisely the
property of being, as in Fermat's [3], the sum of two squares in
two ways. Diophantus may therefore have had an inkling, whether
obtained empirically or otherwise, of some of the properties enunci
ated by Fermat
(4) Still more remarkable is a condition^ possibility of solution
prefixed to the problem V. 9. The object of this problem is " to
divide i into two parts such that, if a given number is added to
either part, the result "will be a square." Unfortunately, the text
of the added condition is uncertain. There is no doubt about the
first few words, " The given number must not be odd" i.e. No number
of the form 4/2 + 3 [or 4/21] can be the sum of two squares.
The text, however, of the latter half of the condition is corrupt.
The true condition is given by Fermat thus : " The given number
must not be odd^ and the double of it increased by one, when divided
by the greatest square which measures it, must not be divisible by a
prime number of the form 4211." (Note upon V. 9; also in a
letter to Roberval 8 .) There is room for any number of conjectures
as to what may have been Diophantus' words 4 .
1 For a fuller account of this note see the Supplement, section i.
2 Novi Commentarii Acad. PetropoL 17523, Vol. rv. (1758), pp. 340, and 1 7545 ,
Vol. V, (1760), pp. 358= Commentationes arifhmetieae, I. pp. 155173 and pp. 210233;
cf. Legendre, Zahlentheorie^ tr. Maser, I. p. 208 ; Weber and Wellstein's Encyckpddie
der EkmentarMathematik) I 2 . pp. 285 sqq.
8 Oeuvres de Penned., n. pp. 2034. See the Supplement, section I,
4 Bachet's text has 8ei Si} TOP 5t86jLievop /tijre irepifforbv efocu, fxfjre 6 8nr\curlw atfroD
q' ju a. jue^ova %# /^pos 8. TJ /terpetrat /7r6 rov a ov . s ov .
He also says that a Vatican MS. reads Mre 6 durXafftw atiroO apidfibv jJLovdda, a.
&XTJ fdpos T^rapro?, 0} jterpetrat ford TOU xpt&rov dpi0/oD.
Neither does Xylander help us much. He frankly tells us that he cannot understand
io8 INTRODUCTION
There would seem to be no doubt that in Diophantus' condition
there was something about "double the number" (i.e. a number of
the form 472), also about " greater by unity " and " a prime number."
It seems, then, not unlikely that, if Diophantus did not succeed in
giving the complete sufficient and necessary condition stated by
Fermat, he made an approximation to it ; and he certainly knew
that no number of the form 4^ + 3 could be the sum of two
squares \
(b) On members which are tfie sum of three squares.
In the problem v. 1 1 a condition is stated by Diophantus re
specting the form of a number which, added to three parts of unity,
makes each of them square. If a be this number, clearly $a+ i
must be divisible into three squares.
Respecting the number a Diophantus says, " It must not be 2
or any multiple of 8 increased by 2."
That is, a number of the form 24^ + 7 cannot be the sum of three
squares. Now the factor 3 of 24 is irrelevant here, for with respect
to three this number is of the form 3z + I, and this, so far as 3 is
concerned, might be a square or the sum of two or three squares.
Hence we may neglect the factor 3 in 24^.
We must therefore credit Diophantus with the knowledge of
the passage. "Imitari statueram bonos grammaticos hoc loco, quorum (ut aiunt) est
multa nescire. Ego ver6 nescio heic non multa, sed paene omnia. Quid enim (ut
reliqua taceam) est fvfyre 6 8ur\a,<rl(av avrov op /to a etc., quae causae huius Trpotrfaopw/AoO,
quae processus? immo qui processus, quae operatic, quae solutio?"
Nesselmann. discusses an attempt made by Schulz to correct the text, and himself
suggests fdjre rbv dtir\afftoj>a aflroC di.pt.Bfji.bv ftovddi fAelotm $x <>v > * jCterpetrac vir6 rivo$
irptirrov &ptfffjioO. But this ignores fjJpos rtraprov and is not satisfactory.
Hankel, however (GescJi. d. Math. p. 169), says: "Ich zweifele nicht, dass die
von den Msscr. arg entstellte Determination so zu lesen ist : Ae? 8i] rbv faSbftevov fje/jre
Trepurobv lva.i f /Jre rbv tinrXacrtova, a^roO &pi>6jj,bv /jLQV&St a fjxl&va /AerpT<r^a6 {/v6 rov
Trpc^rou dpt^/ioO, 8s kv jj.ovd8t a ftetfttw fyv ^P os rfraprw" This correction seems a
decidedly probable one. Here the words /^poy r^ra/rroi' find a place j and, secondly,
the repetition of fj.ov6.St. a fAetfrur might well confuse a copyist, rov for rov is of course
natural enough ; Nesselmann reads TWOS for rov.
Tannery, improving on Hankel, reads A d% rov Bi^evov ivfrre Trepeovrdp elvat, p^re
\rov 5trXct<rtor avrov Ka.1 povifa jiug pelfrva fMcrpefoQat, vir6 rov irpdrrov dpi0/*oC <o5 6
fMvddL fjiig, iulw> #x?7 /w^pos r&raprov t, " the given number must not be odd, and twice
it plus i must not be measured by any prime number which, when increased by i, is
divisible by 4."
1 A discussion of the text and a suggestion as to the considerations which may have
led to the formulation of the condition will be found in Jacobi, "Ueber die Kenntnisse
des Diophantus von der Zusammensetzung der Zahlen" (Berliner Monatsberichte^ 1847;
Gesammelte Werke> vn., 1891, pp. 332344).
THE PORISMS AND OTHER ASSUMPTIONS 109
the fact that no number of the form Sn + 7 can be the sum of
three squares^.
This condition is true, but does not include all the numbers
which cannot be the sum of three squares, for it is not true that
all numbers which are not of the form Sn + ^ are made up of three
squares. Even Bachet remarked that the number a might not be
of the form 32/2 + 9, or a number of the form 9672 + 28 cannot be
the sum of three squares.
Per mat gives the conditions to which a must be subject thus 2 .
Write down two geometrical series (common ratio of each 4),
the first and second series beginning respectively with I, 8,
I 4 16 64 256 1024 4096
8 32 128 512 2048 8192 32768;
then a must not be (r) any number obtained by taking twice any
term of the upper series and adding all the preceding terms, or
(2) the number found by adding to the numbers so obtained any
multiple of the corresponding term of the second series.
Thus a must not be
.4+1
2. 16 + 4+ i = 128/^ + 37,
512^+2.64+16 + 4 + 1 = 512^+149,
and so on, where k = o or any integer.
That is, since i + 4 + 4 2 + . . . + 4 71 " 1 = (4" i), a cannot be either
or 2.4 n
therefore 30 + i cannot be of the form 4" (24^ + 7) or 4* (%k + 7).
Again, there are other problems, e.g. V. 10 and V. 20, in which,
though conditions are necessary for the possibility of solution, none
are mentioned ; but suitable assumptions are tacitly made, without
explanation. It does not follow, from the omission to state the
conditions, that Diophantus was ignorant of even the minutest
points connected with them ; as, however, we have no definite
statements, we must be content to remain in doubt.
1 Legendre proved (Zahhntheoru, tr. Maser, I. p. 386), that numbers of this form are
the only odd numbers which are not divisible into three squares.
3 Note on Diophantus v. u.
no INTRODUCTION
(c) Composition of numbers as the sum of four squares.
Every number is either a square or tlie sum of two } three or four
squares. This wellknown theorem, enunciated by Fermat 1 , and
proved by Lagrange 2 (who followed up results obtained by Euler)
shows at once that any number can be divided into four squares
either integral or fractional, since any square number can be divided
into two other squares, integral or fractional. We have now to look
for indications in the Arithmetica as to how far Diophantus was
acquainted with the properties of numbers as the sum of four squares.
Unfortunately, it is impossible to decide this question with anything
like certainty, There are three problems, IV. 29, 30 and V, 14, in
which it is required to divide a number into four squares, and from
the absence of mention of any condition to which the number must
conform, considering that in both cases where a number is to
be divided into three or two squares, V. n and V. 9, he does
state a condition, we should probably be right in inferring that
Diophantus was aware, at least empirically, that any number could
be divided into four squares. That he was able to prove the
theorem scientifically it would be rash to assert. But we may
at least be certain that Diophantus came as near to the proof of
it as did Bachet, who takes all the natural numbers up to 120
and finds by trial that all of them can actually be expressed
as squares, or as the sum of two, three or four squares in whole
numbers. So much we maybe sure that Diophantus could do, and
hence he might have empirically satisfied himself that it is possible
to divide any number into four squares, integral or fractional, even
if he could not give a rigorous mathematical demonstration of the
general theorem.
1 See note on Diophantus iv. 29 ; cf. also section i. of the Supplement.
2 " Demonstration d'un th6ore"me d'arithmetique " in Nouwaux Mhnoires de FAcad.
royale des sciences de .&?##, anne'e 1770, Berlin 1772, pp. 123133 = Gowns de Lagrtmget
in, pp. 187201 ; cf. Wertheim's account of the proof in his Diophantus, pp. 324330.
CHAPTER VI
THE PLACE OF DIOPHANTUS
IN algebra, as in geometry, the Greeks learnt the beginnings
from the Egyptians. Familiarity on the part of the Greeks with
Egyptian methods of calculation is well attested. Thus (i) Psellus
in the letter published by Tannery 1 speaks of "the method
of arithmetical calculations used by the Egyptians, by which
problems in analysis are handled " ($ /car AlyvTrriovs r&v
SL fj$ oi/covofA6irat, ra Kara rrjv ava\vrt,Krjv
); the details which he goes on to give respecting
the technical terms for different kinds of numbers, including the
powers of the unknown quantity, in use among the Egyptians
are doubtless taken from Anatolius. (2) The scholiast to Plato's
Ckarmides 165 E may be drawing on the same source when he
says that " parts of Xoy urn/erf (the science of calculation) are the
socalled Greek and Egyptian methods in multiplications and
divisions, and the additions and subtractions of fractions..., The
aim of it all is the service of common life and utility for contracts,
though it seems to deal with things of sense as if they were
perfect or abstract/' (3) Plato himself, in the Laws*, says that
freeborn boys should, as is the practice in Egypt, learn, side by
side with reading, simple mathematical calculations adapted to their
age, which should be put into a form such as to give amusement
and pleasure as well as instruction ; e.g. there should be different
distributions of such things as apples, garlands, etc., different
arrangements of numbers of boys in contests of boxing or wrestling,
illustrations by bowls of different metals, gold, copper, silver, etc.,
and simple problems of calculation of mixtures ; all of which are
useful in military and civil life and " in any case make men more
useful to themselves and more wideawake."
1 Dioph. n. pp. 3742. 2 Laws, vn. 819 AC.
ii2 INTRODUCTION
The Egyptian calculations here in point (apart from their
method of writing and calculating in fractions, which differed
from that of the Greeks in that the Greeks worked with ordinary
fractions, whereas the Egyptians separated fractions into sums
of submultiples, with the exception of f which was not decomposed)
are the ^^calculations. Hau, meaning a Jieap, is the term used
to denote the unknown quantity, and the calculations in terms
of it are equivalent to the solutions of simple equations with one
unknown quantity 1 . Examples from the Papyrus Rhind 2 corre
spond to the following equations:
= 19,
) H* + **) IQ.
Before leaving the Egyptians, it is right to mention their
anticipation, though in an elementary form, of a favourite method
of Diophantus, that of the " false supposition " or " regula falsi "
as it is sometimes called. An arbitrary assumption is made as to
the value of the unknown, and the value is afterwards corrected
by a comparison of the result of substituting the wrong value in
the original expression with the actual fact. Two instances
mentioned by Cantor 8 may be given. The first, taken from the
Papyrus Rhind, is the problem of dividing 100 loaves among five
persons in numbers forming an arithmetical progression and such
that one seventh of the sum of the first three shares is equal to
the sum of the other two. If a + 4^, a + 3^, a + zd, a + d, a
are the shares, we have
or d = $%a.
Ahmes merely says, without explanation, " make the difference,
as it is, si>" and then, assuming <2=i, writes the series 23, 17 J,
12, 6, i. The addition of these gives 60, and 100 is if times 60.
Ahmes says simply " multiply if times " and thus gets the correct
values 38^, 29^, 20, lof , i. The second instance (taken from
the Berlin Papyrus 6619) is the solution of the equations
x \y I : , or y \x.
1 For a complete account of the subject the reader is referred to Moritz Cantor's
Geschichte der Mathematik, I 3 , Chapter IL, especially pp. 7481.
2 Eisenlohr, Min mathematisches Handbuch der alten Agypter (Papyrus Rhind des
British Museum) y Leipzig, 1877.
3 Geschickte der Math. I 3 . pp. 789 and p. 95.
THE PLACE OF DIOPHANTUS 113
x is first assumed to be i, and x*+y* is then found to be 25/16.
In order to make 100, 25/16 has to be multiplied by 64 or 8 a . The
true value of x is therefore 8 times i, or 8.
The simple equations solved in the Papyrus Rhind are just the
kind of equations of which we find numerous examples in the
arithmetical epigrams included in the Greek Anthology. Most
of these appear under the name of Metrodorus, a grammarian,
who probably lived about the time of the Emperors Anastasius I.
(491518 A.D.) and Justin I. (518527 A.D.). They were obviously
only collected by Metrodorus, from ancient as well as more recent
sources ; none of them can with certainty be attributed to Metro
dorus himself. Many of the epigrams (46 in number) lead to
simple equations,  with one unknown, of the type of the hau
equations, and several of them are problems of dividing a number
of apples or nuts among a certain number of persons, that is
to say, the very type of problem alluded to by Plato. For
example, a number of apples has to be determined such that, if
four persons out of six receive onethird, oneeighth, onefourth
and onefifth respectively of the total number of apples, while the
fifth person receives ten apples, there remains one apple as the
share of the sixth person, i.e.
We are reminded of Plato's allusion to problems about bowls
(4>id\aL) of different metals by two problems (AnthoL Palat. XIV.
12 and 50) in which the weights of bowls have to be found We
can now understand the allusions of Proclus 1 and the scholiast
on Charmides 165 E to ^\lrat and <f>ia\lTai, apidpol, the adjectives
being respectively formed from fj,fj\ov, an apple, and $40X97, a
bowl. It is clear from Plato's allusions that the origin of such
simple algebraical problems dates back, at least, to the fifth
century B.C.
I have not thought it worth while to reproduce at length the
problems contained in the Anthology 2 , but the following is a
classification of them, (i) Twentythree are simple equations
containing one unknown and of the type shown above ; one of
these is the epigram on the age of Diophantus and incidents
in his life (XIV. 126). (2) Twelve more are easy simultaneous
1 Proclus, Comment, on Eucl. I., ed. Friedlein, p. 40, 5.
2 They are printed in Greek, with the scholia, in Tannery's edition of Diophantus
(II. pp. 4372 and x), and they are included in Wertheim's German translation of
Diophantus, pp. 331343*
H. D. 8
ii 4 INTRODUCTION
equations containing two unknowns, and of the same sort as
Diophantus I. 16 ; or, of course, they can be reduced to a simple
equation with one unknown by means of an easy elimination.
One other (XIV. 51) gives simultaneous equations in three un
knowns
and one (XIV. 49) gives four equations in four unknowns
jr 4^*40, .#+# = 45, x+u = $6, jr+^ + #H = 6o.
With these may be compared Diophantus I. 1621. (3) Six more
are problems of the usual type about the filling of vessels by pipes :
e.g. (xiv. 130) one pipe fills the vessel in one day, a second in two,
and a third in three ; how long will all three running together
take to fill it? Another about brickmakers (XIV. 136) is of the
same sort.
The Anthology contains (4) two indeterminate equations of
the first degree which can be solved in positive integers in an
infinite number of ways (XIV. 48 and 144); the first is a distribution
of apples satisfying the equation x $y =7, where y is not less
than 2, and the original number of apples is 3^ ; the second leads
to the following three equations between four unknown quantities :
the general solution of which is x = 4^, y = k, x^ = 3>fc, jft = 2k. These
very equations, made however determinate by assuming that
x+y***i+jri=* 100, are solved in Diophantus I. 12.
We mentioned above the problem in the Anthology (xiv. 49)
leading to the following four simultaneous linear equations with
four unknown quantities,
x f u
The general solution of any number of simultaneous linear
equations of this type with the same number of unknown quantities
was given by Thymaridas, apparently of Paros, and an early
Pythagorean. He gave a rule, e<o8o9> or method of attack (as
THE PLACE OF DIOPHANTUS 115
lamblichus 1 , our informant, calls it) which must have been widely
known, inasmuch as it was called by the name of the eiravBrj/jLa,
" flower" or "bloom/' of Thymaridas. The rule is stated in general
terms, but, though no symbols are used, the content is pure
algebra. Thymaridas, too, distinguishes between what he calls
aopiaTov, the undefined or unknown quantity, and the &pt,<r/j,evov,
the definite or known, therein anticipating the very phrase of
Diophantus, TrX^^o? (jLovaS&v dopicrTov, "an undefined number of
units," by which he describes his dpiBpo? or x. Thymaridas' rule,
though obscurely expressed, states in effect that, if there are n
equations between n unknown quantities x, x^ ^%i of the
following form,
then the solution is given by
.
n 2,
lamblichus goes on to show that other types of equations can
be reduced to this, so that the rule does not leave us in the lurch
(ov Trapekfcet,) in those cases either. Thus we can reduce to
Thymaridas' form the indeterminate problem represented by the
following three linear equations between four unknown quantities :
^ 4 u = c (y + z).
From the first equation we obtain
from which it follows that, if x, y, z> u are all to be integers,
x+y + z+u must contain a+i as a factor. Similarly it must
contain b f I and c f i as factors.
Suppose now that x+y + z + u = (a+ i)(+ i)(^H 1). There
fore, by means of the first equation, we get
(*+.y)(i+
1 lamblichus, In Nicomachi arithmeticam introductionem (ed. Pistelli), pp. (52,
1 868, 26.
8 2
n6 INTRODUCTION
or x+y = a(b+i)(c+i).
Similarly x + z = b (c+ i)(*+ i),
# + *(* + !)( + i),
and the equations are in the form to which Thymaridas' rule is
applicable.
Hence, by that rule,
a (b + i) (c + i) + . .*  (a + I) ( + (g + i)
*""" 2
In order to ensure that x may always be integral, it is only
necessary to assume
The factor 2 is of course determined by the number of un
knowns. If there are n unknowns, the factor to be put in place
of 2 is n 2.
lamblichus has the particular case corresponding to a = 2,
# = 3, ^ := 4. He goes on to apply the method to the equations
k
~
for. the case where //= f , w/;z = f, //^ = .
Enough has been said to show that Diophantus was not the
inventor of Algebra. Nor was he the first to solve indeterminate
problems of the second degree.
Take, first, the problem of dividing a square number into two
squares (Diophantus II. 8), or of finding a rightangled triangle
with sides in rational numbers. This problem was, as we learn
from Proclus 1 , attributed to Pythagoras, who was credited with
the discovery of a general formula for finding such triangles which
may be shown thus :
where n is an odd number. Plato again is credited, according
to the same authority, with another formula of the same sort,
(2)" + ('!)(*+ if.
1 Comment, on Euclid, Book I. pp. 428, 7sqq.
THE PLACE OF DIOPHANTUS 117
Both these formulae are readily connected with the geometrical
proposition in Eucl II. 5, the algebraical equivalent of which may
be stated as
(abV
_   a fr t
\ 2 I
\ 2
The content of Euclid Book II. is beyond doubt Pythagorean, and
this way of stating the proposition quoted could not have escaped
the Pythagoreans. If we put I for b and the square of any odd
number for a, we have the Pythagorean formula ; and, if we put
a = 2n*, 5 = 2, we obtain Plato's formula. Euclid finds a more
general formula in Book X. (Lemma following X. 28). Starting
with numbers u = c+b and v = c b, so that
uv = c* #,
Euclid points out that, in order that uv may be a square, u and v
must be "similar plane numbers" or numbers of the form
mnq*. Substituting we have
But the problem of finding rightangled triangles in rational
numbers was not the only indeterminate problem of the second
degree solved by the Pythagoreans. They solved the equation
2;tr a y i = I
in such a way as to prove that there are an infinite number of
solutions of that equation in integral numbers. The Pythagoreans
used for this purpose the system of "side" and " diagonal"
numbers 1 , afterwards fully described by Theon of Smyrna *. We
begin with unity as both the first "side" and the first " diagonal ";
thus
#1= I, d l ^i.
We then form (a z> d^ (a s , d^ etc., in accordance with the following
law,
#2 = #! + */i, d% = 2#! + d l ;
#s <h 4 d^ d% 2a% 4 d* ;
and so on. Theon states, with reference to these numbers, the
general* proposition that
and observes that (i) the signs alternate as successive d's and a's
1 See Proclus, In Platonis rempMicam commentarii (Teubner, Leipzig), Vol. II.
c. 27, p. 27, 1118.
3 Theon of Smyrna, ed. Hiller, pp. 43, 44.
n8 INTRODUCTION
are taken, dfza? being equal to  I, 4 2 2rt 2 2 equal to + I,
d  lag equal to  i and so on, (2) the sum of the squares of all
the d's will be double of the sum of the squares of all the a's. For
the purpose of (2) the number of successive terms in each series,
if finite, must of course be even. The algebraical proof is easy.
and so on, while dfza?^ I. Proclus tells us that the property
was proved by means of the theorems of Eucl. II. 9, TO, which
are indeed equivalent to
(2X +JJ/) 2  2 (* +7) a = 2^T a J> 2 .
Diophantus does not particularly mention the indeterminate
equation 2^ 2 i=^ a , still less does he mention "side" and
"diagonal" numbers. But from the Lemma to VI. 15 (quoted
above, p. 69) it is clear that he knew how to find any number of
solutions when one is known. Thus, seeing that # I, ^= i is
one solution, he would put
2 (i 4^r) 3 I = a square
whence x = (4 4 2/)/(/ 2 2).
Take the value / = 2, and we have ^=4, or #+1=5; and
2 . 5 2 i = 49 = 7 2 . Putting x + 5 in place of x, we find a still
higher value, and so on.
In a recent paper Heiberg has published and translated, and
Zeuthen has commented on, still further Greek examples of in
determinate analysis 1 . They come from the Constantinople MS.
(probably of I2th c.) from which Schone edited the Metrica of
Heron. The first two of the thirteen problems had been published
before (though in a less complete form) 3 ; the others are new.
The first bids us find two rectangles such that the perimeter
of the second is three times that of the first, and the area' of the
first is three times that of the second (the first of the two con
ditions is, by some accident, omitted in the text). The number 3
1 Bibliotheca Mathematics vm s , 19078, pp. 118134.
2 Hultsch's Heron, Geepnica t 78, 79. The two problems are discussed by Cantor,
Agrimensoren^ p. 62, and Tannery, M&m. de la sec. des sc. de Bordeaux^ iv 3 , 1882.
THE PLACE OF DIOPHANTUS 119
is of course only an illustration, and the problem is equivalent to
the solution of the equations
u + v = n(x+ff\ ......
xy n.uv J ..................... ^ '*
the solution given in the text is equivalent to
l, y = 2tf )
* 2), ?' #j
( 2 i
u = n( ""  x  "' v Jm
Zeuthen suggests that this solution may have been arrived at
thus. As the problem is indeterminate, it would be natural to
make trial of some hypothesis, e.g. to put v = n. It would follow
from the first equation in (i) that u is a multiple of n, say MS* We
have then
x+y i +#,
*? = *,
whence xy = n*(x +y) ;^ s ,
or (x  ?2 3 ) ( y # s ) = 3 (72 8  i ).
An obvious solution of this is
The second problem is equivalent to the solution of the
equations
and the solution given in the text is
~n*i ..................... (2),
In this case trial may have been made of the assumption
v nx, j/=#X
when the first equation in (i) would give
{n i)x (n z i) #,
a solution of which is #=# 2 i, u n\.
The fifth problem is of interest in one respect. We are asked
to find a rightangled triangle (in rational numbers) with area
of 5 feet. We are told to multiply 5 by some square containing 6
as a factor, e.g. 36. This makes 180, and this is the area of the
triangle (9,jio,til). Dividing each side by 6, we have the triangle
required. The author, then, is aware of the fact that the area
of a rightangled triangle with sides in whole numbers is divisible
120 INTRODUCTION
by 6. If we take the Euclidean formula for a rightangled triangle,
thus making the sides
*' 2
a . mn^ a . ,
2
where # is any number, and m, n are numbers which are both odd
or both even, the area is
2 inn (m n) (m + n)
a _ ,
and, as a matter of fact, the numerator mn (m ;/) (m f n) is
divisible by 24, as was proved later (for another purpose) by
Leonardo of Pisa 1 . There is no sign that Diophantus was aware
of the proposition ; this however may be due to the fact that he
does not trouble as to whether his solutions are integral, but is
satisfied with rational results.
The last four problems (numbered 10 to 13) are of great
interest. They are different particular cases of one problem, that
of finding a rational rightangled triangle such that the numerical
sum of its area and all its three sides is a given number. The
author's solution depends on the following formulae, where a y b
are the perpendiculars, and c the hypotenuse, of a rightangled
triangle, S its area, r the radius of its inscribed circle, and
(The proof of these formulae by means of the usual figure, that
used by Heron to prove his formula for the area of a triangle
in terms of its sides, is easy.)
Solving the first two equations, in order to find a and b, we
have
which formula is actually used by the author for finding a and b.
The method employed is to take the sum of the area and the three
sides 5 + 2s, separated into its two obvious factors s (r + 2), to put
s(r+2) = A (the given number), and then to separate A into
suitable factors to which s and r\ 2 may be equated. They must
obviously be such that sr, the area, is divisible by 6. To take the
first example where A is equal to 280: the possible factors are
1 Scritti, ed, B. Boncompagni, u. (1862), p. 364. Cf. Cantor, Gesch. d. Math, iij ,
p. 40.
THE PLACE OF DIOPHANTUS 121
2 x 140, 4 x 70, 5 x 56, 7 x 40, 8 x 35, 10 x 28, 14 x 20. The
suitable factors in this case are r + 2 = S, ^ = 35, because y is then
equal to 6, and rs is a multiple of 6.
The author then says that
a
and =35 6 = 29.
The triangle is therefore (20, 21, 29) in this case. The
triangles found in the other cases, by the same method, are
(9, 40, 41), (8, 15, 17) and (9, 12, 15).
Unfortunately there is no guide to the date of the problems
just given. The form, however, cannot be that in which the
discoverer or discoverers of the methods indicated originally
explained those methods. The probability is that the original
formulation of the most important of the problems belongs to
the period between Euclid and Diophantus. This supposition best
agrees with the fact that the problems include nothing taken from
the great collection in the Arithmetica. On the other hand, it is
strange that none of the seven problems above mentioned is found
in Diophantus. The five of them which relate to rational right
angled triangles might well have been included by him ; thus he
finds rational triangles such that the area plus or minus one of the
perpendiculars is a given number, but not the rational triangle
which has a given area ; and he finds rational triangles such that
the area plus or minus the sum of two sides is a given number,
but not the rational triangle such that the sum of the area and
the three sides is a given number. The omitted problems might,
it is true, have come in the lost Books ; but, on the other hand,
Book VI. is the place where we should have expected to find
them. Nor do we find in the above problems any trace of
Diophantus' peculiar methods.
Lastly, the famous CattleProblem attributed to Archimedes 1
has to be added to the indeterminate problems propounded before
Diophantus 1 time. According to the heading prefixed to the
epigram, it was communicated by Archimedes to the mathe
maticians at Alexandria in a letter to Eratosthenes. The scholiast
1 Archimedes, ed. Heiberg, Vol. n. p. 450 sqq.
122 INTRODUCTION
on Charmides 165 E also refers to the problem "called by Archi
medes the CattleProblem." Krumbiegel, who discussed the
arguments for and against the attribution to Archimedes, con
cluded apparently that, while the epigram can hardly have been
written by Archimedes in its present form, it is possible, nay
probable, that the problem was in substance originated by
Archimedes 1 . Hultsch 2 has a most attractive suggestion as to
the occasion of it. It is known that Apollonius in his wrcvro/cLov
had calculated an approximation to the value of TT closer than
that of Archimedes, and he must therefore have worked out more
difficult multiplications than those contained in the Measurement
of a circle. Also the other work of Apollonius on the multipli
cation of large numbers, which is partly preserved in Pappus, was
inspired by the Sandreckoner of Archimedes ; and, though we
need not exactly regard the treatise of Apollonius as polemical,
yet it did in fact constitute a criticism of the earlier book. That
Archimedes should then reply with a problem involving such a
manipulation of immense numbers as would be difficult even for
Apollonius is not altogether outside the bounds of possibility. And
there is an unmistakable vein of satire in the opening words of
the epigram, " Compute the number of the oxen of the Sun, giving
thy mind thereto, if thou hast a share of wisdom/' in the tran
sition from the first part to' the second, where it is said that
ability to solve the first part would entitle one to be regarded
as "not unknowing nor Unskilled in numbers, but still not yet
to be counted among the wise," and again in the last lines.
Hultsch concludes that in any case the problem is not much
later than the time of Archimedes and dates from the beginning
of the second century B.C. at the latest.
I have reproduced elsewhere 3 , from Amthor, details regarding
the solution of the problem, and I need do little more than state
here its algebraical equivalent. Eight unknown quantities have
to be found, namely, the numbers of bulls and cows respectively
of each of four colours (I use large letters for the bulls and small
letters for the cows). The first part of the problem connects the
eight unknowns by seven simple equations ; the second part adds
two more conditions.
1 Zeitschrift filr Math. u. Physik (Hist. litt. Abtheilung), xxv. (1880), p. lax sq.
Amthor added (p. i53sq.) a discussion of the problem itself.
2 Art. Archimedes in PaulyWissowa's Reed Encyclopedic, II. r, pp. ^34, 535.
8 The Works of Archimedes, pp. 319326.
THE PLACE OF DIOPHANTUS 123
First part of Problem.
(I) Wti+^X+Y (i),
(2),
(3).
(II) w(i + i)(^T + *) (4),
) (S),
o (6),
) (7).
Second part.
W+X=z square (8),
Y f 2'=a triangular number (9).
The solution of the first part gives
W~ 10366482 n, w = 7206360 ,
JT= 74605 14 n, ^ = 489324672,
F= 4149387 , .7 5439213**
Z = 7358060;^, # = 3515820/2,
where is an integer. The solution given by the scholiast 1 corre
sponds to n 80.
The complete problem would not be unmanageable but for the
condition (8). If for this were substituted the requirement that
W+ X shall be merely a product of two unequal factors (" Wurm's
problem "), the solution in the least possible numbers is
W= 1217263415886, ^ = 846192410280,
x= 876035935422, ^=574579625058,
F= 487233469701, ^ = 638688708099,
Z= 864005479380, ^ = 412838131860.
But, if we include condition (8) and first of all find a solution
satisfying the conditions (i) to (8), we have then, in order to
satisfy condition (9), to solve the equation
q(q+ i)/2 = 51285802909803 . f 2
= 3. 7. n. 29. 353. 4657'. p.
If we multiply by 8, and put
2041=*, 2.4657^=^,.
we have the equation
^1=2.3.7.11.29.353.^,
or t* 4729494 tt*=i.
1 Archimedes, ed. Heiberg, Vol. n. pp. 454, 455.
i2 4 INTRODUCTION
The value of W would be a number containing 206545
digits.
Such are the very few and scattered particulars which we
possess of problems similar to those of Diophantus solved or
propounded before his time. They show indeed that the kind of
problem was not invented by him, but on the other hand they
show little or no trace of anything like his characteristic alge
braical methods. In the circumstances, and in default of discovery
of fresh documents, the question how much of his work represents
original contributions of his own to the subject must remain a
matter of pure speculation. It is pretty obvious that one man
could not have been the author of all the problems contained in
the six Books. There are also inequalities in the work ; some
problems are very inferior in interest to the remainder, and some
solutions may be assumed to be reproduced from other writers
of less calibre, since they reveal none of the mastery of the subject
which Diophantus possessed. Again, it seems probable that the
problem V. 30, which is exceptionally in epigrammatic form, was
taken from someone else. The Arithmetica was no doubt a
collection^ much in the same sense as Euclid's Elements were. And
this may be one reason why so little trace remains of earlier
labours in the same field. It is well known that Euclid's Elements
so entirely superseded the works of the earlier writers of Elements
(Hippocrates of Chios, Leon and Theudius) and of the great
contributors to the body of the Elements, Theaetetus and Eudoxus,
that those works have disappeared almost entirely. So no doubt
would Diophantus' work supersede, and have the effect of con
signing to oblivion, any earlier collections of problems of the
same kind But, if it was a compilation, we cannot doubt that
it was a compilation in the best sense, therein resembling Euclid's
Elements ; it was a compilation by one who was a master of the
subject, who took account of and assimilated all the best that had
been written upon it, arranged the whole of the available material
in due and progressive order, but also added much of his own, not
only in the form of new problems but also (and even more) in the
mode of treatment, the development of more general methods, and
so on.
It is perhaps desirable to add a few words on the previous
history of the theory of polygonal numbers. The theory certainly
goes back to Pythagoras and the earliest Pythagoreans. The
triangle came first, being obtained by first taking I, then adding
THE PLACE OF DIOPHAKTUS 125
2 to it, then 3 to the sum ; each successive number would be
represented by the proper number of dots, and, when each number
was represented by that number of dots arranged symmetrically
under the row representing the preceding number, the triangular
form would be apparent to the eye, thus :
etc.
Next came the Pythagorean discovery of the fact that a similar
successive addition of odd numbers produced
successive square numbers, the odd numbers
being on that account called gnomons, and again
the process was shown by dots arranged to re
present squares. The accompanying figure shows
the successive squares and gnomons.
Following triangles and squares came the figured numbers
in which the "gnomons," or the numbers added to make one
number of a given form into the next larger of the same form,
were numbers in arithmetical progression starting from I, but with
common difference 3, 4, 5, etc., instead of I, 2. Thus, if the
common difference is 3, so that the successive numbers added to
i are 4, 7, 10, etc., the number is a pentagonal number, if the
common difference is 4 and the gnomons 5, 9, 13, etc., the number
is a hexagonal number, and so on. Hence the law that the
common difference of the gnomons in the case of a #gon is
n 2.
Perhaps these facts had already been arrived at by Philippus
of Opus (4th c. B.C.), who is said to have written a work on
polygonal numbers 1 . Next Speusippus, nephew and successor of
Plato, wrote on Pythagorean Numbers, and a fragment of his
book survives 2 , in which linear numbers, polygonal numbers,
triangles and pyramids are spoken of: a fact which leaves no
room for doubt as to the Pythagorean origin of all these con
ceptions 3 .
Hypsicles, who wrote about 170 B.C., is twice mentioned by
Diophantus as the author of a " definition" of a polygonal number,
i, Vitarum scriptores Graeci minores, ed, Westermann, 1845, p. 448.
2 Tkeologumena arithmeticae (ed. Ast), 1817, pp. 61, 62; the passage is translated with
notes by Tannery, Pour Fhistoire de la science keltine^ pp. 386390.
3 Cantor, Geschichte der Mathewatik, I 3 , p. 249.
i 2 6 INTRODUCTION
which is even quoted verbatim 1 . The definition does not mention
any polygonal number beyond the pentagonal ; but indeed this
was unnecessary : the facts about the triangle, the square and the
pentagon were sufficient to enable Hypsicles to pass to a general
conclusion. The definition amounts to saying that the nth agon
(i counting as the first) is
J0 {2 +(!)(* 2)}.
Theon of Smyrna 2 , Nicomachus 8 and lamblichus 4 all devote
some space to polygonal numbers. The first two, who flourished
about 100 A.D., were earlier than Diophantus, and are accordingly
of interest here. Besides a description of the successive polygonal
numbers, Theon gives the theorem that two successive triangular
numbers added together give a square. That is,
( + i) = ^
22
The fact is of course clear if we divide a square
into two triangles as in the figure.
Nicomachus gave various rules for transforming triangles into
squares, squares into pentagons, etc.
1. If we put two consecutive triangles together we get a square
(as in Theon's theorem).
2. A pentagon is obtained from a square by adding to it a
triangle the side of which is I less than that of the square ;
similarly a hexagon from a pentagon by adding a triangle the side
of which is I less than that of the pentagon ; and so on.
In fact,
\n [2 + ( 1)(0 2)} +( i) = \n [2 + (  i) {(a + i)  2}].
Next Nicomachus sets out the first triangles, squares, pentagons,
hexagons and heptagons in a diagram thus :
Triangles ...13 6 10 15 21 28 36 45 55
Squares ...14 9 16 25 36 49 64 81 100
Pentagons ... i 5 12 22 35 51 70 92 117 145
Hexagons ... i 6 15 28 45 66 91 120 153 190
Heptagons... i 7 18 34 55 81 112 148 189 235
and observes that
1 Dioph. I. pp. 470472.
s Exfositio rerum mathematicaruni ad legendum Platonem utilium, ed. Hiller,
pp. 3140.
3 Introductio arithmetic^ ed. Hoche, II. 812, pp. 8799.
4 In Nicomacki arithmeticam introd^ ed. Pistelli, pp. 5861, 6872.
THE PLACE OF DIOPHANTUS 127
3, Each polygon is equal to the polygon immediately above
it in the diagram plus the triangle with I less in its side, i.e. the
triangle in the preceding column.
4. The vertical columns are arithmetical progressions, the
common difference of which is the triangle in the preceding
column.
But Plutarch, a contemporary of Nicomachus, mentioned
another method of transforming triangles into squares: Every
triangular number taken eight times and then increased by \ gives
a square^, That is,
Diophantus generalised this proposition into his theorem for
transforming any polygonal number into a square.
If P be a polygonal number, a the number of angles,
SP (a  2) + (a 4) 2 = a square.
He deduces rules for finding a polygonal number when the
side and the number of angles are given, and for finding the side
when the number and the number of its angles are given. These
fine results and the fragment of the difficult problem of finding
the number of ways in which any given number can be a polygonal
number no doubt represent part of the original contributions by
Diophantus to the theory of that class of numbers.
1 Plat, quaest. V. 2, 4, 1003 F.
THE ARITHMETICA
BOOK I
PRELIMINARY
Dedication.
" Knowing, my most esteemed friend Dionysius, that you are
anxious to learn how to investigate problems in numbers, I have
tried, beginning from the foundations on which the science is
built up, to set forth to you the nature and power subsisting in
numbers.
" Perhaps the subject will appear rather difficult, inasmuch as
it is not yet familiar (beginners are, as a rule, too ready to despair
of success); but you, with the impulse of your enthusiasm and
the benefit of my teaching, will find it easy to master; for
eagerness to learn, when seconded by instruction, ensures rapid
progress."
After the remark that " all numbers are made up of some
multitude of units, so that it is manifest that their formation is
subject to no limit," Diophantus proceeds to define what he calls
the different "species " of numbers, and to describe the abbreviative
signs used to denote them. These "species" are, in the first
place, the various powers of the unknown quantity from the second
to the sixth inclusive, the unknown quantity itself, and units.
Definitions.
A square (=^ 2 ) is Svva/jus (" power "), and its sign is a J with Y
superposed, thus A T .
A cube (=^ 3 ) is #14809, and its sign K Y .
A squaresquare (=# 4 ) is Swa^o^vva^L^, and its sign is A T A.
A squarecube (=^ 5 ) is Swajjioicvftos, and its sign AK T .
A cubecube (=^ 6 ) is /cvffo/cvfios, and its sign K Y K.
1 The term 8vvafAo8foa,/j.is was already used by Heron (Metrica, ed. Schone, p. 48,
u, 19) for the fourth power of a side of a triangle.
H. D. 9
I3 o THE ARITHMETICA
"It is/' Diophantus observes, "from the addition, subtraction
or multiplication of these numbers or from the ratios which they
bear to one another or to their own sides respectively that most
arithmetical problems are formed" ; and "each of these numbers...
is recognised as an element in arithmetical inquiry."
"But the number which has none of these characteristics, but
merely has in it an indeterminate multitude of units (7r\r)6o<$
jjLovdS&v dopurrov) is called dpiffpos, ' number } and its sign is
9 [=*]."
"And there is also another sign denoting that which is in
variable in determinate numbers, namely the unit, the sign being
M with o superposed, thus M."
Next follow the definitions of the reciprocals, the names of
which are derived from the names of the corresponding species
themselves.
Thus
from dptSpo? \x\ we derive the term dpiO/toorov [ i/#]
ov [= !/#*]
ov [= I/**]
ov [= I/*" 4 ]
ov [
Kvj3oKv/3o<rr6v [=
and each of these has the same sign as the corresponding original
species, but with a distinguishing mark which Tannery writes In
the form x above the line to the right.
Thus A* = I/*: 2 , just as 7* = .
Sign of Subtraction (mimts).
"A minus multiplied by a minus makes a plus 1 ; a minus
multiplied by a plus makes a minus ; and the sign of a mimis is a
truncated M* turned upside down^ thus A"
Diophantus proceeds : " It is well that one who is beginning
this study should have acquired practice in the addition, subtraction
and multiplication of the various species. He should know how
to add positive and negative terms with different coefficients to
1 The literal rendering would be "A wanting multiplied by a wanting makes a
forthcoming." The word corresponding to minus is Xei^tf ("wanting"): when it is
used exactly as our minus is, it is in the dative Xetyw, but there is some doubt whether
Diophantus himself used this form (cf. p. 44 above). For the probable explanation of
the sign, see pp. 4244. The word for '* forthcoming" is (brap, from bir&pxw, to exist.
Negative terms are Xeforoj/ra etdij, and positive
BOOK I 131
other terms 1 , themselves either positive or likewise partly positive
and partly negative, and how to subtract from a combination of
positive and negative terms other terms either positive or likewise
partly positive and partly negative.
" Next, If a problem leads to an equation in which certain
terms are equal to terms of the same species but with different
coefficients, it will be necessary to subtract like from like on both
sides, until one term is found equal to one term. If by chance
there are on either side or on both sides any negative terms, it will
be necessary to add the negative terms on both sides, until the
terms on both sides are positive, and then again to subtract like
from like until one term only is left on each side.
" This should be the object aimed at in framing the hypotheses
of propositions, that is to say, to reduce the equations, if possible,
until one term is left equal to one term ; but I will show you later
how> in the case also where two terms are left equal to one term> such
a problem is solved!'
Diophantus concludes by explaining that, in arranging the
mass of material at his disposal, he tried to distinguish, so far as
possible, the different types of problems, and, especially in the
elementary portion at the beginning, to make the more simple lead
up to the more complex, in due order, such an arrangement being
calculated to make the beginner's course easier and to fix what
he learns in his memory. The treatise, he adds, has been divided
into thirteen Books.
PROBLEMS
1. To divide a given number into two having a given
difference.
Given number 100, given difference 40.
Lesser number required x. Therefore
2# + 4O = 100,
^=30.
The required numbers are 70, 30.
2. To divide a given number into two having a given ratio.
Given number 60, given ratio 3:1.
Two numbers x^ $x. Therefore #=15.
The numbers are 45, 15.
species," is the word used by Diophantus throughout.
92
1 32 THE ARITHMETIC A
3. To divide a given number into two numbers such that one
is a given ratio of the other plus a given difference 1 .
Given number 80, ratio 3:1, difference 4.
Lesser number x. Therefore the larger is $x f 4, and
4^+4=80, so that ^= 19.
The numbers are 61, 19.
4. To find two numbers in a given ratio and such that their
difference is also given.
Given ratio 5 : i, given difference 20.
Numbers $x> x. Therefore 4^ = 20, x = 5, and
the numbers are 25, 5.
5. To divide a given number into two numbers such that given
fractions (not, the same) of each number when added together
produce a /given number.
Necessary condition. The latter given number must be such
that it lies between the numbers arising when the given fractions
respectively are taken of the first given number.
First given number 100, given fractions \ and , given
sum of fractions 30.
Second part 5^. Therefore first part = 3 (30 x).
Hence 90 + 2x = 100, and x = 5.
The required parts are 75, 25.
6. To divide a given number into two numbers such that a
given fraction of the first exceeds a given fraction of the other
by a given number.
Necessary condition. The latter number must be less than that
which arises when that fraction of the first number is taken which
exceeds the other fraction.
Given number 100, given fractions \ and \ respectively,
given excess 20.
Second part 6x. Therefore the first part is 4 (x + 20).
Hence larh 80= 100,^ = 2, and
the parts are 88, 12.
1 Literally "to divide an assigned number into two in a given ratio and difference (&
X67<> Kal ^TrepoxS rj 5o9el<ry)." The phrase means the same, though it is not so clear, as
Euclid's expression (Data, Def. ir and passim] 5o06/n pelfuv 77 ev X6y^. According to
Euclid's definition a magnitude is greater than a magnitude " by a given amount (more)
than in a (certain) ratio " when the remainder of the first magnitude, after subtracting
the given amount, has the said ratio to the second magnitude. This means that, if x t y
are the magnitudes, </the given amount, and k the ratio, xdky or x
BOOK I i 33
7. From the same (required) number to subtract two given
numbers so as to make the remainders have to one another a
given ratio.
Given numbers 100, 20, given ratio 3:1.
Required number x. Therefore x 20 = 3 (x 100), and
x = 140.
8. To two given numbers to add the same (required) number so
as to make the resulting numbers have to one another a given ratio.
Necessary condition. The given ratio must be less than the
ratio which the greater of the given numbers has to the lesser.
Given numbers 100, 20, given ratio 3 : 1.
Required numbers. Therefore 3^ + 60 = ^+ 100, and
9. From two given numbers to subtract the same (required)
number so as to make the remainders have to one another a given
ratio.
Necessary condition. The given ratio must be greater than the
ratio which the greater of the given numbers has to the lesser.
Given numbers 20, 100, given ratio 6 : I.
Required number x. Therefore 120 6x = 100 x, and
*=4.
10. Given two numbers, to add to the lesser and to subtract
from the greater the same (required) number so as to make the
sum in the first case have to the difference in the second case
a given ratio.
Given numbers 20, 100, given ratio 4:1.
Required number x. Therefore (20 +x) 4 ( too  x\ and
11. Given two numbers, to add the first to, and subtract the
second from, the same (required) number, so as to make the
resulting numbers have to one another a given ratio.
Given numbers 20, 100, given ratio 3:1.
Required number x. Therefore yc 300 = x+ 20, and
x = 160.
12. To divide a given number twice into two numbers such
that the first of the first pair may have to the first of the second
pair a given ratio, and also the second of the second pair to the
second of the first pair another given ratio,
i 3 4 THE ARITHMETIC A
Given number 100, ratio of greater of first parts to lesser
of second 2:1, and ratio of greater of second parts
to lesser of first parts 3:1.
x lesser of second parts.
The parts then are
f and
IOO 2X)
Therefore 300 $x = 100, x = 40, and
the parts are (80, 20), (60, 40).
1 3. To divide a given number thrice into two numbers such that
one of the first pair has to one of the second pair a given ratio,
the second of the second pair to one of the third pair another
given ratio, and the second of the third pair to the second of the
first pair another given ratio.
Given number 100, ratio of greater of first parts to lesser
of second 3:1, of greater of second to lesser of
third 2 : I, and of greater of third to lesser of
first 4:1.
x lesser of third parts.
Therefore greater of second parts = 2x, lesser of second
= 100 2x y greater of first = 300 6x
Hence lesser of first = 6# 200, so that greater of third
= 242* 800.
Therefore 2$x 800 = 100, x = 36, and
the respective divisions are (84, 16), (72, 28), (64, 36).
14. To find two numbers such that their product has to their
sum a given ratio. [One is arbitrarily assumed.]
Necessary condition. The assumed value of one of the two
must be greater than the number representing the ratio 1 .
Ratio 3 : I, # one of the numbers, 12 the other (> 3).
Therefore 1 2x = 3^ + 36, x = 4, and
the numbers are 4, 12.
15. To find two numbers such that each after receiving from
the other a given number may bear to the remainder a given
ratio.
Let the first receive 30 from the second, the ratio being
then 2:1, and the second 50 from the first, the ratio
being then 3:1; take x + 30 for the second.
* Literally * f th rmmber homonynious with the given ratio. 1 '
BOOK I i 35
Therefore the first = 2;tr 30, and
(*+8o) = 3(2^80).
Thus ^=64, and
the numbers are 98, 94.
16. To find three numbers such that the sums of pairs are
given numbers.
Necessary condition. Half the sum of the three given numbers
must be greater than any one of them singly.
Let (i) 4 (2) = 20, (2) + (3) 30, (3) + (i) = 40.
x the sum of the three. Therefore the numbers are
x  30, x  40, x  20.
The sum x = ix 90, and x = 45.
The numbers are 15, 5, 25.
17. To find four numbers such that the sums of all sets of three
are given numbers.
Necessary condition. Onethird of the sum of the four must be
greater than any one singly.
^Sums of threes 22, 24, 27, 20 respectively.
x the sum of all four. Therefore the numbers are
X22) X2/(., X2'J) X2Q.
Therefore 4^ 93 =#, x = 3 1, and
the numbers are 9, 7, 4, n.
1 8. To find three numbers such that the sum of any pair
exceeds the third by a given number.
Given excesses 20, 30, 40.
2x the sum of all three.
We have (i) + (2) = (3) + 20.
Adding (3) to each side, we have : twice (3) f 20 = 2x> and
(3) = ^ 10.
Similarly the numbers (i) and (2) are #15, ^20
respectively.
Therefore 3^ 45 = 2x, #=45, and
the numbers are 30, 25, 35.
\Otherwise thus 1 . As before, if the third number (3) is x,
Next, if we add the equations
(I) + (2) ( 3 ) = 20
(2) + (3) (i) =30
Tannery attributes the alternative solution of I. 18 (as of i. 19) to an old scholiast.
136 THE ARITHMETICA
we have (2) = J (20 4 30) = 25.
Hence (i)=^S.
Lastly (3) + (i) (2) = 40,
or 2^525=40.
Therefore ^=35
The numbers are 30, 25, 35.]
19. To find four numbers such that the sum of any three
exceeds the fourth by a given number.
Necessary condition. Half the sum of the four given differences
must be greater than any one of them.
Given differences 20, 30, 40, 50.
2x the sum of the required numbers. Therefore the
numbers are
^15, x 20, ^"25, x 10.
Therefore 4#  70 = zx, and x = 35.
The numbers are 20, 15, 10, 25.
[Otherwise thus 1 . If the fourth number (4) is x t
(I) + (2) + ( 3 ) = *+20.
Put (2) 4 (3) equal to half the sum of the two excesses 20
and 30, i.e. 25 [this is equivalent to adding the two
equations
(2) + (3) + ( 4 ) (i) = 30],
It follows by subtraction that (i) = ;r 5.
Next we add the equations beginning with (2) and (3)
respectively, and we obtain
(3) 4 (4) = i (30 440) = 35,
so that (3) = 354;.
It follows that (2) = x 10.
Lastly, since (4) + (i) + (2)  (3) = 50,
3#~ IS (35*) = 50, and ^ = 25.
The numbers are accordingly 20, 15, 10, 25.]
20. To divide a given number into three numbers such that the
sum of each extreme and the mean has to the other extreme a
given ratio,
Given number 100; and let (i) + (2)3.(3) and (2) + (3)
= 4 (i).
1 Tannery attributes the alternative solution of I. 19 (as of i. 18) to an old scholiast.
BOOK I 137
x the third number. Thus the sum of the first and second
= 3^ and the sum of the three = 4^ = 100.
Hence #= 25, and the sum of the first two = 75.
Let y be the first 1 . Therefore sum of second and third
= 4?> 57=100 andj = 2O.
The required parts are 20, 55, 25.
21. To find three numbers such that the greatest exceeds the
middle number by a given fraction of the least, the middle exceeds
the least by a given fraction of the greatest, but the least exceeds
a given fraction of the middle number by a given number.
Necessary condition. The middle number must exceed the
least by such a fraction of the greatest that, if its denominator 2 be
multiplied into the excess of the middle number over the least, the
coefficient of x in the product is greater than the coefficient of
x in the expression for the middle number resulting from the
assumptions made 3 .
Suppose greatest exceeds middle by of least, middle
exceeds least by \ of greatest, and least exceeds
of middle by 10. [Diophantus assumes the three
given fractions or submultiples to be one and the
same.]
x + 10 the least. Therefore middle = 3^r, and greatest
= 6* 30.
Hence, lastly, 6x 30  3# = (x+ 10),
or x+ 10 = 9^90, and x= 12^.
.The numbers are 45, 37 J,
1 As already remarked (p. 52), Diophantus does not use a second symbol for the
second unknown, but makes dpcfyi6s do duty for the second as well as for the first.
2 "Denominator," literally the "number homonymous with the fraction," i.e. the
denominator on the assumption that the fraction is, or is expressed as, a submultiple.
3 Wertheim points out that this condition has reference, not to the general solution of
the problem, but to the general applicability of the particular procedure which Diophantus
adopts in his solution. Suppose Jf, Y, Z required such that X Y~Z\m, YZ=XJn,
Za=Ylp. Diophantus assumes Z=x+a, whence Y=px, X=n(pxxd). The
condition states that np  n>p. If we solve for x by substituting the values of X, Y, Z
in the first equation, we in fact obtain
m {(npnp) x na}=xta,
or x (mnp  mn  mp  \} = a(mn+i).
In order that the value of x may be positive, we must have
that is,.
np>n+p\
^ r m
or (if *, n t p are positive integers] np>n+p.
I3 8 THE ARITHMETICA
[Another solution 1 .
Necessary condition. The given fraction of the greatest must
be such that, when it is added to the least, the coefficient of x in
the sum is less than the coefficient of x in the expression for the
middle number resulting from the assumptions made 2 .
Let the least number be x+ 10, as before, and the given
fraction J ; the middle number is therefore 3^
Next, greatest = middle + (least) = 3^ + 3$.
Lastly, 3*=* * + IO + J (3i* + 3 J)
Therefore x\2\, and
the numbers are, as before, 45, 37^, 22^.]
22. To find three numbers such that, if each give to the next
following a given fraction of itself, in order, the results after each
has given and taken may be equal.
Let first give \ of itself to second, second of itself to
third, third \ of itself to first.
Assume first to be a number of jfs divisible by 3, say
3#, and second to be a number of units divisible by
4, say 4.
Therefore second after giving and taking becomes #+3.
Hence the first also after giving and taking must become
#+3; it must therefore have taken #+3 2#, or
3 x\ 3~* must therefore be of third, or third
= 155*.
Lastly, 15  5* (3 x) + I =*+ 3,
or 134^ = ^ + 3, and # = 2.
The numbers are 6, 4, 5.
23. To find four numbers such that, if each give to the next
following a given fraction of itself, the results may all be equal.
Let first give of itself to second, second \ of itself
to third, third \ of itself to fourth, and fourth of
itself to first.
Assume first to be a number of ^s divisible by 3, say 3^,
and second to be a number of units divisible by 4,
say 4.
1 Tannery attributes this alternative solution, like the others of the same kind, to an
ancient scholiast
2 Wertheim observes that the scholiast's necessary condition comes to the same thing
as Diophantus' own,
BOOK I i 39
The second after giving and taking becomes #+3.
Therefore first after giving x to second and receiving
J of fourth =x+ 3 ; therefore fourth
= 6(3:43 2#)=l86#.
But fourth after giving 3^ to first and receiving of
third =4rh 3 ; therefore third = 30^ 60.
Lastly, third after giving 6x~i2 to fourth and receiving
i from second =^43.
That is, 243; 47=^+3, and ^=ff.
The numbers are therefore ^, 4, ^, ^L;
or, after multiplying by the common de
nominator, 150, 92, 120, 114.
24. To find three numbers such that, if each receives a given
fraction of the sum of the other two, the results are all equal.
Let first receive of (second 4 third), second } of
(third f first), and third  of (first 4 second).
Assume first =.#, and for convenience 1 sake(roi) vrpoxeipov
eveicev) take for sum of second and third a number of
units divisible by 3, say 3.
Then sum of the three = ^13,
and first 4 \ (second f third) = #+!.
Therefore second + ^ (third I first) = x+ I ;
hence 3 times second 4 sum of all = 4^r 4 4,
and therefore second = ;rHJ.
Lastly, third 4 1 (first 4 second) = x 4 1 ,
or 4 times third 4 sum of all
and third =
Therefore #+(*+ ) + (*+ J)
and ^Tf*
The numbers, after multiplying by the common
denominator, are 13, 17, 19.
25. To find four numbers such that, if each receives a given
fraction of the sum of the remaining three, the four results are
equal
Let first receive J of the rest, second J of the rest,
third \ of rest, and fourth of rest.
Assume first to be x and sum of rest a number of units
divisible by 3, say 3.
Then sum of all =x 4 3.
Now first 4 4 (second 4 third 4 fourth) = x f I .
r 4 o THE ARITHMETICA
Therefore second f J (third + fourth + first) =#+ I,
whence 3 times second + sum of all =
and therefore second =.*
Similarly third =*
and fourth
Adding, we have 4^ + $$ = x + 3,
and ^ = w
The numbers, after multiplying by a common
denominator, are 47, 77, 92, 101.
26. Given two numbers, to find a third number which, when
multiplied into the given numbers respectively, makes one product
a square and the other the side of that square.
Given numbers 200, 5 ; required number x.
Therefore 2OO^==(S4r) 2 , and
#=8.
27. To find two numbers such that their sum and product are
given numbers.
Necessary condition. 'The square of half the sum must exceed
the product by a square number. &m Se TOVTO TrX.acrjjLarLKov 1 .
Given sum 20, given product 96.
2% the difference of the required numbers.
Therefore the numbers are iof^, 10 x.
Hence 1 00^ = 96.
Therefore #== 2, and
the required numbers are 12, 8.
28. To find two numbers such that their sum and the sum of
their squares are given numbers.
Necessary condition. Double the sum of their squares must
exceed the square of their sum by a square, eorrt 8e teal TOVTO
1 There has been controversy as to the meaning of this difficult phrase. Xylander,
Bachet, CossaH, Schulz, Nesselmann, all discuss it. Xylander translated it by "effictum
aliunde." Bachet of course rejects this, and, while leaving the word untranslated,
maintains that it has an active rather than a passive signification ; it is, he says, not
something "made up" (effictum) but something "a quo aliud quippiam effmgi et
plasmari potest," " from which something else can be made up,'* and this he interprets as
meaning that from the conditions to which the term is applied, combined with the
solutions of the respective problems in which it occurs, the rules for solving mixed
quadratics can be evolved. Of the two views I think Xylander's is nearer the mark.
v\aff^anKf>v should apparently mean "of the nature of a 7rXd<TAca," just as ffpa/taruc&r
means something connected with or suitable for a drama ; and TrXaVjwa means something
BOOK I 141
Given sum 20, given sum of squares 208.
Difference 2x.
Therefore the numbers are lo+^r, lOx
Thus 2OO + 2;tr 2 = 208, and #= 2.
The required numbers are 12, 8.
29. To find two numbers such that their sum and the difference
of their squares are given numbers.
Given sum 20, given difference of squares 80.
Difference 2x.
The numbers are therefore io+^r, 10 x.
Hence (lof xf (10 ^r) 2 =8o,
or ^ox 80, and x 2.
The required numbers are 12, 8.
30. To find two numbers such that their difference and product
are given numbers.
Necessary condition. Four times the product together with
the square of the difference must give a square, ecrrt, e /cal rovro
Given difference 4, given product 96.
2x the sum of the required numbers.
Therefore the numbers are jr+2, x2\ accordingly
#2 4=96, and x= 10.
The required numbers are 12, 8.
31. To find two numbers in a given ratio and such that the
sum of their squares also has to their sum a given ratio.
Given ratios 3 : I and 5 : 1 respectively.
Lesser number x.
Therefore I or 2 =5.4^, whence x = 2, and
the numbers are 2, 6.
32. To find two numbers in a given ratio and such that the
sum of their squares also has to their difference a given ratio.
Given ratios 3 : i and 10 : I.
Lesser number x, which is then found from the equation
!Otr 2 = IO. 2x.
Hence ^=2, and
the numbers are 2, 6.
"formed" or " moulded." Hence the expression would seem to mean "this is of the
nature of a formula," with the implication that the formula is not difficult to make up
or discover. Nesselmann, like Xy lander, gives it much this meaning, translating it "das
lasst sich aber bewerkstelligen." Tannery translates vXaff/jMTiKov hy "formativum."
i 42 THE ARITHMETICA
33. To find two numbers in a given ratio and such that the
difference of their squares also has to their sum a given ratio.
Given ratios 3 : I and 6 : I.
Lesser number x, which is found to be 3.
The numbers are 3, 9.
34. To find two numbers in a given ratio and such that the
difference of their squares also has to their difference a given
ratio.
Given ratios 3 : I and 12 : i.
Lesser number x> which is found to be 3.
The numbers are 3, 9.
Similarly by the same method can be found two numbers in
a given ratio and (i) such that their product is to their sum in a
given ratio, or (2) such that their product is to their difference in a
given ratio.
35. To find two numbers in a given ratio and such that the
square of the lesser also has to the greater a given ratio.
Given ratios 3 : i and 6 : i respectively.
Lesser number x, which is found to be 18.
The numbers are 18, 54.
36. To find two numbers in a given ratio and such that the
square of the lesser also has to the lesser itself a given ratio.
Given ratios 3 : 1 and 6 : i.
Lesser number x, which is found to be 6.
The numbers are 6, 18.
37. To find two numbers in a given ratio and such that the
square of the lesser also has to the sum of both a given ratio.
Given ratios 3 : I and 2:1.
Lesser number #, which is found to be 8.
The numbers are 8, 24.
38. To find two numbers in a given ratio and such that the
square of the lesser also has to the difference between them a
given ratio.
Given ratios 3 : i and 6 : 1.
Lesser number x, which is found to be 12.
The numbers are 12, 36.
BOOK II 143
Similarly can be found two numbers in a given ratio and
(1) such that the square of the greater also has to the
lesser a given ratio, or
(2) such that the square of the greater also has to the
greater itself a given ratio, or
(3) such that the square of the greater also has to the sum
or difference of the two a given ratio.
39. Given two numbers, to find a third such that the sums of
the several pairs multiplied by the corresponding third number
give three numbers in arithmetical progression.
Given numbers 3, 5.
Required numbers.
The three products are therefore 3^+15, 5^f 15, &r.
Now 3# 4 1 5 must be either the middle or the least of
the three, and 5^+15 either the greatest or the
middle.
(1) 5#+ 1 5 greatest, 3^+ 15 least.
Therefore 5^+ 15 +3^+ 152, &r, and
*=*.
4
(2) $x hi 5 greatest, 3* + 1 5 middle.
Therefore (5*+ 1 5)  (ye f 1 5) = y+ 1 5  &F, and
(3) &* greatest, 3^+15 least.
Therefore &tr 4 $x 41 5 = 2 ($x + 1 5), and
BOOK II
[The first five problems of this Book are mere repetitions of problems in
Book I. They probably found their way into the text from some 'ancient
commentary. In each case the ratio of one required number to the other
is assumed to be 2 : 1. The enunciations only are here given.]
1. To find two numbers such that their sum is to the sum of
their squares in a given ratio [cf. I. 3 1].
2. To find two numbers such that their difference is to the
difference of their squares in a given ratio [cf. I. 34].
144 THE ARITHMETICA
3. To find two numbers such that their product is to their sum
or their difference in a given ratio [cf. I. 34],
4. To find two numbers such that the sum of their squares is to
their difference in a given ratio [cf. I. 32].
5. To find two numbers such that the difference of their squares
is to their sum in a given ratio [cf. I. 33].
6 1 . To find two numbers having a given difference and such
that the difference of their squares exceeds their difference by a
given number.
Necessary condition. The square of their difference must be
less than the sum of the said difference and the given excess
of the difference of the squares over the difference of the
numbers.
Difference of numbers 2, the other given number 20.
Lesser number x. Therefore x f 2 is the greater, and
4^44 = 22.
Therefore x = 4^, and
the numbers are 4^, 6.
7 1 . To find two numbers such that the difference of their
squares is greater by a given number than a given ratio of
their difference 2 . {Difference assumed^
Necessary condition. The given ratio being 3 : i, the square of
the difference of the numbers must be less than the sum of three
times that difference and the given number.
Given number 10, difference of required numbers 2.
Lesser number x. Therefore the greater is x f 2, and
4^ + 4=3.2 + 10.
Therefore #= 3, and
the numbers are 3, 5.
8. To divide a given square number into two squares 3 .
1 The problems n. 6, 7 also are considered by Tannery to be interpolated from some
ancient commentary.
2 Here we have the identical phrase used in Euclid's Data (cf. note on p. 132 above) :
the difference of the squares is rifr inrepoxrjs aOrw^ So0frri dpifyy ttetfav 77 w Xdyy,
literally "greater than their difference by a given number (more) than in a (given) ratio,"
by which is meant ?' greater by a given number .than a given proportion or fraction
of their difference."
3 It is to this proposition that Fermat appended his famous note in which he
enunciates what is known as the " great theorem " of Fermat. The text of the note is
as follows :
"On the other hand it is impossible to separate a cube into two cubes, or a
BOOK II 145
Given square number 16.
x* one of the required squares. Therefore i6# 2 must
be equal to a square.
Take a square of the form 1 (inx$\ m being any
integer and 4 the number which is the square root
of 1 6, e.g. take (2^4)*, and equate it to 16 # 2 .
Therefore 4** 1 6x 4 1 6 = 1 6  .r 3 ,
or S^r 2 = i6^r, and ^r ^.
The required squares are therefore , ~,
, ,
9. To divide a given number which is the sum of two squares
into two other squares 2 .
biquadrate into two bi quadrates, or generally any power except a square into two powers
with the same exponent. I have discovered a truly marvellous proof of this, which
however the margin is not large enough to contain."
Did Fermat really possess a proof of the general proposition that x m +^ m a m cannot
be solved in rational numbers where m is any number >2? As Wertheim says, one
is tempted to doubt this, seeing that, in spite of the labours of Euler, LejeuneDirichlet,
Kummer and others, a general proof has not even yet been discovered. Euler proved
the theorem for ^ = 3 and w=4, Dirichlet for 02 = 5, and Kummer, by means of the
higher theory of numbers, produced a proof which only excludes certain particular
values of m, which values are rare, at all events among the smaller values of m ; thus
there is no value of m below 100 for which Rummer's proof does not serve. (I take
these facts from Weber and Wellstein's Encyclopedia der ElementarMothematik, I 2 ,
p. 284, where a proof of the formula for m=4 is given.)
It appears that the Gottingen Academy of Sciences has recently awarded a prize
to Dr A. Wieferich, of Minister, for a proof that the equation j^+^zP cannot be 
solved in terms of positive integers not multiples of /, if 2*  2 is not divisible by / 2 ,
" This surprisingly simple result represents the first advance, since the time of Kummer,
in the proof of the last Fermat theorem " (Bulletin of the American Mathematical Society,
February 1910).
Fermat says ("Relation des nouvelles decouvertes en la science des nombres,"
August 1659, Oeuvres, II. p. 433) that he proved that no cube is divisible into two cubes by
a variety of his method of infinite diminution (descente infinie or indefinie) different from
that which he employed for other negative or positive theorems; as to the other cases, see
Supplement, sections I., n.
1 Diophantus' words are: "I form the square from any number of Apricot minus
as many units as there are in the side of i6V' It is implied throughout that m must
be so chosen that the result may be rational in Diophantus' sense, i.e. rational and
positive.
2 Diophantus' solution is substantially the same as Euler's (Algebra, tr. Hewlett,
Part n. Art. 219), though the latter is expressed more generally.
Required to find x, y such that
If x^
Put therefore <x f+pz, ygqz\
H, D, I0
146 THE ARITHMETICA
Given number 13 = 2 2 + 3 2 .
As the roots of these squares are 2, 3, take (x + 2f as the
first square and (mx$f as the second (where m is
an integer), say (2x 3) 2 .
Therefore (x* + 4* + 4) + (4^ 2 + 9  * **) = * 3
or 5^+138^=13.
Therefore # = , and
324 i
the required squares are , .
10. To find two square numbers having a given difference.
Given difference 60.
Side of one number x, side of the other x plus any
number the square of which is not greater than 60,
say 3.
Therefore (x + 3) 2  x* = 60 ;
;r=8i, and
the required squares are 72 J, 132 J.
it. To add the same (required) number to two given numbers
so as to make each of them a square.
(i) Given numbers 2, 3 ; required numbers.
X + 2 )
Therefore \ must both be squares.
) 4
r called a doubleequation (S
To solve it, take the difference between the two expressions
and resolve it into two factors 1 ; in this case let us say
4, i
Then take either
(a) the square of half the difference between these factors
and equate it to the lesser expression,
or (V) the square of half the sum and equate it to the
greater.
hence
in which we may substitute all possible numbers for/, q.
1 Here, as always, the factors chosen must be suitable factors, i.e. such as will lead to
a "rational " result, in Diophantus' sense.
BOOK II 147
In this case (a) the square of half the difference is ^f.
Therefore #+ 2 =^f, and x= ?, the squares being ^ 5 , ^.
Taking () the square of half the sum, we have x+ 3=2^^
which gives the same result.
(2) To avoid a doubleequation 1 ,
first find a number which when added to 2, or to 3,
gives a square.
Take e.g. the number x? 2, which when added to 2 gives
a square.
Therefore, since this same number added to 3 gives a
square,
x* f 1 = a square = (x 4)*, say,
the number of units in the expression (in this case 4)
being so taken that the solution may give x* > 2.
Therefore x = 1 g, and
the required number is ?, as before.
04.
1 2. To subtract the same (required) number from two given
numbers so as to make both remainders squares.
Given numbers 9, 21.
Assuming 9 .r 2 as the required number, we satisfy one
condition, and the other requires that 12 +#* shall be
a square.
Assume as the side of this square x minus some number
the square of which > 12, say 4.
Therefore (x  4)* =* 1 2 + x\
and ^=J.
The required number is then 8f .
[Diophantus does not reduce to lowest terms, but says
% =  and then subtracts f from 9 or ^>]
1 This is the same procedure as that of Euler, who does not use doubleequations.
Euler (Algebra, tr. Hewlett, Part n. Art. 214) solves the problem
Suppose
therefore ^=/ 2 ~4, and
Suppose that ^ + 3 =
therefore / = (3 
Thus * = (9 
or, if we take a fraction r\s instead of ,
10  2
148 THE ARITHMETICA
13. From the same (required) number to subtract two given
numbers so as to make both remainders squares.
Given numbers 6, 7.
(i) Let x be the required number.
Therefore \ are both squares.
*7J
The difference is i, which is the product of, say, 2 and ;
and, by the rule for solving a double equation,
(2) To avoid a doubleequation, seek a number which exceeds
a square by 6, say x* 4 6.
Therefore x*  i must also be a square  (x 2) 2 , say.
Therefore x = \, and
the required number is ^.
14. To divide a given number into two parts and to find a
square which when added to each of the two parts gives a square
number.
Given number 20.
Take two numbers 1 such that the sum of their squares
< 20, say 2, 3.
1 Diophantus implies here that the two numbers chosen must be such that the sum of
their squares <2o. Tannery pointed out (Bibllotheca Mathematics 1887, p. 103) that
this is not so and that the condition actually necessary to ensure a real solution in
Diophantus' sense is something different. We have to solve the equations
We assume #=0+w, z>=2 + , and, eliminating *, y, we obtain
2 (m + n)
In order that z may be positive, we must have m z + n z <a; but z need not be positive
in order to satisfy the above equations. What is really required is that x, y shall both be
positive.
Now from the above we derive
__ (m n) (a + iwn)
m + n
Solving for ,#, y, we have
_m(a + mn~n 2 ) _ n (a + mn 
~
m+n
If, of the two assumed numbers, m>n, the condition necessary to secure that JP, y shall
both be positive is a+mn>ufi.
BOOK II 149
Add x to each and square.
We then have
jr 2 4 4*44)
x* 4 6x 49J'
and, if jt i are respectively subtracted, the remainders
are the same square.
Let then x* be the required square, and we have only to
make ^ \ the required parts of 20.
Thus 10^413 = 20,
and x=zfQ.
The required parts are then ( ?3?) and
\io 10 /
the required square is .
15. To divide a given number into two parts and to find a
square which, when each part is respectively subtracted from it,
gives a square.
Given number 20.
Take (#+ m)* for the required square 1 , where m* is not
greater than 20,
e.g. take (x 4 2)*.
This leaves a square if either 4^4 4) . , t , ,
H ^ , hs subtracted.
or 2^43]
Let these then be the parts of 20.
1 Here again the implied condition, namely that m* is not greater than 20, is not
necessary ; the condition necessary for a real solution is something different.
The equations to be solved are x +y= a, z*x=u 2 , z*yiP.
Diophantus here puts (+#*) 2 for s 2 , so that, if #=2/+#& 2 , the second equation is
satisfied. Now (+ #/) 2 y must also be a square, and if this square is equal to (+ m  ) 2 ,
say, we must have
y = 2 f imn  2 .
Therefore, since ,
imn  n=*a
. am* + 
whence =  ;  r  ,
2 (m + )
and it follows that
m (a  mn+tfi) n(a
 i  i. ~ i
m+n
If m>n, it is necessary, in order that x, y may both be positive, that
which is the true condition for a real solution*
tSo THE ARITHMETICA
Therefore 6x + 7 20, and x ^.
The required parts are therefore fg, rj , and
the required square is g.
1 6. To find two numbers in a given ratio and such that each
when added to an assigned square gives a square.
Given square 9, given ratio 3:1.
If we take a square of side (mx+$) and subtract 9
from it, the remainder may be taken as one of the
numbers required.
Take, e.g., (x + 3) 3 9, or x z + 6x, for the lesser number.
Therefore 3^ 3 f \%x is the greater number, and 3^r 2 + 18#+9
must be made a square = (2x 3) 2 , say.
Therefore x = 30, and
the required numbers are 1080, 3240.
17. To find three numbers such that, if each give to the next
following a given fraction of itself and a given number besides,
the results after each has given and taken may be equal 1 .
First gives to second \ of itself 4 6, second to third \ of
itself f 7, third to first \ of itself 4 8.
Let first and second be $#, 6x respectively.
When second has taken x+ 6 from first it becomes 7^4 6,
and when it has given ^r+7 to third it becomes
But first when it has given x + 6 to second becomes
4:r  6 ; and this too when it has taken f of third
+ 8 must become 6x i.
Therefore f of third f 8 = 2x 4 5, and
third = 14^21.
Next, third after receiving \ of second +7 and giving \ of
itself + 8 must become 6x i.
Therefore 13^19 = 6^1, and x ^.
The required numbers are , ?,  5 .
1 Tannery is of opinion that the problems II. 17 and 18 have crept into the text
from an ancient commentary to Book r. to which they would more appropriately belong.
Cf. I. 22, 7.3,
BOOK II 151
1 8. To divide a given number into three parts satisfying the
conditions of the preceding problem 1 .
Given number 80.
Let first give to second of itself f 6, second to third
of itself + 7, and third to first f of itself f 8.
[What follows in the text is not a solution of the problem
but an alternative solution of the preceding. The
first two numbers are assumed to be $x and 12, and
the numbers found are iZ?, , ^1.1
19' 19* 19 J
19. To find three squares such that the difference between the
greatest and the middle has to the difference between the middle
and the least a given ratio.
Given ratio 3:1.
Assume the least square = x*, the middle x* + 2x + I .
Therefore the greatest = x* + 8# + 4 = square = (x 4 3) 3 , say.
Thus ^=2^, and
the squares are 30^, 12^, 6J.
20. To find two numbers such that the square of either added
to the other gives a square 2 .
1 Though the solution is not given in the text, it is easily obtained from the general
solution of the preceding problem, which again, at least with our notation, is easy.
Let us assume, with Wertheim, that the numbers required in u. 1 7 are 5^, 6y, *jz.
Then by the conditions of the problem
4# 6 + z+S=ty7+x+6=6z 8 + x + 7,
from which two equations we can find JP, z in terms of y.
In fact Jf=(26>i8)/i9 and *=(
and the general solution is
5(26j/i8)/i9, 6y
In his solution Diophantus assumes x=*y 9 whence y= I .
Now, to solve II. 18, we have only to equate the sum of the three expressions to 80,
and so find jf.
We have .
^(5.26 + 6. 19 + 7. 17) 5. 187. 3=80. 19,
and the required numbers are
9440 9786 9814
3^3 ' 363 ' 363 *
2 Euler (Algebra^ Part II. Art. 239) solves this problem more generally thus.
Required to find x, y such that &+y and y*+x are squares.
If we begin by supposing A: 2 +j>/=/ 2 , so that y=>px\ and then substitute the value
of y in terms of x in the second expression, we must have
/*  2/ 3 .r 2 + jc 4 + x = square.
But, as this is difficult to. solve, let us suppose instead that
J52 THE ARITHMETICA
Assume for the numbers x, 2x+i t which by their form
satisfy one condition.
The other condition gives
4JT 2 f 5#{ 1 = square = (2% 2) 2 , say.
Therefore ^=^, and
the numbers are ^, ^.
21. To find two numbers such that the square of either minus
the other number gives a square.
x+ I, 2^+1 are assumed, satisfying one condition.
The other condition gives
4^3 + yc . square = 9^, say,
Therefore # = f, and
the numbers are , ^
22. To find two numbers such that the square of either added
to the sum of both gives a square.
Assume x> z + i for the numbers. Thus one condition is
satisfied.
It remains that
x? + 4x+2 = square = (x 2) 2 , say.
Therefore x = , and
the numbers are , $.
4 4
[Diophantus has f, ^.]
23. To find two numbers such that the square of either minus
the sum of both gives a square.
Assume x, x 4 1 for the numbers, thus satisfying one
condition.
Then x?  2x  I = square = (x 3) 2 , say.
Therefore #= 2^, and
the numbers are 2^, 3^.
and that j
It follows that
whence
4^1  41
Suppose, for example, /=, ^=3, and we have *= , ^= ; and so on. We
must of course choose /, such that #, ^ are both positive. Diophantus' solution is
obtained by putting ^=1,^=3.
BOOK II 153
24. To find two numbers such that either added to the square
of their sum gives a square.
Since ^hj^, x^ + Sx* are both squares, let the numbers
be 3:r 2 , &Z 2 and their sum x.
Therefore I2ix* = j? t whence n&*=x 9 and x = fa*
The numbers are therefore 3 , .
25. To find two numbers such that the square of their sum
minus either number gives a square.
If we subtract 7 or 12 from 16, we get a square.
Assume then 12%*, 7^ for the numbers, and 163? for the
square of their sum.
Hence 19^=4^, and x=^.
The numbers are 9?, *.
301 301
26. To find two numbers such that their product added to
either gives a square, and the sides of the two squares added
together produce a given number.
Let the given number be 6.
Since x(^x i)+4r is a square, let#,4Jtr i be the numbers.
Therefore 4# 2 f jar i is a square, and the side of this
square must be 6 2# [since 2# is the side of the
first square and the sum of the sides of the square
is 6].
Since 4^ f y i = (6  2xf y
we have x =ff, and
the numbers are 2, ^.
27. To find two numbers such that their product minus either
gives a square, and the sides of the two squares so arising when
added together produce a given number.
Let the given number be 5.
Assume 42 + i, x for the numbers, so that one condition
is satisfied.
Also 4^ 3# i=(52z) 2 .
Therefore x = ff , and
the numbers are , .
154 THE ARITHMETICA
28. To find two square numbers such that their product added
to either gives a square.
Let the numbers 1 be # 2 , jj> 2 .
__ $\
Therefore , n \ are both squares.
* ?) ^
To make the first expression a square we make x* + i a
square, putting
& + i = (x  2j> say.
Therefore x = f , and x? = &.
We have now to make ^(y* + i) a square [and y must be
different from x\
Put gf + 9 = (3?  4) 3 , say,
and y = &.
Therefore the numbers are ^, ^.
29. To find two square numbers such that their product mimes
either gives a square.
Let ^,y be the numbers.
_
Then /, fl > are both squares.
x*y* j?) ^
A solution of x* i =(a square) is ^ 2 = f.
We have now to solve
^f  ft = a square.
Put yi=(j/4) 2 , say.
Therefore jj/ = ^, and
, 1 280 100
the numbers are ^, ^.
30. To find two numbers such that their product their sum
gives a square.
Now m* f # 2 2; is a square.
Put 2, 3, say, for ?;z, ;z respectively, and of course
2 s + 3 2 2 . 2 . 3 is a square.
Assume then product of numbers = (2 2 + 3 2 )^ or 13^, and
sum = 2 . 2 . yc* or I2# 2 .
The product being 13^, let JIT, 13^ be the numbers.
Therefore their sum 14^= I2^r 2 , and # = J.
The numbers are therefore ?, ^?.
6 J o
1 Diophantus does not use two unknowns, but assumes the numbers to be ^t 2 and i
until he has found x. Then he uses the same unknown (x] to find what he had first taken
to be unity, as explained above, p. 52. The same remark applies to the next problem.
BOOK II 155
31. To find two numbers such that their sum is a square and
their product + their sum gives a square.
2 . 2m .ma. square, and (zmj + m* + 2 . 2m . m = a square.
If m = 2, 4 2 + 2 2 2 . 4 . 2 = 36 or 4.
Let then the product of the numbers be (4* + 2 z )x* or 2cutr 3 ,
and their sum 2.4.2** or i6# 2 , and take 2x, IQX for
the numbers.
Then 1 2x = 1 6;tr 3 , and # = f .
j
The numbers are , ^.
4 4
32. To find three numbers such that the square of any one of
them added to the next following gives a square.
Let the first be x, the second 2x 4 i, and the third
2(2#i)f i or 4*rf 3, so that two conditions are
satisfied.
The last condition gives (4^ f 3) 3 + x = square = (^x  4) 2 ,
say.
Therefore x ^, and
the numbers are ^ ,3 529,
33. To find three numbers such that the square of any one of
them mimts the next following gives a square.
Assume x f i, 2x + i, 4*+ i for the numbers, so that two
conditions are satisfied.
Lastly, 1 6x* + ?x= square = 2 5# 2 , say,
and # = $
The numbers are , 3, 2.
9' 9 s 9
34. To find three numbers such that the square of any one
added to the sum of all three gives a square.
fy(m n)}*+mn Is a square. Take a number separable
into two factors (m, n) in three ways, say 12, which is
the product of (i, 12), (2,6) and (3, 4).
The values then of % (m  n) are 5 > 2, \.
Take i\x, 2x, %x for the numbers, and for their sum \2x\
Therefore &tr= izar 2 , and # = f.
The numbers are ", ^ I.
333
[Diophantus says , and ^, f, .]
156 THE ARITHMETICA
35. To find three numbers such that the square of any one
minus the sum of all three gives a square.
{J (m 4 n)} z mn is a square, Take, as before, a number
divisible into factors in three ways, as 12.
Let then 6^r, 4^, $\x be the numbers, and their sum
Therefore 14^= I2x* 9 and.r = .
The numbers are
BOOK III
1. To find three numbers such that, if the square of any one
of them be subtracted from the sum of all three, the remainder
is a square 1 .
Take two squares .ar 2 , 4# 2 ; the sum is 5# 2 .
If then we take $x* as the sum of the three numbers, and
x, 2x as two of them, we satisfy two conditions.
Next divide 5, which is the sum of two squares, into two
other squares ^ ^ [n. 9], and assume a? for the
third number.
Therefore x + 2x + \x = &*, and x f
The numbers are ^, * &.
[Diophantus writes ^ for ^r and ^\, ^, ^ for the numbers.]
2. To find three numbers such that the square of the sum of
all three added to any one of them gives a square.
Let the square of the sum of all three be x* 9 and the
numbers yc*, &r 2 , \$x\
Hence 26# 2 = x y # = &$> anc *
the numbers are ^, ^ J* .
3. To find three numbers such that the square of the sum of
all three minus any one of them gives a square.
Sum of all three 4^, its square i6# 2 , the numbers 7# 2 ,
Then 34** = 43;, # = $, and
the numbers are , , .
1 The fact that the problems in. 14 are very like n, 34, 35 makes Tannery suspect
that they have found, their way into the text from some ancient commentary.
BOOK III 157
4. To find three numbers such that, if the square of their sum
be subtracted from any one of them, the remainder is a square.
Sum x, numbers 2^r 2 , 5^r 2 , io;r 2 .
Then 1 7^r 2 = x, x = ^, and
the numbers are  , ^~, ^.
5. To find three numbers such that their sum is a square and
the sum of any pair exceeds the third by a square.
Let the sum of the three be (#+ i) 2 ; let first + second
= third 41, so that third = ^ + x ; let second f third
= first f x*, so that first = x f J.
Therefore second = ^x* + .
It remains that first + third = second 4 a square.
Therefore 2x = square = 16, say, and x = 8.
The numbers are 8J, 32 \, 40.
Otherwise thus 1 .
First find three squares such that their sum is a square.
Find eg. what square number + 4 + 9 gives a square,
that is, 36 ;
4 3 36? 9 are therefore squares with the required property.
Next find three numbers such that the sum of each pair =
the third i a given number ; in this case suppose
first f second third 4,
second f third  first = 9,
third f first  second = 36.
This problem has already been solved [I. 18].
The numbers are respectively 20, 6, 22^.
1 We should naturally suppose that this alternative solution, like others, was inter
polated. But we are reluctant to think so because the solution is so elegant that it
can hardly be attributed to a scholiast. If the solution is not genuine, we have here
an illustration of the truth that, however ingenious they are, Diophantus' solutions are not
always the best imaginable (Loria, Le scienze esatte nelf antica Grecia, Libro v. pp. 1389).
In this case the more elegant solution is the alternative one. Generalised, it is as follows.
We have to find x, y, z so that
S=SL square \
z=a squared ,
/s=a square)
and also x +y t z = a square.
We have only to equate the first three expressions to squares a z , ^, c* such that
fl 2+2_j. f 2 =a square, JP say, since the sum of the first three expressions is itself
x+y+z.
The solution is then
158 THE ARITHMETICA
6. To find three numbers such that their sum Is a square and
the sum of any pair is a square.
Let the sum of all three be x + 2x + I, sum of first and
second x\ and therefore the third 2x ti ; let sum of
second and third be (x i) 2 .
Therefore the first = 4^ and the second = x*  4#.
But first + third = square,
that is, 6>+ i = square =121, say.
Therefore x = 20, and
the numbers are 80, 320, 41.
[An alternative solution, obviously interpolated, is practically
identical with the above except that it takes the square 36 as
the value of 6^rf r, so that x and the numbers are ^
_ 840 385 456 

7. To find three numbers in A.P. such that the sum of any
pair gives a square.
First find three square numbers in A.P. and such that half
their sum is greater than any one of them. Let
x\ (x+ if be the first and second of these ; therefore
the third is x* + 4# + 2 = (x 8) 2 , say.
Therefore 4r = forf;
and we may take as the numbers 961, 1681, 2401.
We have now to find three numbers such that the sums
of pairs are the numbers just found.
The sum of the three = ^^ = 252 1, and
the three numbers are i2o, 840^, 1560^,
8. Given one number, to find three others such that the sum
of any pair of them added to the given number gives a square, and
also the sum of the three added to the given number gives a
square.
Given number 3.
Suppose first required number h second = x* + ^x + i,
second + third = x* + 6x + 6,
sum of all three = x* 4 8# 4 1 3.
Therefore third =43; +12, second == x* +2X 6, first =
Also first i third f 3 = a square,
that is, 6x h 22 = square = 100, suppose.
Hence #13, and
the numbers are 33, 189, 64.
BOOK III 159
9. Given one number, to find three others such that the sum
of any pair of them mimis the given number gives a square, and
also the sum of the three minus the given number gives a square.
Given number 3.
Suppose first of required numbers + second =.# 2 f 3,
second 4 third = x z 4 2x f 4,
sum of the three = x* + 42 f 7.
Therefore third =4^ + 4, second = JF 2 24:, first = 24: + 3.
Lastly, first + third 3=dar}4 = a square = 64, say.
Therefore x = 10, and
(23, 80, 44) is a solution.
10. To find three numbers such, that the product of any pair
of them added to a given number gives a square.
Let the given number be 12. Take a square (say 25)
and subtract 12. Take the difference (13) for the
product of the first and second numbers, and let these
numbers be i^x, ifx respectively.
Again subtract 12 from another square, say 16, and let the
difference (4) be the product of the second and third
numbers.
Therefore the third number = 4^.
The third condition gives 52^ 2 + 12 = a square ; now
52 = 4. 13, and 13 is not a square; but, if it were a
square, the equation could easily be solved 1 .
Thus we must find two numbers to replace 13 and 4 such
that their product is a square, while either f 12 is
also a square.
Now the product is a square if both are squares ; hence we
must find two squares such that either + 12 = a square.
" This is easy 2 and, as we said, it makes the equation easy
to solve."
The squares 4, J satisfy the condition.
1 The equation 52^+ 12 = 2 can in reality be solved as it stands, by virtue of the fact
that it has one obvious solution, namely x = i . Another solution is found by substituting
y+ i for #, and so on. Cf. pp. 69, 70 above. The value .r= i itself gives (13, i, 4) as
a solution of the problem.
z The method is indicated in II. 34. We have to find two pairs of squares differing
by 12. (a) If we put 12 = 6. 2, we have
and 1 6, 4 are squares differing by 12, or 4 is a square which when added to 12 gives a
square, (b) If we put 12 = 4.3, we find j(43)[ or  to be a square which when
added to 12 gives a square.
160 THE ARITHMETICA
Retracing our steps, we now put 4^, i\x and x/4 for the
numbers, and we have to solve the equation
x* + 1 2 = square = (x f 3) 2 , say.
Therefore # = ^, and
t 2, is a solution 1 .
II. To find three numbers such that the product of any pair
minus a given number gives a square.
Given number 10.
Put product of first and second = a square + 10 = 4+ 10,
say, and let first = 14*, second = i/ar.
Let product of second and third = a square +10=19, say ;
therefore third = IOJIT.
By the third condition, 266> 2  10 must be a square ; but
266 is not a square 2 .
Therefore, as in the preceding problem, we must find two
squares each of which exceeds a square by 10.
The squares 30^, \2\ satisfy these conditions 3 .
Putting now 30^, I/*, \2\x for the numbers, we have,
by the third condition, 37O^r 2  10 = square [for
370^ Diophantus writes 37oJ T V]j
therefore 5929^  160 = square = (77^  2) 2 , say.
Therefore # = ff , and
the numbers are SE*. B, SSi.
1 Euler (Algebra, Part II. Art. 232) has an elegant solution of this problem in whole
numbers. Let it be required to find x, y^ z such that xy + a, yz + a, zx + a are all squares.
Suppose xy+a=jP, and make z?=.x+y+q\
therefore x& + a =x* + xy + qy, f a =
and yz +a xy +y* + qy+
and the right hand expressions are both squares if q=* 2/, so that z = x+y ip.
We can therefore take any value for p such that #?>#, split ja into factors,
take those factors respectively for the values of x and y, and so find z.
E.g. suppose a 11 and ^=25, so that ^=13; let #=1,^=13, and we have
z= i4 10=24 or 4, and (i, 13, 4), (r, 13, 24) are solutions.
2 As a matter of fact, the equation 266V io=# 2 can be solved as it stands, since it
has one obvious solution, namely x=i. (Cf. pp. 69, 70 above and note on preceding
problem, p. 159.) The value x= r gives (14, r, 19) as a solution of the problem.
3 Tannery brackets the passage in the text in which these squares are found, on
the ground that, as the solution was not given in the corresponding place of in. TO, there
was no necessity to give it here. 10 and i being factors of 10, '
thus 3oJ is a square which exceeds a square by 10. Similarly ji (5 + 2) I or i2j is such
a square. The latter is found in the text by putting a* 2  10= square =:(m  2) 2 , whence
w=3^, and tn 2 =u.
BOOK III 161
12. To find three numbers such that the product of any two
added to the third gives a square.
Take a square and subtract part of it for the third number ;
let ;tr 2 H6> + 9 be one of the sums, and 9 the third
number.
Therefore product of first and second = x* + 6x\ let first
= x, so that second = x f 6.
By the two remaining conditions
IQX + 54) , .
Z.Y are both squares.
lorH 6J H
Therefore we have to find two squares differing by 48 ;
" this is easy and can be done in an infinite number
of ways."
The squares 16, 64 satisfy the condition. Equating these
squares to the respective expressions, we obtain
x i, and
the numbers are i, 7, 9.
13. To find three numbers such that the product of any two
minus the third gives a square.
First x, second x\ 4 ; therefore product = x* +4#, and we
assume third = ^x.
Therefore, by the other conditions,
^_ [ are both squares.
4jr 2 ^4J M
The difference = i6x + 4 = 4 (4^ 4 1), and we put
Therefore x ^ } and
the numbers are 3 J55 !.
20 20 zo
14. To find three numbers such that the product of any two
added to the square of the third gives a square 1 .
1 Wertheim gives a more general solution, as follows. If we take as the required
numbers X=~ax, Y=ajc+ZP t Z=IP, two conditions are already satisfied, namely
X y+ Z* = a square, and YZ + X s = a square.
It only remains to satisfy the condition ZATt 7 2 =a square, or
a*x? + ~ afix + & = a square.
Put
where /& remains undetermined.
H. D.
i6i THE ARITHMETICS
Firsts, second 4^ + 4, third i. Two conditions are thus
satisfied.
The third condition gives
* + (4* + 4) 2 = a square = (4* 5) 2 , say.
Therefore # = 7 %, and
the numbers (omitting the common denominator)
are 9, 328, 73.
15. To find three numbers such that the product of any two
added to the sum of those two gives a square 1 .
[Lemma.] The product of the squares of any two con
secutive numbers added to the sum of the said
squares gives a square 2 .
Let 4, 9 be two of the required numbers, # the third.
Therefore l ^ L are both squares.
5^ + 4) 4
The difference = 5*+ 5 = 5 (x + 1).
Equating the square of half the sum of the factors to
i or + 9, we have
Therefore x 28,, and (4, g, 28) is a solution.
1 The problem can of course be solved more elegantly, with our notation, thus. (The
same remark applies to the next problem, in. 16.)
If x, y, z are the required numbers, xy+x+y, etc. are to be squares. We may
therefore write the conditions in the form
)= a square +i,
a square +i,
(jc 4 i)(y+ 1) =a square + 1.
Assuming a 2 , ^, fi for the respective squares, and putting
we have to solve
[This is practically the same problem as that in the Lemma to Dioph. v. 8.]
Multiplying the second and third equations and dividing by the first, we have
with similar expressions for T;,
JT, y, z are these expressions minus i respectively, a 2 , ^, ^ must of course be so
chosen that the resulting values of , 97, f may be rational. Cf. Euler, Commentationes
arithmetical^ II. p. 577.
2 In fact,
BOOK III 163
Otherwise thus^.
Assume first number to be x, second 3.
Therefore 4^ f 3 = square = 25 say, whence x = 5^, and 5 ,
3 satisfy one condition.
1 This alternative solution would appear to be undoubtedly genuine.
Diophantus has solved the equations
Fermat shows how to solve the corresponding problem with four numbers instead of
three. He uses for this purpose Diophantus 9 solution of V. 5, namely the problem
of finding x\ y\ z 2 , such that
Diophantus finds f , ^, ^ J as a solution of the latter problem. Fermat takes
these as the first three of the four numbers which are to satisfy the condition that the
product of any two plus the sum of those two gives a square, and assumes x for the
fourth. Three relations out of six are already satisfied, and the other three require
9 9 99
64 64 73 64
x  , or
196 196 205* 196
Zjc+jct Z~, or  I  z 
9 9 99
to be made squares: a "tripleequation" to be solved by Fermat's method. (See the
Supplement, section v.)
Fermat does not give the solution, but I had the curiosity to work it out in order to
verify to what enormous numbers the method of the tripleequation leads, even in such
comparatively simple cases.
We may of course neglect the denominator 9 and solve the equations
73* + 64=2/2,
The method gives
x _ 459818598496844787200
631629004828419699201 '
the denominator being equal to (25132230399)2.
Verifying the correctness of the solution, we find that, in fact,
?7V
64
o5j. _ n = ( 12275841601 V
196 \25i32230399/
Strictly speaking, as the value found for x is negative, we ought to substitute y  A
for it (where A is the value found) in the three equations and start afresh. The
portentous numbers which would thus arise must be left to the imagination.
II  2
1 64 THE ARITHMETIC A
Let the third be x, while 5 J, 3 are the first two.
Therefore * * 3 i must both be squares ;
otjX f Si J
fo//, $/# //* coefficients in one expression are respectively
greater than those in the other, but neither of the ratios
of corresponding coefficients is that of a square to a
square^ our suppositions will not serve the purpose ; we
cannot solve by our method.
Hence (to replace 5^, 3) we must find two numbers such
that their product + their sum = a square, and the
ratio of the numbers increased by i respectively is
the ratio of a square to a square.
Let these be y and 4? + 3, which satisfy the latter con
dition ; and, in order that the other may be satisfied,
we must have
4? 2 + 87 + 3 = square = (2y  3)*, say.
Therefore y = fa .
Assume now ft, 4j, # for the three numbers.
Therefore *** ^ \ are both squares,
iTJ2" r TO 1 )
or, if we multiply by 25 and 100 respectively,
105 i are both squares .
130*+ 30] H
The difference is 75 = 3 . 25, and the usual method of
solution gives #" = &.
The numbers are , *?, ^.
id To find three numbers such that the product of any two
minus the sum of those two gives a square.
Put x for the first, and any number for the second; we
then fall into the same difficulty as in the last
problem.
We have to find two numbers such that
(a) their product minus their sum = a square, and
(V) when each is diminished by i, the remainders
have the ratio of squares.
Now 4/y+ i, y + i satisfy the latter condition.
The former (a) requires that
4^ i = square = (27 2) 2 , say,
which gives y = .
Assume then ^, %j,x for the numbers.
BOOK III 165
Therefore ^ ~ ~ I are both squares,
or, if we multiply by 4, 16 respectively,
IO^T 14) , 
J } are both squares.
!O;tr 20) ^
The difference is 12 = 2.6, and the usual method gives
*=3
The numbers are^p, 3i = f , S^y
17. To find two numbers such that their product added to
both or to either gives a square.
Assume #,4^1 for the numbers, since
xi^x i) + ;r = 4^, a square.
A%& J_ 'Ig J ^
Therefore also \ are both squares.
4^ ~r 4^ ~~ i j
The difference is x^^x. \, and we find
*=*
The numbers are ^, ^.
1 8. To find two numbers such that their product minus either,
or mimis the sum of both, gives a square 1 .
3 With this problem should be compared that in paragraph 42 of Part I. of the
Inventwm Novum of Jacobus de Billy (Oeuvres de Fermat, ill. pp. 3512), where three
conditions correspond to those of the above problem, and there is a fourth in addition.
The problem is to find , y (>ty) such that
fcrt
. *'&> ( are all squares.
+ >77[ H
>?frJ
Suppose 17= ^ =!.#; the first two conditions are thus satisfied. The other
two give
Separating the difference 2J? into the factors ?, T, we put, as usual,
whence *= I > and the numbers are  , f .
O O O
To find another value of x by means of the value thus found, we put y+^t in place of
O
x in the doubleequation, whence
Multiplying the lower expression by 49, we can solve in the usual way. Our expressions
1 66 THE ARITHMETICA
Assume x + i, 4^ for the numbers, since
^x (x 4 1) Ax = a square.
Therefore also 4 ^ 3 * " j  are both squares.
The difference is 4^ = 4^. I, and we find
The numbers are 2^, 5.
19. To find four numbers such that the square of their sum
plus or minus any one singly gives a square.
Since, in any rightangled triangle,
(sq. on hypotenuse) (twice product of perps.) = a square,
we must seek four rightangled triangles [in rational
numbers] having the same hypotenuse,
or we must find a square which is divisible into two
squares in four different ways ; and "we saw how to
divide a square into two squares in an infinite
number of ways." [IL 8.]
Take rightangled triangles in the smallest numbers,
(3> 4. 5) and (5, 12, 13); and multiply the sides of
are now ,7 s  y + ~ and 49^  Qy + Z , and the difference between them is 48^  1 1 oy.
The solution next mentioned by De Billy was clearly obtained by separating this
difference into factors such that, when the square of half their difference is equated to
y 2  y+ P , the absolute terms cancel out. The factors are ^y, y   , and we put
'T TT / 55 "T
This gives y=  4 4 j )I95 , whence #= 22 ^ 59<z7 , and the numbers are
71362992 7i3 62 99*
48647065 22715927
71362992' 71362992*
A solution in smaller numbers is obtained by separating 48y 2  noy into factors such
that the terms in x? in the resulting equation cancel out. The factors are 6y, ty  S , and
we put
whence ,*Z29, ^d *= 4Z2S2 + f , .
10416 10416 8 10416
This would give a negative value for IJP; but, owing to the symmetry of the
original doubleequation in x, since *=! satisfies it, so does ~^ ; hence the
numbers are ^g and ^4j& : a solution also mentioned by De Billy.
Cf. note on IV. 23.
BOOK III 167
the first by the hypotenuse of the second and vice
versa.
This gives the triangles (39, 52, 65) and (25, 60, 65); thus
65 2 is split up into fwo squares in two ways.
Again, 65 is "naturally" divided into two squares in two
ways, namely into 7 2 + 4 2 and 8 2 + i 2 , "which is due
to the fact that 65 is the product of 13 and 5, each of
which numbers is the sum of two squares."
Form now a rightangled triangle 1 from 7, 4. The sides
are (; 3  4 2 , 2 . 7 . 4, 7* 4 4 2 ) or (33, 56, 65).
Similarly, forming a rightangled triangle from 8, i, we
obtain (2.8.1, 8 2  1 2 , 8 3 + I s ) or 16, 63, 65.
Thus 6s 2 is split into two squares in four ways.
Assume now as the sum of the numbers. 65^ and
as first number 2 . 39 . 52^ = 4056^,
second 2.25.60^ = 3000^,
third 2.33.56^ = 3696^,
fourth 2.16. 633? = 20i6x* t
the coefficients of x? being four times the areas of the
four rightangled triangles respectively.
The sum 12768^ = 65^, and Z"= : T$%F
The numbers are
17136600 12675000 15615600 8517600
163021824 ' 163021824* 163021824* 163021824*
20. To divide a given number into two parts and to find a
square which, when either of the parts is subtracted from it, gives
a square 2 .
Given number 10, required square & 4 2;r + i.
Put for one of the parts 2#+ I, and for the other 4*.
The conditions are therefore satisfied if
= 10.
Therefore ^r= i^;
the parts are (4, 6) and the square 6J.
1 If there are two numbers/, y 9 to "form a rightangled triangle" from them means
to take the numbers / 2 +^ 2 , p  g 2 , zp?. These are the sides of a rightangled triangle,
2 This problem and the next are the same as II. 15, 14 respectively. It may therefore
be doubted whether the solutions here given are genuine, especially as interpolations
from ancient commentaries occur most at the beginning and end of Books,
1 68 THE ARITHMETICA
21. To divide a given number into two parts and to find a
square which, when added to either of the parts, gives a square.
Given number 20, required square ^+ 2x 4 I.
If to the square there be added either 2^ + 3 or 4^+8,
the result is a square.
Take 2^ + 3, $x + 8 as the parts of 20, and 6x+ 11=20,
whence x=\\.
Therefore the parts are (6, 14) and the square 6.
BOOK IV
1. To divide a given number into two cubes such that the sum
of their sides is a given number 1 .
Given number 370, given sum of sides 10.
Sides of cubes 5 f .#, 5 x> satisfying one condition.
Therefore 30^ + 250= 370, x= 2,
and the cubes are 7 3 , 3 8 , or 343, 27.
2. To find two numbers such that their difference is a given
number, and also the difference of their cubes is a given number.
Difference 6, difference of cubes 504.
Numbers x+ 3, x 3.
Therefore 1 8^ + 54 = 504, ;*r 2 = 25, and x = 5.
The sides of the cubes are 8, 2 and the cubes 512, 8.
3. To multiply one and the same number into a square and
its side respectively so as to make the latter product a cube and
the former product the side of the cube.
Let the square be x*. Its side being x^ let the number
be 8/JF.
Hence the products are 8#, 8, and
(8*)* = 8.
Therefore 2 = 8^, x= \> and the number to be multiplied
is 32.
The square is ^ and its side i.
1 It will be observed that Diophantus chooses, as his given numbers, numbers such
as will make the resulting "pure" quadratic equation give a " rational " value for x. If
the given numbers are 20, 2^, respectively, we assume +#, bx as the sides of the
cubes, and we have
so that jP=*(a &)fa6 m , x is therefore " irrational" unless (alP)l3& is a square. In
Diophantus' hypothesis a is taken as 185, and b as 5, and the condition is satisfied. He
shows therefore incidentally that he knew how to find two numbers a, b such that
(a&}\$b is a square (Loria, Lc scienze esatte nel? antica Grecia^ Libro V. pp. 12930).
A similar remark applies to the next problem, iv. a.
BOOK IV 169
4. To add the same number to a square and its side re
spectively and make them the same 1 [i.e. make the first product a
square of which the second product is the side],
Square x*, with side x.
Let the number added to x* be such as to make a square
say yc\
Therefore yc* + x = side of 4# 2 = 2x, and x = .
The square Is , its side \, and the number .
"3 3
5. To add the same number to a square and its side and make
them the opposite 2 .
Square # 2 , the number a square number of times &
minus x, say 43? x.
Hence $x* #= side of 4^ 2#, and x = f .
The square is ^, its side , and the number ~.
3 O )
6. To add the same square number to a cube and a square
and make them the same.
Let the cube be & and the square any square number of
times x*, say gx\
We want now a square which when added to gx~ makes
a square. Take two factors of 9, say 9 and I, sub
tract i from 9, take half the difference and square.
This gives 1 6.
Therefore i6x* is the square to be added.
Next, x* + 16^= a cube = 81?, say; and x = J^.
The cube is therefore *8_, the square ^, and
343 49
the added square number ^_.
49
1 In this and the following enunciations I have kept closely to the Greek partly
for the purpose of showing Diophantus' mode of expression and partly for the brevity
gained thereby.
In Prop. 4 to "make them the same" means what I have put in brackets ; to " make
them the opposite" in Prop. 5 means to make the first product a side of which the second
product is the square.
3 Nesselmann solves the problem generally, thus (Notes in Zeitschrift fur Math, u.
Physik, xxxvn. (1892), Hist. litt. Abt p. 162).
x*+y=*J(x+y)\ therefore #*f tx*y+y
Solving for y^ we obtain, as one of the solutions,
To make the expression under the radical a square we put +^rje 2 =f mx j ,
m+i
^. T Al * . ,
whence x 2 , y= , 2 ^ . Diophantus solution corresponds to m ~ 2,
1 70 THE ARITHMETICA
7. To add the same square number to a cube and a square
respectively and make them the opposite.
For brevity call the cube (i), the second square (2) and
the added square (3).
Now, since (2) f (3) = a cube, suppose (2) + (3) = (i).
Since a? + &2a& is a square, suppose (i) = (a 2 + 6 2 ),
(3) = 2ab, so that the condition that (i) 4 (3) = square
is satisfied.
But (3) is a square, and, in order that 2ab may be a square,
we put a x, 6 = 2x.
Suppose then (i) = ^f (2xf = s* 2 , (3) = 2, #. 2^ = 4^;
therefore (2) =^ 3 , by subtraction.
But 5# 2 is a cube; therefore x = 5,
and the cube (1)5=125, the square (2) = 25, the
square (3) = 100.
Otherwise thus,
Let (2) + (3)= (i).
Then, since (i) + (3) = a square, we have to find two squares
such that their sum 4 one of them = a square.
Let the first of these squares be # 2 , the second 4.
Therefore 2tf + 4 = square = (2% 2) 2 , say ; thus x = 4,
and the squares are 16, 4,
Assume now (2) = 4jr 2 , (3) = 16^.
Therefore 2O# 3 is a cube, so that #= 20;
the cube (i) is 8000, the square (2) is 1600, and the
added square (3) is 6400.
8. To add the same number to a cube and its side and make
them the same 1 .
Added number x, cube 8^, say. Therefore second sum
= 3^, and this must be the side of 8^h^r.
That is, 8^ + ^= 27^, and 19^ = ^, or 19^= i.
1 Nessdmann (pp. cit. p. 163) gives a more general solution.
We have 3?+y=(x+y)$, whence i
Solving for.?, we find
, jw#a3# .. .
311(1 y~  3+ ^2  " the P oslt ive sign be taken, then, in order that y may
always be positive, m]n must be >3+Vis; Diophantus' solution corresponds to * = 7,
BOOK IV 171
But 19 is not a square. Hence we must find, to replace
it, some square number. Now 193? arises from
27 x*  Sx 8 , where 27 is the cube of 3, and 8 the cube
of 2. And the $x comes from the assumed side 2.x,
by increasing the coefficient by unity.
Thus we must find two consecutive numbers such tJtat their
cubes differ by a square.
Let them bej,ji/f I.
Therefore &  $y 4 I = square = (i 2yf, say, and y = 7.
Going back to the beginning, we assume added number
.#, side of cube = *jx.
The side of the new cube is then Sx, and
Therefore x* = 3^, and x = 
The cube is ^, its side ? , and the added
number .
9. To add the same number to a cube and its side and make
them the opposite 1 .
Suppose the cube, is 8^, its side being 2^ and the added
number is 27^ 2^. (The coefficients 8, 27 are
chosen as cube numbers.)
Therefore 35^ 2^ = side of cube 27^=3^, or 35# 2 = 5
This gives no rational value.
But 35 = 27 + 8, and 5 = 3+2.
Therefore we have to find two cubes such that their sum
has to the sum of their sides the ratio of a square
to a square 2 .
Let sum of sides = any number, 2 say, and side of first
cube = #, so that the side of the other cube is 2 z.
1 Nesselmann (op. cit f p. 163) solves as follows. The equation being x+y=(x?+y} 3 ,
putj/=5f^c 3 , and the equation becomes x ^z a; 3 =2^, or jfi + z*=x+s.
Dividing by x + s, we have jfi  jcz + s 3 = r .
Solving for x, we obtain x= {z =t/v/(4  3 2 ) }.
To make 43 2 a square, equate it to f 22) ; therefore z= 5  2 , so that
x  ^  5^  , and y  z  jc 3 . If the positive sign be taken, Diophantus' solution
*
corresponds to *w=2, = i.
2 It will be observed that here and in the next problem Diophantus makes no use of
the fact that
Cf. note on IV. n below.
172 THE ARITHMETICA
Therefore 8  1 2s + 6sr 2 must be twice a square.
That 'is, 4  &r f 3* 2 = square = (2  4^) 2 , say; * = {, and
the sides are , .
Neglecting the denominator and the factor 2 in the
numerators, we take 5, 8 for the sides.
Starting afresh, we put for the cube 125^ and for the
number to be added 512^5^; we thus get
637^ 5^=8^, and x = \.
The cube is ^25, its side , and the added number^.
343 / OTO
10. To find two cubes the sum of which is equal to the sum
of their sides.
Let the sides be 2;tr, $x.
This gives 35^= $x\ but this equation gives an irrational
result,
We have therefore, as in the last problem, to find two
cubes the sum of which has to the sum of their sides
the ratio of a square to a square 1 .
These are found, as before, to be 5 8 , 8 s .
Assuming then $x t 8x as the sides of the required cubes,
we obtain the equation 637^ = 13^, and x\.
The cubes are gf, gf.
1 Here, as in the last problem, Diophantus could have solved his auxiliary problem
of making (jr 3 +^ 3 )/(jr+^) a square by making sPxy+jP a square in the same way as in
Lemma I. to v. 7 he makes a? + xy+jp a square.
The original problem, however, of solving
can be more directly and generally solved thus. Dividing out by (x +y), we must have
This can be solved by the method shown in the note to the preceding problem.
Alternatively, we may (with Wertheim) put ^#j/+^==(,r+j/) 2 , and at the same
time i = (#+j/).
Thus we have to solve the equations
which give ar==
where k remains undetermined.
Diophantus' solution is obtained by taking the positive sign and putting k= or by
taking the negative sign and putting k~ ~
BOOK IV 173
II, To find two cubes such that their difference is equal to
the difference of their sides.
Assume 2x, $x as the sides.
This gives 193? = x, and.r is irrational.
We have therefore to find two cubes such that their
difference has to the difference of their sides the
ratio of a square to a square 1 . Let them be (#+ i) 3 ,
#3, so that the difference of the sides may be a square,
namely i.
Therefore 3# 2 h 3^ + i = square = (i  2z)\ say, and z = 7.
Starting afresh,, assume 7^, %x as the sides; therefore
169^ .#, and ^r = 1 ^.
The sides of the two cubes are therefore ^, J.
1 Nesselmann (Die Algebra der Griechen, pp. 4478) comments on the fact that
Diophantus makes no use here of the formula (x?jP)l(x~y)=x*+xy+jP, although he
must of course have known it (it is indeed included in Euclid's much more general
summation of a geometrical progression, ix. 35). To solve the auxiliary problem in
IV. ii he had only to solve the equation
x. 2 +xy+jfl=a, square,
which equation he does actually solve in his Lemma i. to v. 7.
The whole problem can be more simply and generally solved thus. We are to have
or
Nesselmann's method of solution (cf. note on iv. 9) gives x= { y^*/ (4 
and hence y=   5, x  7/?/ 9 ( l 3 7 ) ^ Diophantus' solution is obtained by
J 2  2 2 2 v J
putting mi) n = i and taking the lower sign.
Wertheim's method (see note on preceding problem) gives in this case
where k is undetermined.
If we take the negative sign and put k  3, we obtain Diophantus' solution,
Bachet in his notes to iv. 10, 1 1 solves the problems represented by
a? db j/ 3 = m (x y)
subject to the condition that m is either a square or the third part of a square. His method
corresponds to that of Diophantus. He does not divide out by xy 9 and he reduces the
problem to the subsidiary one of finding , t\ such that the ratio of ^rfciT 3 to 17 is the
ratio of a square to a square. His assumptions for the ** sides," , 97, are of the same kind
as those made by Diophantus; in the first problem he assumes x, 6x and in the second
x, x+i. In fact, it being given that (x 3 y s )l(xy)=a, Bachet assumes xy=z and
thus obtains
which equation can easily be solved by Diophantus' method if a is a square or triple of a
square.
Fermat observes that the 8iopi<r/jt,6s of Bachet is incorrect because not general. It
should be added that the number (m) may also be the product of a square number into a
prime number of the form $n + 1, as 7, 13, 19, 37 etc. or into any number which has no
factors except 3 and prime numbers of the form 3+ 1, as 21, 91 etc. " The proof and
the solution are to be obtained by my method."
174
THE ARITHMETICA
12. To find two numbers such that the cube of the greater
4 the less = the cube of the less 4 the greater 1 .
Assume 2^, 3^ for the numbers.
Therefore 27^42^=8^+3^, or 19^ = ^, and x is
irrational.
But 19 is the difference of two cubes, and I the difference
of their sides. Therefore, as in the last problem,
we have to find two cubes such that their difference
has to the difference of their sides the ratio of a
square to a square 2 .
The sides of these cubes are found, as before, to be 7, 8.
Starting afresh, we assume *jx, 8x for the numbers; then
343z 3 4 %x = 5 1 2JF 3 + TX> and x = T V
The numbers are ^, .
13. To find two numbers such that either, or their sum, or
their difference added to unity gives a square.
Take for the first number any square less i ; let it be,
say, gx* + 6x But the second + I = a square , and
first f second 4 I also = a square. Therefore we must
find a square such that the sum of that square and
ojtr 2 4 6x = a square.
Take factors of the difference gx* 4 6x, say gx 4 6, x\
the square of half the difference between these factors
16^424^49.
Therefore, if we put for the second number this expres
sion minus i, or 16^4 24^48, three conditions are
satisfied.
The remaining condition gives difference 4 I  square,
or 7^ 4 18^4 9 = square = (3 3^r) 2 , say.
Therefore x 18, and (3024, 5624) is a solution.
14. To find three square numbers such that their sum is equal
to the sum of their differences.
Sum of differences = (greatest) (middle) 4 (middle) 
(least) 4(greatest) least = twice difference of greatest
and least.
This is .equal to the sum of all three, by hypothesis.
Let the least square be I, the greatest x* 4 2^r4 i ;
1 This problem will be seen to be identical with the preceding problem.
2 See note, p. 173.
BOOK IV 175
therefore twice difference of greatest and least = sum of
the three = 2x? t 4^.
But least f greatest x* f 2x + 2, so that
middle = jr 2 + 2X2.
Hence x* + 2;tr 2 square = (jr  4) 2 , say, and ;tr = f .
The squares are (S, H5, ^ O r (196, 121, 25).
15. To find three numbers such that the sum of any two
multiplied into the third is a given number.
Let (first + second) x third = 35, (second f third) x first
= 27 and (third f first) x second = 32.
' Let the third be x.
Therefore (first + second) 3 s/ar.
Assume first = lojx, second = 25/2; then
250
~
T/tese equations are inconsistent ; but they would not be if
25 10 were equal to 32 27 or 5.
Therefore we have to divide 35 into two parts (to replace
25 and 10) such that their difference is 5. The parts
are 15, 20. [Cf. I. i.]
We take therefore \$\x for the first number, 2O/*r for the
second, and we have
300
~r + 2o=3
Therefore x= 5, and (3, 4, 5) is a solution 1 .
! As Loria says (Le scienze esatte neir antica Grecia^ Libro v. p. 131), this method of
the "false hypothesis," though somewhat indirect, would not be undeserving of a place
in a modern textbook.
Here again, as in iv. i, 2, Diophantus tacitly chooses, for his given numbers, numbers
which will make the resulting " pure" quadratic equation give a rational value for x,
We may put the solution more generally thus. We have to solve the equations
Diophantus takes z for his principal unknown and, writing the third equation in the
form x+y=clz, he assumes x=alz, y=filz> where a, j8 have to be determined. One
equation connecting a, ]8 is a i 13=^. Next, substituting the values of x, y in the first two
equations, we have
0=0,
i 7 6 THE ARITHMETICA
1 6. To find three numbers such that their sum is a square,
while the sum of the square of each added to the next following
number gives a square.
Let the middle number be any number of #>s, say 4^;
we have therefore to find what square + 4^ gives
a square. Split 4^ into two factors, say 2x> 2, and
take the square of half their difference, (x if. This
is the square required.
Thus the first number is x  I.
Again, 16^ + third number = a square. Therefore, if we
subtract i6^ 2 from a square, we shall have the third
number. Take as the side of this square the side of
i6^r 2 , or tyc,plus i.
Therefore third number = (4* + i) 2  i6x* = &r + I.
Now the sum of the three numbers = a square ; therefore
13^= a square = i6oj/ 2 , say 3 .
The numbers are then 13^* I, 527*, 104^+ I.
Lastly, (third) 2 f first = a square.
Therefore 108 loj/ 4 + 22 ij/ 2 = a square,
or io8i6> 2 4 221 = a square = (i04jj>+ i) 2 , say.
17. To find three numbers such that their sum is a square, while
the square on any one minus the next following also gives a square.
The solution is precisely similar to the last.
whence it follows that a  j3=#  b. From this condition and a+/J=, we obtain
a.=(ab + c), p=l(
a / ((gt+cHa + lc)} B_ I ((* + } + $( a +l> c )\
z V 1 2(a + d + c) J' y z~ \/ \ 3 ( a ~3if) I*
Now x y y, is must all be rational, and this is the case if
where/, ^, r are any integers.
This gives a=j>(q+r), b=q(r+p], c=
a fact which can hardly have been unknown to Diophantus, since his values 0=27, =32,
=35 correspond to the values/ = 3, ^=4, r=5 (Loria, loc. cit.).
1 Diophantus uses the same unknown s for y as for x, writing actually /cai ytvercu 6
$ A r 17, literally "and # becomes is^ 2 ."
BOOK IV I?7
The middle number is assumed to be ^.x. The square
which exceeds this by a square is (x + i) y and we
therefore take x + i for the first number.
For the third number we take i6x s (4.x  i) 2 or 8x i.
The sum of the numbers being a square,
13:1: a square = i6gy*, say.
The numbers are then i3^ 2 h i, $2?*, iO4j/ 2 i.
Lastly, since (third) 2 first = a square,
io8i6y22y> 2 = a square,
or io8i6j 2 221 = a square = (1047 i) 2 >say.
1 8. To find two numbers such that the cube of the first added
to the second gives a cube, and the square of the second added to
the first gives a square.
First number x. Therefore second is a cube number
minus x*, say 8 .z 8 .
Therefore x* 1 6x* + 64 4 x = a square = (z 8 + 8) 2 , say,
whence 32^=^, or 32^=1.
This gives an irrational result; x would however be
rational if 32 were a square.
But 32 comes from 4 times 8. We must therefore sub
stitute for 8 in our assumptions a cube which when
multiplied by 4 gives a square. If j 3 is the cube,
4J 8 = a square = i6y s say; whence y = 4.
Thus we must assume x> 64 X* for the numbers.
Therefore x 6 1 282? + 4096+* = a square = (x* + 64) 2 , say ;
whence 256^ =.#, and ^ ^.
The numbers are ,
19. To find three numbers indeterminately 1 such that the
product of any two increased by I is a square.
Take for the product of first and second some square
minus i, say x* + 2#; this satisfies one condition.
Let second =.r, so that first x+ 2.
Now product of second and third f I =a square; let the
1 The expression is 6> r<? &opi<rr<p, which is defined at the end of the problem to mean
in terms of one unknown (and units), so that the conditions of the problem are satisfied
whatever value is given to the unknown.
H. D. 12
I 7 8 THE ARITHMETICA
square be (3^+ i)~, so that product of second and
third = 9# 2 + 6x ;
therefore third = 9^+6.
Also product of third and first f I = a square; therefore
9# 2 + 24% f 13 = a square.
Now, if 13 ze/r # square, and the coefficient of x were
twice the product of the side of this square and tJie
side of the coefficient of x?, the problem would be
solved indeterminately,
But 13 comes from 2.6+1, the 2 in this from twice i,
and the 6 from twice 3. Therefore we want two
coefficients (to replace i, 3) such that the product
of their doubles + i = a square, or four times their
product f i = a square.
Now four times the product of any two numbers plus the
square of their difference gives a square. Thus the
requirement is satisfied by taking as coefficients any
two consecutive numbers, since the square of their
difference is i. [The assumption of two consecutive
numbers for the coefficients simultaneously satisfies
the second of the two requirements indicated in the
italicised sentence above.]
Beginning again, we take (x+ i) 2 i for the product of
first and second and (2x + i) 2 I for the product of
second and third.
Let the second be x, so that first = x+ 2, third = 4^ 4 4.
[Then product of first and third + i = 43? f 1 2x + 9, and
the third condition is satisfied.]
Thus the required indeterminate solution 1 is J
(x + 2, x, 4X + 4)
1 The result obtained by Diophantus really amounts to the more general solution
aPx+ia, x, (a+i)*x+i(a+i).
With this solution should be compared that of Euler (Algebra, Part II. Art. 231).
I. To determine x, y, & so that
xy+i, yz+i, zx+i are all squares.
Suppose zx +i=/ s , yz+i=?*,
so that *=(^i)/*, y=(fi)lz.
Therefore xy + 1 = (riJfe**) + r =a square ,
or (/ 2 i)(? 2 i)+2 2 =a square
= (z  r)*, say ; [Euler has (z + r) 2 ]
whence ' ^~^~ ^^~^,
ir *
where any numbers may be substituted for^, ^, r.
BOOK IV 179
20. To find four numbers such that the product of any two
increased by unity is a square.
For the product of first and second take a square minus i,
say (x+ i) 2  i = x* + 2x.
Let first = #, so that second =^r+ 2.
For example, if rpq+ 1, we shall have
')(?* i
II. But, if whole numbers are required ', we put xy+ 1 *= /^ and assume *=:
We then have
and
These expressions are both squares if ^= =t ip.
Thus a solution is obtained from xy^jP ~ I combined with either
z=zx+y\2p, or z=zx+yip.
We take a certain value for fi 9 split ^  i into two factors, take these factors for the
values of j?, y respectively, and so find 2.
For example, let ^=3, so that^ 2  1 = 8; if we make #=2, j>=4, we find 2= either 12
or o ; and in this case x=i t y=+ 9 a= 12 is the solution.
If we put^=(^+ 1) 2 , we have ^=^+2!; and if we put #=+2, .y=, we have
or o.
The solution is then (+2, , 4^+4), as in Diophantus.
Fermat in his note on this problem shows how to find three numbers satisfying not
only the conditions of the problem but three more also, namely that each of the numbers
shall itself when increased by i give a square, *.. to solve the equations
Solve, he says, the present problem, of Diophantus in such a way that the terms
independent of x in the first and third of the numbers obtained by his method shall be
such as when increased by i give a square. It is easy to find a value for a such that
20+ 1 and 2 (a+ 1)+ 1 are both squares. Fermat takes the value 2<z=, which satisfies
the conditions, and the general expressions for the three numbers sought are therefore
Each of these has, when increased by i, to become a square, that is, we have to solve the
tripleequation
' %
X+l = 2
Fermat does not give the solution ; but it is effected as follows.
Multiplying the third expression by 36 and the first by . 36 (in order that the
absolute terms in the two may be equal), we have to solve
!8o THE ARITHMETICA
For the product of first and third take (2*+ i) 2  I, or
4^+4, the coefficient of x being the number next
following the coefficient (i) taken in the first case,
for the reason shown in the last problem;
thus third number = 4^ + 4.
Similarly take (3*+ i) 2  i, or 9^ + 6^, for the product of
first and fourth; therefore fourth = 9^+6.
And product of third and fourth + I
= (4#+ 4)(9^r + 6) + i = 36^ + 60*425,
which is a square 1 .
(
\7
In order to solve by the method of the tripleequation, we make x + i a square by
putting x=)P + 2jy.
Substitute this value in the other two expressions, and for convenience multiply each
by 1445 this gives
2 =a square]
(85) 2 (y* + w) + (i32) 2 =a square
738 / 45 3 .2. 45 2\
7 '\ 7 ' 7 /
The square of half the difference of the factors equated to the smaller expression gives
whence j>= ^ 4 ^. 3 ; and we find that
05085
It is easily verified that
so that the value of x satisfies the three equations.
The numbers satisfying Fermat's six conditions are then
^ ^3^100604981 50193^44576 , 7$ x 85^48192621
5184 36 171348100' 7 2 39457225* 5184 36 4008004
1 This results from the fact that, if we have three numbers x, y, z such that
xy+i=;(mx+i.)* and xz+i
then ;'+! = {m (m + t)x+ (im + 1)} 2 .
BOOK IV 181
Lastly, product of second and fourth Hi = 9^ f 24^4 13 ;
therefore gx? h 24^ + 13 = square = (3^ 4)2, say ;
which gives ^" = ^5.
All the conditions are now satisfied 1 ,
and 3' H 1? is the solution3 '
1 The remaining condition was : product of second and third + 1 =a square. That this
is satisfied also follows from the general property stated in the last note. In fact
which is a square.
2 With this solution should be compared Euler's solution (Algebra^ Part II. Art. 233)
of the problem of finding x, y t z, v such that the six expressions
xy+a, yz+a, zx+a, xv+a t yv+a t w+a
are all squares. The solution follows the method adopted to solve the corresponding
problem with three unknowns x, y, z only. See note on in. 10 above.
If we begin by supposing xy+a=^ f and take z=x+y + i#, the second and third
expressions become squares (vide note on in. 10, p. 160).
If we further suppose v=x+y ip, the fourth and fifth expressions also become
squares (vide the same note).
Consequently we have only to secure that the sixth expression zv + a shall be a square ;
that is,
x 2 + ixy +y*  4/5 2 + a = a square,
or (since xy + a f] x*  ixy +jP  $a = a square.
Suppose that (x y)* $a=(xy ?) 2 ;
therefore x y = (q* + 30) ( '2$,
Consequently J
If we put / =y + r, we have
iry+** ?
and y=<
from which/, x, and therefore 2, v also, are found in terms of g t r, where q, r may have
any values provided that x 9 y, g, v are all positive.
Euler observed that this method is not suited for finding integral solutions, and,
pursuing the matter farther, he gave the following very elegant solution of Didphantus'
actual problem (the case where <z i) in integers ("Miscellanea analytica" in Com
mentationes ariihmeticae^ II. pp. 456).
Six conditions have to be satisfied. If x t y t 2, z/ are the required numbers, let xm 9
y=n, where m, n are any integers such that mn+ i=P.
Then put z^m + +/, and three conditions are already satisfied, for
xy + 1 = mn + i = / 2 , by hypothesis,
xz+ 1 = m (m + n + it] f i = (/+ m)\
yz+ 1 = n (m + n f il) + 1 (/+ n)\
The three conditions remaining to be satisfied are
xv + i =i/v+ i =a square,
yv + 1 = nv + 1 = a square,
Let us make the continued product of these expressions a square.
182 THE ARITHMETICA
21. To find three numbers in proportion and such that the
difference of any two is a square.
Assume x for the least, x + 4 for the middle (in order
that the difference of middle and least may be a
square), x+ 13 for the greatest (in order that differ
ence of greatest and middle may be a square).
This product will be found to be
1 + 2 (
Let us equate this to \i + (m + n+/)vzP> , in order that the terms in z>, z> 2 as well
as the absolute term may vanish ; therefore

= (mn + 1 ) (m + n f /) + /*
whence
=/(/+)(/+),
and therefore w = 4/ (/+ m) (1+ n),
It is true that we have only made the product of the three expressions mv+ i, nv+ i,
(w + wliOz'+ia square ; but, as the value of v has turned out to be an integral number,
so that all three formulae are prime to one another, we may conclude that each of the
expressions is a square.
The solution is therefore
where
In fact, while three of the conditions have been above shown to be satisfied, we find,
as regards the other three, that
i = 4/w (l+m) (/+) + 1 = (iP + ilm i) 2 ,
i=4?n (l+m) (/+) + 1=(2^+2/~ i) 2 ,
1 = 4/ (m + + it) (/+ m} (/+ ) + 1 = (4^ + ilm + iln  i) 2 .
It is to be observed that / may be either positive or negative.
Ex. Let *=3, w=8, so that /= 5.
If /= +5, the solution is 3, 8, 21, 2080 ; if /=  5, the solution is 3, 8, i, 120.
Fermat shows how to solve this problem, alternatively, by means of the "tripleequation. "
Take three numbers with the required property, e.g. 3, i, 8. Let x be the fourth, and
we have then to satisfy the conditions
Put xz=y*+<iy, so as to make the second expression a square, and then substitute the
value of x in the other two. We have then the doubleequation
1 = z
The difference = 5 (f H iy) = 5^ (y + a) .
We put then (3^+ T )2= g (y* + iy) + 1,
whence j/= 10, and ^=y+ 2j^= 120, which value satisfies the tripleequation.
The four numbers are then 3, i, 8, 120, which solution is identical with one of those
obtained by Euler as above.
BOOK IV 183
If now 13 were a square, we should have an indeterminate
solution satisfying three of the conditions.
We must therefore replace 13 by a square which is the
sum of two squares. Any rational rightangled
triangle will furnish what is wanted, say 3, 4, S ;
we therefore put for the numbers #, ^49, #+25.
The fourth condition gives
jtr(>425) = 0r4 9) 2 , and x = ^.
Thus *, ?tf S* is a solution.
777
22. To find three numbers such that their solid content 1 added
to any one of them gives a square.
Assume continued product x? + 2x, first number i, second
number ^x 4 9, so that two conditions are satisfied.
The third number is then (x* 4 2x)\(A f x 4 9).
This cannot be divided out unless x* : 4^=2^ 19 or,
alternately, # 3 : 2# = ^x : 9 ; but it could be done if 4
were half of 9.
Now 4^* comes from 6^r 2x, and the 6^r in this from
twice 3^:; the 9 comes from 3 2 .
Therefore we have to find a number m to replace 3 such
that 2m  2 s= w 2 : thus #z 2 = 47^  4, whence 2 m = 2.
We put therefore for the second number (4:4 2) a  (# 2 4 2x\
or 2X + 4', the third number is then
(#* 4 2x)l(2x 4 4) or J^r.
Lastly, the third condition requires
x* 4 2X+ \x=*z. square = 4^, say.
Therefore^ f,
and (i, ^, ^J is a solution 3 .
1 6 e a^rwy orepeis, " the solid (number formed) from them "= the continued product
of the three numbers.
8 Observe the solution of a mixed quadratic.
8 Fermat gives a solution which avoids the necessity for the auxiliary problem.
Let the solid content be x*~ 2*, the first number i, and the second number 2.*; two
conditions are thus satisfied.
The third number is now &  ix divided by wr . r , or  x  i ; and the third condition
gives
3 x i=a square.
Now JT must be greater than 2 ; we therefore put
*a*l = (*w)
where m is greater than 2.
1 84 THE ARITHMETIC A
23. To find three numbers such that their solid content minus
any one gives a square 1 .
First numbers, solid content x* + x; therefore product of
' second and third =x+i.
Let the second be I, so that the third is x 4 I.
Jhe two remaining conditions require that
X ~ \ shall both be squares. [Doubleequation.]
X " I J
The difference = x = \ . 2x, say ;
thus (^+J) 2 =^ 2 f^~ i, and;r = Jf.
The numbers are (S, i, &).
24. To divide a given number into two parts such that their
product is a cube minus its side.
Given number 6. First part x\ therefore second = 6 x y
' and 6x x*= a cube minus its side.
Form a cube from a side of the form mx i, say 2x i,
and equate 6x x* to this cube minus its side.
Therefore Sx 3  1 2x* + 4^= 6x x*.
1 A remarkable problem of this kind (in respect of the apparent number of conditions
satisfied) is given by De Billy in the Invention Novum, Part I. paragraph 43 (Oeuvres
d& Format^ III. p. $52) i : To find three numbers , 17, f (, f, >; being in ascending order
of magnitude) such that the following nine expressions may become squares :
(1) ffctf, ( 4 ) iffftS (7) hW,
(2) ufi7f, (5) rEW, (8) tfW,
(3) fw, (6) iirw, (9) i? a ^r
Take Jf, i, i x as the values of ^, 17, f respectively. Then six conditions^ namely,
(!)> (3)> (4)> (6)> (7) ( 8 )> are a11 automatically satisfied.
By conditions (2) and (9) alike,
i  x + y? = a square.
And, by (5) , i  3* + j^ = a square.
Solving this doubleequation in the usual way, we get #=,and the numbers are
8
3 . 5
8' ' 8"
Another solution can be obtained by putting y+\ in place of x in the two expressions,
and so on. Cf. note on m. 18 above.
It would appear from a letter from Fermat to De Billy of 26 Aug. 1659 (Oeuvres, 11.
pp. 4368) that this problem and the above single solution were De Billy's own. De Billy
had supposed that this was the only solution, but Fermat observed that there were any
number, as the above doubleequation has any number of solutions. Fermat gave
10416 4i
.,
**
BOOK IV 185
Now, if the coefficient of x were the same on both sides,
this would reduce to a simple equation, and x would
be rational.
In order that this may be the case, we must put m for 2
in our assumption, where 3 w 7/2=6 (the 6 being
the given number in the original hypothesis). Thus
#2 = 3.
We therefore assume
or
and
The parts are 2$, H$.
c 27 27
25. To divide a given number into three parts such that their
continued product gives a cube the side of which is equal to the
sum of the differences of the parts.
Given number 4.
Since the product is a cube, let it be Sx 5 , the side of
which is 2;ir.
Now (second part)  (first) + (third)  (second) + (third)
(first) = twice difference between third and first
Therefore difference between third and first = half sum of
differences = x.
Let the first be any multiple of x, say 2x\ therefore the
third = $x.
Hence second = SxP/Go? = x\ and, if the second had lain
between the first and third> the problem would have been
solved.
Now the second came from dividing 8 by 2 . 3, and the
2 and 3 are not two numbers at random but con
secutive numbers.
Therefore we have to find two consecutive numbers such
that, when 8 is divided by their product, the quotient
lies between the numbers.
Assume m, m+i] therefore 8/(w 2 + #z) lies between
m and m f I.
o
Therefore   + i > m + I,
m* + m
so that m*  m h 8 > m* + 2 w 2
or 8 > m* + m\
r86 THE ARITHMETICA
I form a cube such that it has aw 8 , m* as terms, that is, the
cube (m + ) 3 , which is greater than m* + m 2 , and I put
therefore w 4 = 2, and ?;* = f .
Assume now for first number \x\ the third is \x> and
the second is \x.
Multiplying throughout by 15, we take 25^, 27^, 40^,
and the product of these numbers is a cube the
side of which is the sum of their differences.
The sum = 2x = 4, by hypothesis.
Therefore x = ^,
and (!' H' ll) are the parts rec l uired
[N,B. The condition S/(m? + m)<m + i is ignored in
the work, and is incidentally satisfied.]
26. To find two numbers such that their product added to
either gives a cube.
Let the first number be of the form m*x, say %x.
Second X s  I. Therefore one condition is satisfied, since
8^  8# + Sx = a cube.
Also 84^8^4^ I =a cube = (2^. i) 3 , say.
Therefore 13^ = 14^, and r = J .
The numbers are y?, ^.
27. To find two numbers such that their product minus either
gives a cube.
Let the first be of the form w$x, say 8#, and the second
i (since 8^ + $>x  8^r= a cube).
r^ 2 i must be a cube, "which is impossible 1 ."
1 Diophantus means that, if we are to get rid of the third power and the absolute
term, we can only put the expression equal to (ixif, which gives a negative and
therefore "impossible *' value for x. But the equation is not really impossible, for we can
get rid of the terms in ^ and j^ by putting
ji=(or Y, whence jrssiliL,
V "/ W52
or we can make the term in x and the absolute term disappear by putting
S^farj^isJ.a: iV whence *=^
Diophantus has actually shown us how to do the former in IV. 25 just preceding.
BOOK IV 187
Accordingly we assume for the first number an expression
of the form m*x+ I, say 8x\ 1, and for the second
number x* (since &z 3 + & & = a cube).
Also S^f^S^r I = a cube = (2x i) s , say.
Therefore # = !,
and the numbers are ^5 ^.
13 IO9
28. To find two numbers such that their product their
sum gives a cube.
Assume the first cube (product + sum) to be 64, and the
second (product sum) to be 8.
Therefore twice sum of numbers = 64 8 = 56, and the
sum = 28, while the product + the sum = 64; therefore
the product = 36.
Therefore we have to find two numbers such that their
sum is 28 and their product 36. If 14+ #, 14 # are
the numbers 1 , we have 196 ^=36, or^ 2 = 160; and,
if 1 60 were a square, we should have a rational
solution.
Now 160 arises from 14* 36, and 14 = ^.28 = ^.56
= i (difference of two cubes); also 36 = J (sum of
the cubes).
Therefore we have to find two cubes such that
(J of their difference) 2  their sum = a square.
Let the sides of the cubes be (z h i), (# i);
therefore J of difference = I \& f , and the square of this
\ the sum of the cubes is
therefore 2 \^ + i%z* h J &  3^  a square,
or 9#* + 6*?+ i 43? 128 a square =(3^ + i dz) 2 , say;
whence 32^ = 36^, and # = f .
The sides of the cubes are therefore ^, ^, and the cubes
512 "5T'
Put now product of numbers f their sum = ^^ , and pro
duct sum = 5^2
Therefore their sum = ^^, and their product =
Now let the first number = x I half sum
and the second = half sum x
therefore ^^ # = %^>
and 262144^ = 250000,
i Cf. i, n '
X 88 THE ARITHMETICA
Therefore #=$},
and gj, g) is a solution.
Otherwise thus.
If any square number is divided into two parts one of
which is its side, the product of the parts added to
their sum gives a cube.
[That is, x(x*x) + x*x + x=* cube.]
Let the square be x^ and be divided into the parts x, x*x.
Then, by the second condition of the problem,
x* x* x=3? zr 2 = a cube (less than^ 3 )=(^) 3 , say.
Therefore S^ 3  i6x 2 =**, so that x= ^,
and f , ^\ is a solution.
29. To find four square numbers such that their sum added to
the sum of their sides makes a given number 1 .
Given number 12.
Now x* f x 4 \ = a square.
Therefore the sum of four squares f the sum of their sides
f i=the sum of four other squares = 13, by hypothesis.
Therefore we have to divide 13 into four squares; then, if
we subtract  from each of their sides, we shall have
the sides of the required squares.
1 On this problem Bachet observes that Diophantus appears to assume, here
and in some problems of Book v., that any number not itself a square is the sum of
two or three or four squares. He adds that he has verified this statement for all
numbers up to 325, but would like to see a scientific proof of the theorem. These
remarks of Sachet's are the occasion for another of Fermat's famous notes : " I have
been the first to discover a most beautiful theorem of the greatest generality, namely this :
Every number is either a triangular number or the sum of two or three triangular
numbers; every number is a square or the sum of two, three, or four squares; every
number is a pentagonal number or the sum of two, three, four or five pentagonal
numbers; and so on ad infinitum^ for hexagons, heptagons and any polygons whatever,
the enunciation of this general and wonderful theorem being varied according to the
number of the angles. The proof of it, which depends on many various and abstruse
mysteries of numbers, I cannot give here ; for I have decided to devote a separate and
complete work to this matter and thereby to advance arithmetic in this region of inquiry
to an extraordinary extent beyond its ancient and known limits."
Unfortunately the promised separate work did not appear. The theorem so far as it
relates to squares was first proved by Lagrange (Nouv. M&moires de FAcad. de Berlin,
anne"e 1770, Berlin 1772, pp. 123133; Oeuvres, ill. pp. 189201), who followed up
results obtained by Euler. Cf. also Legendre, Zahlentheorie, tr. Maser, I. pp. 212 sqq.
Lagrange's proof is set out as shortly as possible in Wertheim's Diophantus, pp. 324330.
The theorem of Fermat in all its generality was proved by Cauchy (Oeuvres, n e serie,
Vol. vi. pp. 320353) ; cf. Legendre, Zvhlentheorie, tr. Maser, II. pp. 332 sqq.
BOOK IV 189
Now i3
and the sides of the required squares are i, $, g, 1$,
the squares themselves being ^, A, g, gj.
30. To find four squares such that their sum minus the sum of
their sides is a given number.
Given number 4.
Now x* x 4 J = a square.
Therefore (the sum of four squares) (sum of their sides)
f i =the sum of four other squares = 5, by hypothesis.
Divide 5 into four squares, as ^, f , ff , ff .
The sides of these squares///^ in each case are the sides
of the required squares.
Therefore sides of required squares are j, {^, fj, {$,
and the squares themselves , ^52 441 5??.
^ zoo' 100* ioo' 100
31. To divide unity into two parts such that, if given numbers
are added to them respectively, the product of the two sums gives
a square.
Let 3, 5 be the numbers to be added ; x, I x the parts of I.
Therefore (x+ 3)(6 x) = 18 f 3^^ 2 =a square = 4^, say;
thus 1 8 + 3# = 5^, which does not give a rational result*
Now 5 comes from a square + 1 ; and, in order that the
equation may have a rational solution, we must sub
stitute for the square taken (4) a square such that
(the square + i) . 18 + (f J = a square.
Put (m* + I ) 1 8 + 2 J = a square,
or J2m* h 8 1 = a square = (Bm + 9) 2 , say,
and m 18, ^ = 324.
Hence we must put
(x+ 3) (6 *) = 18 + 3^r^ 2 = 324**.
Therefore 1 325^ 3^18 = 0,
and lj\ is a solution.
Otherwise thus.
The numbers to be added being 3, 5, assume the first of
the two parts to be x 3 ; the second is then 4 x.
Therefore x (9  x) = a square = 4^, say,
and l **=
5/ / cannot take 3 /r0w , and x must be > 3 and < 4.
1 Observe the solution of a mixed quadratic equation.
1 9 o THE ARITHMETICA
Now the value of x comes from 9/(a square + i), and, since
x> 3, this squared 1 should be<3, so that the square
must be less than 2 ; but, since x < 4, the square + i
must be > f , so that the square must be > f .
Therefore I must find a square lying between and 2, or
between fj and ^.
^Q or f satisfies the condition.
Put now x (g x) =
therefore
is a solution.
/ 29^ is
V4 1 4 1 /
32. To divide a given number into three parts such that
the product of the first and second the third gives a square.
Given number 6.
Suppose third part =#, second = any number less than 6,
say 2; therefore first part = 4 x
The two remaining conditions require that 8 2.x x = a
square,
8 x )
or c f are k ot h sc l uares  [Doubleequation.]
This does not give a rational result ("is not rational"), since
tfie ratio of tlie coefficients ofx is not a ratio of a square
to a square.
But the coefficients of x are 2 i and 2 4 1 ; therefore we
must find a number^ to replace 2 such that
(y + l )l(y ~~ J ) = rat i f square to square f , say.
Therefore y + I = 4 j  4, and ^ = f .
Put now second part = f ; therefore first =  x.
Therefore ^ f x x = a square.
That is, ,. I are both squares,
26024^] , .
or ^ r are both squares.
65  H
The difference = 195 = 1 5 . 1 3; '
we put therefore J (i 5 1 3)* = 65 24^, and x = f .
Therefore the required parts are (3, 5, 5\\
\ J O O/
i Fermat observes: ( *The following is an easier method of solution. Divide the
number 6 into two in any manner, e.g. into 5 and i. Divide their product less i, that is
4, by 6, the given number: the result is  . Subtracting this first from 5 and then from i,
3
BOOK IV 191
33. To find two numbers such that the first with a fraction
of the second is to the remainder of the second in a given ratio,
and also the second with the same fraction of the first is to the
remainder of the first in a given ratio.
Let the first with the fraction of the second = 3 times the
remainder of the second, and the second with the
same fraction of the first = 5 times the remainder of
the first.
[The fraction may be either an aliquot part or not, TO
avro fiepo? or ra avra pepy as Diophantus says,
following the ordinary definition of those terms ("the
same part" or "the same farts"): cf. Euclid VII.
Deff. 3, 4]
Let the second =x+ I, and let the part of it received by
the first = i ;
therefore the first =3^1 (since $x I + i = 3^).
Since the second plus the fraction of the first = 5 times
the remainder of the first,
the second + the first = 6 times the remainder of the first
And first i second = 4^ ; therefore remainder of first
= !#, and hence the second receives from the first
3#  i  far or fri.
We have therefore to secure that \x I is the same
fraction of ye i that i is of x+ i.
This requires that (%x i) (x + i) = (3^ i). I ;
therefore $x? + f# i 3^ i, and #=f.
Accordingly the numbers are f , ^; and I is ^ of the
second.
we have as remainders and  , which are the first two parts of the number to be
o 5
divided; the third is therefore ."
That is, if , 17, f be the required parts of the number a, Fermat divides a into two
parts x, a  x and then puts
, . ,.. x
whence ^= a  ( + ij) =
The three general expressions in x satisfy the conditions, and x may be given any
value <.a.
192 THE ARITHMETICA
Multiply by 7 and the numbers are 8, 12, and the fraction
is ^ ; but 8 is not divisible by 12: so multiply by 3,
and (24, 36) is a solution.
Lemma to the next problem.
To find two numbers indeterminately such that their product
together with their sum is a given number.
Given number 8.
Assume the first number to be.^r, the second 3.
Therefore 3^+^+3 = given number = 8; # = f, and the
numbers are f , 3.
Now f arises from (8 3)/(3 + i), where 3 is the assumed
second number.
We may accordingly put for the second number (instead
of 3) any (undetermined) number whatever 1 ; then,
substituting this for 3 in the above expression, we
have the corresponding first number.
For example, we may take;? I for the second number;
the first is then 9 x divided by x, or  I.
34. To find three numbers such that the product of any two
together with the sum of those two makes a given number 2 .
1 The Greek phrase is &v apa T<W/Z> rbv jP v s olovSfjirore (oiovdtfiroTc s in Lemma
to iv. 36), "If we make the second" (literally "put the second at"} " any ^whatever."
But the s is not here, as it is in the Lemma to iv. 36, the actual x of the problem, for
Diophantus goes on to say "E.g. let the second be x i." In the Lemma to iv. 34 the
corresponding expression is "any quantity whatever" (ooouS^Trore without s). The
present Lemma amounts to saying that, if xy+x+y~a t then x(a ~y}\ (y + 1 ).
 3 This determinate set of equations can of course be solved, with our notation, by
a simple substitution.
The equations yz +y +z=a\
are equivalent to
where
The solution is {+ 1 = ./ j WTWrin ^
In order that the result may.be rational, it is only necessary that (a + 1) (b + 1) (c + 1)
should be a square; it is not necessary that each of the expressions #+r, 3+r, ^(i
should be a square, as Diophantus says.
BOOK IV I93
Necessary condition. Each number must be I less than some
square \
Let (product 4 sum) of first and second = 8.
second and third =15.
third and first = 24.
Bythefirst equation,if we divide(8second)by(secondf i),
we have the first number.
Let the second number be x i.
Therefore the first= ?==  i.
x x
Similarly the third number = i.
The third equation remains, which gives
144
=1=24, and ^ = 4.
The numbers are 33, 2 ,
or, when reduced to a common denominator, ^, g, 34?
Leinma to the following problem.
To find two numbers indeterminately such that their product
minus their sum is a given number.
Given number 8.
First number x> second 3, suppose; therefore
(product) (sum) = 3^~^r 3 = 2^r 3 = 8, and ^=5^.
The first number is therefore sj, the second 3.
But Si comes from (8 4 3)/(3  i), and we may put for 3
any number whatever.
E.g. put the second number =x+ i ; the first is then #+9
divided by x, or I 4.
35. To find three numbers such that the product of any two
minus the sum of those two is a given number 2 .
Necessary condition. Each of the given numbers must be i less
than some square 2 .
Let (product sum) of first and second = 8.
second and third =15.
third and first = 24.
1 See last paragraph of preceding note.
3 The notes to iv. 34 above apply, mutatis mutandis^ to this problem as well.
H. D. X 3
i 9 4 THE ARITHMETIC A
By the first equation, if we divide (8 f second) by
(second i ), we have the first number.
Assuming x + I for the second number, we have I f 
x
for the first.
Similarly I i  is the third number, and two conditions
are satisfied.
The third gives ~  I = 24, and x= J.
The numbers are , ?, ??
12 5 ^
or, with a common denominator,^, ~^ t *_.
Lemma to the following problem.
To find two numbers indeterminately such that their product
has to their sum a given ratio.
Let the given ratio be 3:1, the first number x, the
second 5.
Therefore 5^=3(5+^), # = 7i; and the numbers are
7i 5
But J\ arises from 1 5 divided by 2, while the 1 5 is the
second number multiplied by the given ratio, and
the 2 is the excess of the second number over the
ratio.
Putting therefore x (instead of 5) for the second number,
we have, for the first number, $x divided by x  3.
The numbers are therefore ^x\(x 3), x.
36. To find three numbers such that the product of any two
bears to the sum of those two a given ratio.
Let product of first and second be 3 times their sum.
second and third be 4 times their surn.
third and first be 5 times their sum.
Let second number be x\ the first is therefore yel(x 3),
by the Lemma, and similarly the third is 4z/(# 4).
Lastly
J
.
x 3 #4 \jir 3 #
or i2or 2 = 3S^ 2  I20JF.
Therefore x = *$$>
and the numbers are 3$?, ^, &.
51 23 Zo
BOOK IV 195
37. To find three numbers such that the product of any two
has to the sum of the three a given ratio 1 .
Let product of first and second = 3 times sum of the three,
of second and third =4
of third and first = 5
First seek three numbers such that the product of any two
has to an arbitrary number (say 5) the given ratio.
Then product of first and second = 15; and, if x be the
second, the first is IS/JF.
The product of second and third = 20; therefore third
= 20/ar.
It follows that 20. lS/r 2 = 25.
And, if the ratio of 20 . 15 to 25 were that of a square to a
square, the problem would be solved.
Now 15 = 3.5, an< 3 20 is 4. 5, the 3 and 4 being fixed by
the original hypothesis, but 5 being an arbitrary
number.
We must therefore find a number m (to replace 5) such
that i2#/ 2 /5;# = ratio of a square to a square.
Thus 1 2m* . 5 m = 6o;;/ J = a square = QOO/;/ 2 , say ; and m 1 5 .
Let then the sum of the three numbers be 15.
Product of first and second is therefore 45, and first = 45/ar.
Similarly third = 6o/r.
Therefore 45 . 6o/x* = 75, and x = 6.
Therefore the numbers are 7^, 6, 10, and the sum of
these = 23^.
Now, if this sum were 1 5 instead, the problem would be
solved.
1 Loria (op. cit. p. 130) quotes this problem as an instance of Diophantus* ingenious
choice of unknowns. Here the equations are, with our notation,
and Diophantus chooses as his principal unknown the sum of the three numbers,
x+y + z~?u 9 say.
We may then write x=^cw\y^ z~aw\yi so that zjc=ac i w 2 ly s =&w, and y^u^
Putting w=f ?, we have
from which, by eliminating x, y, a, we obtain
Hence x=*(bc + ca+ab)la, y^(bc+ca+ab)lb, z=(bc+ca+at>)\c.
132
196 THE ARITHMETICA
Put therefore for the sum of the three numbers I $x*, and
for the numbers themselves J\x, 6x t
Therefore 23^= I5^ 2 , so that * = $,
and 32*, ?^, 47 is a solution.
30 30 30
38. To find three numbers such that their sum multiplied
into the first gives a triangular number, their sum multiplied into
the second a square, and their sum multiplied into the third a cube.
Let the sum be x~, and let the numbers be m/x z , n/x*, pjx*,
where m, n,p are a triangular number, a square and
a cube respectively ;
say first number = 6/x*, second 4/x*, third 8/z*.
But the sum is x*\ therefore iS/x 2 = x? y or 18 =x*.
Therefore we must replace 18 by some fourth power.
JBut 1 8 = sum of a triangular number, a square and a cube.
Let & be the required fourth power, which must therefore be
the sum of a triangular number, a square and a cube.
Let the square be xt  2x* + i ;
therefore the triangular number + the cube= 2x* I.
Let the cube be 8; therefore the triangular number is
2x* g.
But 8 times a triangular number 4 I = a square \ therefore
i6^ 3 ~ 71 = a square (4^ i) 3 , say; thus ^ = 9, the
triangular number is 153, the square 6400 and the
cube 8.
Assume then as the numbers i53/# 2 , 640O/x\
Therefore 6^61 jx 2 x*, or ^ = 6561, and x=g.
1 The procedure may be shown more generally thus.
Let f , ^ f be the required numbers ; suppose
and
It follows that
i
Suppose now that fi^jpz* [Diophantus and Bachet assume z= i].
Then
Eight times the left hand side plus i gives a square (by the property of triangular
numbers) ; that is,
2 = i6*W &r*~ 873+ i =a square
= (<*), say,
BOOK IV i 97
39. To find three numbers such that the difference of the
greatest and the middle has to the difference of the middle and the
least a given ratio, and also the sum of any two is a square.
Ratio 3:1. Since middle number f least = square, let the
square be 4.
Therefore middle > 2 ; let it be JT + 2, so that least = 2  x.
Therefore difference of greatest and middle = 6x, whence
the greatest = 7^+2.
Therefore , . f are Dot ^ squares. [Doubleequation^
Take the difference 2x, split it into factors, say \x^ 4, and
proceed by the rule ; therefore #= 1 12.
But I cannot take 1 12 from 2; therefore x must be found
to be < 2, so that 6x + 4 < 16,
Thus there are to be three squares 8^+4, 6^44 and 4
(the 4 arising from 2 . 2), and the difference of the
greatest and middle is of the difference of the
middle and least
We have therefore tofind three squares having this property
and such that the least = 4 and the middle < 16.
Let side of middle square be $ f 2 ; therefore excess of
middle over least = z z + 42, whence excess of greatest
over middle = J# 3 f !#, and therefore the greatest
This must be a square; therefore, multiplying by 9, we have
1 22* f 48^ f 36 a square,
,
whence
 ~  .
OKS
But    must be integral, and therefore a integral, so that (^zxA i) must be
integral; that is, 8 ** + 87 *J^' l " 1 ^must be integral.
Bachet assumes that it is necessary, with Diophantus, to take =i, observing that
trial will show that the problem can hardly be solved otherwise. On this Fermat remarks
that Bachet*s trial had not been carried far enough. We may, he says, put for y 3 any
cube, for instance, with side of the form 3 + i. Suppose, for example, we take 7 3 .
Then \z being i] we have to make
2^ 344 a triangle,
and therefore 16^ 2751 a square, and we may take, if we please, 4* 3 as the side of
this square [so that k is in this case 3].
By varying the cubes we may use an unlimited variety of odd numbers, besides 3,
as values for k which will satisfy the required condition.
Loria (op. cit* p. 138) points out that the problem could have been more simply
solved by substituting x fcr x 2 and z for a 2 in the above assumptions. The ultimate
expression to be made a square would then have been T6sy?8 2 ~87 3 + i, and we could
have equated this to X 2 , thus finding jr.
i g8 THE ARITHMETIC A
or 3s 2 + 1 25 4 9 = a square = (MS  3) 2 , say.
It follows that .?= (6m + i2)/(; a  3), which must be < 2.
Therefore 6w + 12 < 2?^ 2  6, or 2;;/ 2 > 6;# f 18.
"When we solve such an equation 1 , we multiply half the
coefficient of x into itself this gives 9 then multiply
the coefficient of x into the units 2 .18 = 36 add
this last number to the 9, making 45, and take the
side [square root] of 45, which is not less than 7;
add half the coefficient of x making a number not
less than 10 and divide the result by the coefficient
of x*\ the result is not less than 5."
[3 2 +18.2 = 45, and \/45 +} is not less than f + .]
We may therefore put m = f + f , or 5, and we thus have
Therefore # = f , and the side of the middle square is
ff , the square itself being ^j^.
Turning to the original problem, we put 6x+ 4 ^^f, and
x = J^6^ w hich is less than 2.
The greatest of the required numbers = Jx + 2 = ^22,
the middle = ;r + 2 = ^2,
and the least = 2 ^ ^.
The denominator not being a square, we can make it
a square by dividing out by 6; the result is
121 ' 121 ' 121 '
or again, to avoid the \ in the numerators, we may
multiply numerators, and denominators, by 4; thus
1 I have quoted Diophantus' exact words here, with the few added by Tannery,
** making a number not less than 10... coefficient of jc 3 ," in order to show the precise
rule by which Diophantus solved a complete quadratic.
When he says ^45 is not less than 7, Diophantus is not seeking exact limits. Since
^/45 is between 6 and 7 we cannot take a smaller integral value than 7 in order to
satisfy the conditions of the problem (cf. p. 65 above).
2 A note in the Inventum Novum (Partu, paragraph 26) remarks upon the prolix and
involved character of Diophantus' solution and gives a shorter alternative. The problem
is to solve
and w = 3, say)
BOOK IV
199
40. To find three numbers such that the difference of the
squares of the greatest and the middle numbers has to the differ
ence of the middle and the least a given ratio, and also the sum of
each pair is a square.
Ratio 3:1.
Let greatest f middle number = the square i6jir 2 ; therefore
greatest >8.ar 2 : let it be 8.ar 2 + 2 ; hence middle=S# 2 2.
And, since greatest + middle > greatest + least,
i6^ 2 > (greatest f least) > 8x* ;
let greatest + least = c^tr 2 , say; therefore least = ^ 2.
Now difference of squares of greatest and middle = 64^,
and difference of middle and least = /jr 2 .
But 64 is not equal to 3 . 7 or 21.
Now 64 comes from 32 . 2; therefore we must find a
number m (in place of 2) such that 327/2 = 21.
Therefore m = fj.
Assume now greatest number = &# 2 f f, middle = 8# 2 f ,
least = & f.
[And difference of squares of greatest and middle
The only condition left is: middle + least = square; that is,
9^ ff = a square = (y 6) 3 , say.
Therefore
Take an arbitrary square number, say 4, for the sum of ^ ; suppose 2 + x= 17, 2  JT
so that 17  f = 2Jc ; therefore   1\ = 3 (?  fl = ^JT , whence = 2 + 70:.
The last two conditions require that
4 " f "^l shall be squares.
4+64 H
Replace Jt: by ^ jr +  y This will make 4 + 6x a square. It remains that
^+^= a square
3 3
say.
and >, that ^
^ , ., r 1362 402 82
The numbers are therefore TTr' 77; ' VTT
121 131 121
200 THE ARITHMETICA
BOOK V
r. To find three numbers in geometrical progression such that
each of them minus a given number gives a square.
Given number 12.
Find a square which exceeds 12 by a square. "This is
easy [II. 10]; 42 \ is such a number."
Let the first number be 42 J, the third x*\ therefore the
middle number = 6^x.
3? 12)
Therefore ^, [ are both squares;
6j^I2j ^
their difference = ^ 6\x = x(x 6J) ; half the difference
of the factors multiplied into itself^; therefore,
putting 6\x 12=:^, we have ^^ffj,
and (4**, 2 ff , *) is a solution.
2. To find three numbers in geometrical progression such that
each of them when added to a given number gives a square.
Given number 20.
Take a square which when added to 20 gives a square,
say 1 6.
Put for one of the extremes 16, and for the others 3 , so
that the middle term = 4^.
. c o]
1 herefore x are both squares.
4^420] ^
Their difference is ^ ^x = # (^r 4), and the usual method
gives 4^ + 20=4, which is absurd^ because the 4
ought to be some number greater than 20.
But the 4 = I (16), while the 16 is a square which when
added to 20 makes a square; therefore, to replace 16,
we must find some square greater than 4.20 and
such that when increased by 20 it makes a square.
Now 8 1 > 80; therefore, putting (w + 9) 2 for the required
square, we have
(m + 9) 2 f 20 = square = (m 1 1 ) 2 , say ;
therefore m = , and the square = (9i) 2 = 90 J.
BOOK V 201
Assume now for the numbers 90^, g\x, jp, and we have
20
, .
both squares.
4
The difference =#(# 9 J), and we put Qi
Therefore JIT = ^, and
solution.
3. Given one number, to find three others such that any one of
them, or the product of any two of them, when added to the given
number, gives a square.
Given number 5.
" We have it in the Porisms that if, of two numbers, each,
as well as their product, when added to one and the
same given number, severally make squares, the
two numbers are obtained from the squares of con
secutive numbers 1 ."
Take then the squares (x+ 3), (#+4), and, subtracting
the given number 5 from each, put for the first
number x z +6x+4., and for the second tf+Sx+n,
and let the third 2 be twice their sum minus I, or
29.
1 On this Porism, see pp. 99, 100 ante.
2 The Porism states that, if a be the given number, the numbers x*a,
satisfy the conditions.
In fact, their product + = {#(* +i)} a ~ a (
Diophantus here adds, without explanation, that, if AT", Y denote the above two numbers,
we should assume for the third required number Z~ 2 (X+ Y) i. We want three numbers
such that any two satisfy the same conditions as X, Y. Diophantus takes for the third
Zi(X\ y) i because, as is easily seen, with this assumption two out of the three
additional conditions are thereby satisfied.
For Z=i(X+ Y)i=2(iJc i 'i ix hi) 43 i
therefore
while
= (x H) 2 (ix Hi) 2  a (Sx 3 hi ix+ 4) + 4 <z 3
The only condition remaining is then
Z+a=z square,
or (ix + 1) 2  $a = a square =(ix ) 2 , say,
and x is found.
Cf. pp. ioo, 104 above.
202 THE ARITHMETICA
Therefore 4** + 2%x + 34 = a square = (2x  6) 2 , say.
Hence ,*> and (*, ^ 6 ) is a solution*.
4. Given one number, to find three others such that any one
of them, or the product of any two, minus the given number gives
a square.
Given number 6.
Take two consecutive squares .#*, x* + 2# + I .
Adding 6 to each, we assume for the first number & + 6,
and for the second & + 2^r + 7.
For the third 2 we take twice the sum of the first and
second minus i, or 4^r 2 f 4^f 25.
Therefore third minus 6 43? f 4^4 I9=square =(2^: 6) 2 ,
say.
Therefore # = ,
[The same Porism is assumed as in the preceding problem
but with a minus instead of a, plus. Cf. p. 99 above.]
5. To find three squares such that the product of any two
added t"o the sum of those two, or to the remaining square, gives
a square.
" We have it m the Porisms" that, if the squares on any two
consecutive numbers be taken, and a third number
be also taken which exceeds twice the sum of the
squares by 2, we have three numbers such that the
product of any two added to those two or to the
remaining number gives a square 3 .
1 Diophantus having solved the problem of finding three numbers , 17, satisfying
the six equations
, ,
Fermat observes that we can deduce the solution of the problem
To find four numbers such that the product of any pair added to a given number
produces a square.
Taking three numbers, as found by Diophantus, satisfying the above six conditions,
we take x+ 1 as the fourth number. We then have three conditions which remain to be
satisfied. These give a " tripleequation " to be solved by Fermat's method.
2 Diophantus makes this assumption for the same reason as in the last problem, v. 3.
The second note on p. 201 covers this case if we substitute  a for a throughout.
8 On this Porism, see pp. roor ante,
BOOK V 203
Assume as the first square x z + 2x+ i, and as the second
x? + ^x j 4, so that third number = 4^ M 2 # f 1 2.
Therefore x" + $x f 3 = a square = (#  3) 2 , say, and x = $.
Therefore (, &, ^) is asolution.
6. To find three numbers such that each minus 2 gives a
square, and the product of any two minus the sum of those two,
or minus the remaining number, gives a square.
Add 2 to each of three numbers found as in the Porism
quoted in the preceding problem.
Let the numbers so obtained be
All the conditions are now satisfied 1 , except one, which
gives
4^ f 4# i 6 2 = a square.
Divide by 4, and 3? + x + I = a square = (^r 2)', say.
Therefore ^ = f ,
Lemma / #? /f/^ following problem.
To find two numbers such that their product added to the
squares of both gives a square.
Suppose first numbers, second any number (;/z), say i.
Therefore x. i +x? + I ~x*+x+ i = a square = (x 2) 3 , say.
Thus^r=f, and
($, i\ is a solution, or (3, 5).
Lemma II to the following problem.
To find three rightangled triangles (i.e. three rightangled
triangles in rational numbers*) which have equal areas.
We must first find two numbers such that their product
hthe sum of their squares = a square, e.g. 3, 5, as in
the preceding problem.
1 The numbers are # 3 + 2, (x + 1) 3 + 2,2 [ j: 2 + (jf t 1) 3 + 1 } + * ; and if X, Y, Z denote
these numbers respectively, it is easily verified that
xz(x+z) =(3^2+^+3)2, ^r^ y =
and KZ
3 All Diophantus' rightangled triangles must be understood to be rightangled
triangles with sides expressible in rationa.1 numbers.. In future I shall say " rightangled
triangle " simply, for brevity,
2o 4 THE ARITHMETICA
Now form rightangled triangles from the pairs of
numbers *
(7,3), (7,5), (7,3 + 5)
[i.e. the rightangled triangles (7 2 + 3 2 > 7 2 ~ 3 a > 2.7.3), etc.].
The triangles are (40, 42, 58), (24, 70, 74), (15, 112, 113),
the area of each being 840.
1 Diophantus here tacitly assumes that, if ab + a? + $ = <?, and rightangled triangles
be formed from (^, a), (r, b) and (r, a + b) respectively, their areas are equal. The
areas are of course (ra?)ca, (c z fi)cb and {{a + fy' 2 <?} (a + b) c, and it is easy to
see that each abc(a j rb).
Nesselmann suggests that Diophantus discovered the property as follows. Let the
triangles formed from (n, m}, (g, m), (r, m) have their areas equal ; therefore
n (m 2  2 ) = q (m* f}~r (r*  m 2 ).
It follows, first, since m*n  n*=m < *g  3 ,
that w 2 = (n 3  <f}l(n  q) = n 2 + nq + q z 
Again, given (q, m, ), to find r.
We have q (m* q^r^ m 2 ),
and w 2  q* = tfi + nq^ from above ;
therefore q ( 3 4 nq) = r(r* w 2  nq  q\
or q (n 2 + nr) + g f2 (n + r)=^r(r 2 # 2 ).
Dividing by r + #, we have qn 4 q* = r 2  m ;
therefore (^ + r) w = r 2  g*>
and ^ss^lw.
Feraiat observes that, given any rational rightangled triangle, say z, b 9 d> where s
is the hypotenuse, it is possible to find an infinite number of other rational rightangled
triangles having the same area. Form a rightangled triangle from z 2 , 2&f ; this gives
the triangle j& 4 f4^ 3 *f 2 , s 4 ^*/ 2 , ypbd. Divide each of these sides by iz (fid z ) t
(j \
bd\ as the original triangle.
Trying this method with Diophantus' first triangle (40, 42, 58), we obtain as the new
triangle 1412881 1:412880 1681
1189 ' 1189 ' iTSp*
The method gives ( , , J as a rightangled triangle with area equal
to that of (3, 4, 5).
Another method of finding other rational rightangled triangles having the same area as
a given rightangled triangle is explained in the Inventum Novum, Part I, paragraph 38
(CEitvres de Fermat, in. p. 348).
Let the given triangle be 3, 4, 5, so that it is required to find a new rational right
angled triangle with area 6.
Let 3, jf 44 be the perpendicular sides ; therefore
the square of the hypotenuse =x z + $x +25= a square.
Again, the area is x+6 ; and, as this is to be 6, it must be six times a certain square ;
that is, ~*+6 divided by 6 must be a square, and this again multiplied by 36 must
be a square; therefore
gx + 36= a square.
Accordingly we have to solve the doubleequation
9^+36=3
BOOK V 205
7. To find three numbers such that the square of any one
the sum of the three gives a square.
Since, in a rightangled triangle,
(hypotenuse) 2 twice product of perps. = a square,
we make the three required numbers hypotenuses and the
sum of the three four times the area.
Therefore we must find three rightangled triangles having
the same area, e.g., as in the preceding problem,
(40,42, 58), (24, 70, 74), (15, 112, 113).
Reverting to the substantive problem, we put for the
numbers 5&F, 742, 113^; their sum 245 # = four times
the area of any one of the triangles = 3360^.
Therefore #=^,
} s a solution,
and ( 4
?
Lemma to the following problem.
Given three squares, it is possible to find three numbers such
that the products of the three pairs shall be respectively equal to
those squares.
. . 672560x5
This gives x= 3? ,
& 2405601
whence x + 4 =
* 2405601
The triangle is thus found to *be
2896804 7776485
** 2405601 * 2405601 *
The area is 6 times a certain square, namely the root of which is .
2405001 1551
Dividing each of the above sides by ~ , we obtain a triangle with area 6, namely
4653 34PJ 7776485
851* 1551' i3 I 99 I *
Another solution of the doubleequation, or=   giving jr44 = , leads to
3000 40
the same triangle ( , , i^j as that obtained by Fermafs rule (see above).
^ \io 7 70 /
The method of the Inventum Novum has a feature in common with the procedure in
the ancient Greek problem reproduced and commented on by Heiberg and Zeuthen
(Bibliotheca Mathematica, viiT 3 , 1907/8, p. 122), where it is required to find a rational
rightangled triangle, having given the area, 5 feet, and where the 5 is multiplied by a
square number containing 6 as a factor and such that the product " can form the area of
a rightangled triangle." 36 is taken and the area becomes 180, which is the area of
(9, 40, 41). The sides of the latter triangle are then divided by 6, and we have the
required triangle (cf. p. 119, ante}.
206 THE ARITHMETICA
Squares 4, 9, 16.
First number #, so that the others are 4/ar, gjx\ and 36/^=16.
Therefore #=f, and the numbers are (if, 2, 6).
We observe that;t'= , where 6 is the product of 2 and 3,
and 4 is the side of 16.
Hence the following rule. Take the product of two sides
(2, 3), divide by the side of the third square 4 [the
result is the first number]; divide 4, 9 respectively
by the result, and we have the second and third
numbers.
8. To find three numbers such that the product of any two +
the sum of the three gives a square.
As in Lemma II to the 7th problem, we find three right
angled triangles with equal areas ; the squares of
their hypotenuses are 3364, 5476, 12769.
Now find, as in the last Lemma, three numbers such that
the products of the three pairs are equal to these
squares respectively, which we take because each
4 . (area) or 3360 gives a square ; the three numbers
then are
W> *W X Pfttl 1 * Tannery],
^&x PifH** Tannery].
It remains that the sum of the three = 3360^.
Therefore SfffffS 4 * [ 1 \V&W 4 * Tannery] = 3360*=.
Therefore * = $^$$& [tttMftfflfr or jf}Mh Tannery],
and the numbers are *
9. To divide unity into two parts such that, if the same given
number be added to either part, the result will be a square.
Necessary condition. The given number must not be odd and
the double of it f I must not be divisible by any prime number
which, when increased by I, is divisible by 4 \_Le. any prime number
of the form 472 i] 1 .
Given number 6. Therefore 13 must be divided into two
squares each of which > 6. If then we divide 13 into
two squares the difference of which < i, we solve the
problem.
1 For a discussion of the text of this condition see pp. 1078, ante.
BOOK V 207
Take half of 13 or 6, and we have to add to 6 a small
fraction which will make it a square,
or, multiplying by 4, we have to make  ^ 1 26 a square,
i.e. 2&x 2 + i = a square = (5.3:4 i) 2 , say, whence #= 10.
That is, in order to make 26 a square, we must add y^j, or,
to make 6 a square, we must add ^j, and
Therefore we must divide 13 into two squares such that their
sides may be as nearly as possible equal to f. [This
is the 7ra/H<roT?;T09 dycoyrj described above, pp. 95~8.]
Now i3 = 2 2 f3 2 . Therefore we seek two numbers such
that 3 minus the first f, so that the first = ^, and
2 plus the second = f, so that the second = .
We write accordingly (n^r + 2) 2 , (3 gxf for the required
squares [substituting x for ^j].
The sum =202^ IQX+ 13 = 13.
Therefore #= ^ and the sides are fjg, f^f .
Subtracting 6 from the squares of each, we have, as the
parts of unity,
443 5353
I020I' I020I*
10. To divide unity into two parts such that, if we add different
given numbers to each, the results will be squares.
Let the numbers 1 be 2, 6 and let them and the unit be
represented in the figure, where DA 2, ABi,
BE 6, and G is a point in AS so chosen that DG,
GE are both squares.
p A GB _
Now Z? = 9. Therefore we have to divide g into two
squares suck that one of them lies between 2 and 3.
Let the latter square be ^, so that the other is gx*>
where 3>^r 2 >2.
Take two squares, one >2, the other <3 [the former
being the smaller], say f, fj.
1 Loria (pp. at. p. 150 .), as well as Nesselmann, observes that Diophantus omits to
state the necessary condition, namely that the sum of the two given numbers $lus i must
be the sum of two squares.
208 THE ARITHMETICA
Therefore, if we can make x* lie between these, we shall
solve the problem.
We must have x>^ and < .
Hence, in making Q^ 2 a square, we must find
x>)& and < ff.
Put 9 ^ = (3  mxf, say, whence x 6mj(m" + i).
Therefore
The first inequality gives J2m > i?m* + 17 ; and
36*  17. 17 =1007,
the square root of which 1 is not greater than 31 ;
therefore m $ 3I + 3 , i.e. m^^.
Similarly from the inequality igm*+iQ>72m we find a
Let m = 3^. Therefore 9  # s = (3  3j#) 2 , and x
Therefore .* = H$,
and the segments of i are (^? ^ V
\25O9 23O9 /
ii. To divide unity into three parts such that, if we add the
same number to each of the parts, the results are all squares.
Necessary condition*. The given number must not be 2 or any
multiple of 8 increased by 2.
Given number 3. Thus 10 is to be divided into three
squares such that each > 3.
Take \ of 10, or 3^, and find x so that 4 + 3j may be a
yX
square, or 3Otr 2 h i a square = (5^+ i) 2 , say.
Therefore ^=2, # 2 = 4, i/^ 3 = i, and
Therefore we have to divide 10 into three squares each of
which is as near as possible to ^. [Tra/noor^To?
076)777.]
Now 10= 3 2 + i 2 = the sum of the three squares 9, f , ^.
Comparing the sides 3, .f , with ^,
or (multiplying by 30) 90, 24, 18 with 55, we must
make each side approach 55.
1 I.e. the integral part of the root is * 31. The limits taken in each case are a fortiori
limits as explained above, pp. 613.
2 See p. 6 1, ante.
3 See pp. 1089, ant e.
BOOK V 209
[Since then MSfiHf + fit + tJlwe put for the
sides of the required numbers
335*. 3i^+i 37^+i
The sum of the squares = 3555# 2 i i6x + 10 10.
Therefore ^=^VA>
and this solves the problem*
12. To divide unity into three parts such that, if three different
given numbers be added to the parts respectively, the results are
all squares.
Given numbers 2, 3, 4. Then I have to divide 10 into
three squares such that the first > 2, the second > 3,
and the third > 4.
Let us add J of unity to each, and we have to find three
squares such that their sum is 10, while the first lies
between 2, 2\, the second between 3, 3, and the
third between 4, 4^.
It is necessary, first, to divide 10 (the sum of two
squares) into two squares one of which lies between
2, 2,\\ then, if we subtract 2 from the latter square,
we have one of the required parts of unity.
Next divide the other square into two squares, one of
which lies between 3, 3J;
subtracting 3 from the latter square, we have the second
of the required parts of unity.
Similarly we can find the third part 1 .
1 Diophantus only thus briefly indicates the course of the solution. Wertheim solves
the problem in detail after Diophantus' manner j and, as this is by no means too easy,
I think it well to reproduce his solution.
I. It is first necessary to divide 10 into two squares one of which lies between 2
and 3. We use the vapi<r6rqros d7wyjJ
The first square must be in the neighbourhood of i\ ; and we seek a small fraction
^ which when added to 2^ gives a square: in other words, we must make 4 / 2j+ 5 J
a square. This expression may be written 10 + (  J , and, to make this a square, we put
ro y i +i = (3^+i) 2 , say,
whence ^=6,^=36, ^=144, so that 2 i+^ =: J = (jf) which is an approximation
to the first of the two squares the sum of which is 10.
The second of these squares approximates to 7^, and we seek a small fraction , such
that 7i+Js is a square, or 30+ (^J =30 W J , say =a square.
H. D. 14
210 THE ARITHMETICA
13. To divide a given number into three parts such that the
sum of any two of the parts gives a square.
Given number 10.
Put sojr + i = ($y + i ) 2 , say ;
= 2,y= 4 , **=i6 f so that 7 J+ 1= 'V = ^)* = (ff )*
Now, since 10= i 2 t3 2 , and ^=r+^> while ^ = 3^,
12 12 12 12
we put (i i 7*) 2 h (3  3*) 3 = TO, [Cf. v. 9]
so that #= ,
29
Therefore the two squares into which 10 is divided are ~z an ^ 2, and the first of
041 041
these lies in fact between 2 and 2^.
II. We have next to divide the square ~ into two squares, one of which, which
841
we 'will call .r 2 , lies between 3 and 4. [The method of v. 10 is here applicable.]
Instead of 3, 4 take 4z, ^ as the limits.
Therefore ' ,
16 16
And ~  jf 2 must be a square = (  kx \ . say,
8 4 r 4 \ig ) **
. , .
wluch^ves
k has now to be chosen such that
/ \
(I)
from which it follows that
whence
We may therefore put
, ,.
Therefore
and
84!
The three required squares into which 10 is divided are therefore
1849 2624400 2893401
841 ' 707281 ' 707281 '
And if we subtract 2 from the first, 3 from the second and 4 from the third, we obtain
as the required parts of unity
140447 503557 64277
707281* 707281* 707281"   . '
BOOK V 211
Since the sum of each pair of parts is a square less than
10, while the sum of the three pairs is twice the
sum of the three parts or 20,
we have to divide 20 into three squares each of which
is < 10.
But 20 is the sum of two squares, 16 and 4 ;
and, if we put 4 for one of the required squares, we
have to divide 16 into two squares, each of which is
< 10, or, in other words, into two squares, one of which
lies between 6 and 10. This we learnt how to do 1
[v. 10].
We have, when this is done, three squares such that
each is < 10, while their sum is 20 ;
and by subtracting each of these squares from 10 we
obtain the parts of 10 required.
14. To divide a given number into four parts such that the
sum of any three gives a square.
Given number 10.
Three times the sum of the parts = the sum of four squares.
Therefore 30 has to be divided into four squares, each of
which is < 10.
(i) If we use the method of approximation
we have to make each square approximate to
1 Wertheim gives a solution in full, thus.
Let the squares be jc 2 , i6x? 9 of which one, x?, lies between 6 and 10.
2S
Put instead of 6 and IP the limits and 9, so that
To make 16  x? a square, we put
i6
whence
p.
These conditions give, as limits for , 284... and 2'2i..
We may therefore e,g* put =2j.
_, 80 9 6400 .,
Then *=;' ^=*T' ri
29 041 041
The required three squares making up 20 are 4, ^ , ^ .
Subtracting these respectively from 10, we have the required parts of the given number
. c 2010 1354
10, namely 6, gjj, gjj.
142
212 THE ARITHMETICA
then, when the squares are found, we subtract each
from 10, and so find the required parts.
(2) Or, observing that 30 = 16 + 9 + 4 + I, we take 4, 9 for
two of the squares, and then divide 17 into two squares,
each of which < 10.
If then we divide 17 into two squares, one of which lies
between 8 and 10, as we have learnt how to do 1
[cf. V. ro], the squares will satisfy the conditions.
We shall then have divided 30 into four squares, each of
which is less than 10, two of them being 4, 9 and the
other two the parts of 17 just found.
Subtracting each of the four squares from 10, we have the
required parts of 10, two of which are I and 6.
15. To find three numbers such that the cube of their sum
added to any one of them gives a cube.
Let the sum be x and the cubes jx* s sdar 3 , 63^.
Therefore g6x*=x, or g6x* = I.
But 96 is not a square \ we must therefore replace it by a
square in order to solve the problem.
1 Wertheim gives a solution of this part of the problem.
As usual, we make SJ+p, , or 34 + f 2 V a square.
Putting  = , we must make 34+ f \ a square.
Let 34T 2 + 1 = a square = (6y  i) 2 ,
and we obtain ^=6, ^=36, #2=144.
Thus 8 + L = I^=(35Y
144 H4 VV
and is an approximation to the side of each of the required squares.
Next, since 17 = 12+42, and ^=1+^=4 15,
*
and we obtain
The squares are then (i + 23*)^
121801 '
349 / I2i8oi '
Subtracting each of these from 10, we have the third and fourth of the required parts
of 10, namely ^ *
l 79649
121801* 121801"
BOOK V 213
Now 96 is the sum of three numbers, each of which is i less
than a cube;
therefore we have to find three numbers such that each
of them is a cube less i, and the sum of the three
is a square.
Let the sides of the cubes 1 be m +1,2 m, 2, whence the
numbers are m 9 + yn* + 3M, 7  1 2? I 6m*  m s , 7 ;
their sum = gm z gm f 14 = a square = (yu 4)*, say ;
therefore ^ = ^,
and the numbers are ^ff , ^V 7 , 7
Reverting to the original problem, we put x for the sum of
the numbers, and for the numbers respectively
whence
that is (if we divide out by 15 and by x),
2916*2=: 22 5, and;r=^.
The numbers are therefore found.
16. To find three numbers such that the cube of their sum
minus any one of them gives a cube.
Let the sum be x, and the numbers $&, $ x 3 , $f &.
Therefore fff^=^>
and, if ff were the ratio of a square to a square, the
problem would be solved
But ffff = 3  (the sum of three cubes).
Therefore we must find three cubes, each of which < i,
and such that (3  their sum) =* a square.
If, a fortiori, the sum of the three cubes is made < I, the
square will be > 2. Let 2 it be 2\+
1 If a*, ft, c* are the three cubes, so that <$ + ^ + **  3 has to be a square, Diopbantus
chooses <? arbitrarily (8) and then makes such assumptions for the sides of a 3 , tf 3 , being
linear expressions in m, that, in the expression to be turned into a square, the coefficient
of ;;/ 3 vanishes, and that of w 2 is a square. If <z=w, the condition is satisfied by
putting 6 = $&tn, where k is any number.
3 Bachet, finding no way of hitting upon 2} as the particular square to be taken
in order that the difference between it and 3 may be separable into three cubes, and
observing that he could not solve the problem if he took another arbitrary square between
2 and 3, e.g. aj, instead of zj, concluded that Diophantus must have hit upon sj,
which does enable the problem to be solved, by accident.
Fermat would not admit this and considered that the method used by Diophantus for
finding 2j as the square to be taken should not be difficult to discover. Fermat accord
ingly suggested a method as follows.
Let x  i be the side of the required square lying between i and 3. Then 3  (x  1)
 2 + 2*  #2, and this has to be separated into three cubes. Fermat assumes for the sides
2i 4 THE ARITHMETICA
We have therefore to find three cubes the sum of which
forM;
that is, we have to divide 162 into three cubes.
But 162=125 + 6427;
and "we have it in the Porisms" that the difference of
two cubes can be transformed into the sum of two
cubes 1 .
Having thus found the three cubes 2 , we start again, and
x2\x*, so that x f.
The three numbers are thus determined.
of two of the required cubes two linear expressions in x such that, when the sum of their
cubes is subtracted from 2 + ix  # 2 , the result only contains terms in x 2 and tf 3 or in x
and units.
The first alternative is secured if the sides of the first and second cubes are j   x and
3
i f x respectively ; for
This latter expression has to be made a cube, for which purpose we put
26
, say.
27 27 y
which gives a value for x. We have only to see that this value makes x less than i,
and we can easily choose m so as to fulfil this condition.
* suppose w=5, and we find x= , so that
i 13 i 20 72
jt=2, i  x , ifjc = i,
3 33 3 33 33
and the side of the third cube is  .
"""(A)' Kg)'*' (I) fc)'
We then have three cubes which make up the excess of 3 over a certain square ; but,
while the first of these cubes is < i, the second is > i and the third is negative. Hence
we must, like Diophantus, proceed to transform the difference between the two latter
cubes into the sum of two other cubes.
It will, however, be seen by trial that even Format's method is not quite general, for
it will not, as a matter of fact, give the particular solution obtained by Diophantus in
which the square is 2j.
1 On the transformation of the difference of two cubes into the sum of two cubes, see
. pp. 1013, an te*
2 Vieta's rule gives 4 3  3 3 = f 33* Y + ^V. It follows that
and, since jc 3 = , the required numbers are
11. A 4998267 _8^ 20338417 _8^
216*27* 6028568 '27' 20346417*27'
BOOK V 215
17. To find three numbers such that each of them minus the
cube of their sum gives a cube.
Let the sum be x and the numbers 2r*, 9^, 28^.
Therefore 39# 2 = i ;
and we must replace 39 by a square which is the sum
of three cubes + 3 ;
therefore we must find three cubes such that their sum
+ 3 is a square.
Let their sides 1 be m, ^ m, and any number, say i.
Therefore gm* f 31  27;^ = a square = (37/2  7) a , say, so
that m = f , and the sides of the cubes are f, f , i.
Starting again, we put x for the sum, and for the numbers
whence 1445^=125, # 2 =^, and xfa
The required numbers are thus found.
1 8. To find three numbers such that their sum is a square and
the cube of their sum added to any one of them gives a square.
Let the sum be x* and the numbers 3^ 6 , 8^r 6 , i$x 6 .
It follows that 26x* = i ; and, if 26 were a fourth power, the
problem would be solved.
To replace it by a fourth power, we have to find three
numbers such that each increased by i gives a square,
while the sum of the three gives a fourth power.
Let these numbers be #z 4 2#z 2 , m* + 2m> m* 2m [the sum
being ^]; these are indeterminate numbers satisfying
the conditions.
Putting any number, say 3, for m, we have as the required
auxiliary numbers 63, 1 5, 3,
Starting again, we put^r 2 for the sum and 3^, I5# 8 , 63^ for
the required numbers,
and we have 81^ = ^, so that x .
The numbers are thus found (^ , ^ , ) .
19. Tofind three numbers such that their sum is a square and
the cube of their sum mimis any one of them gives a square.
[There is obviously a lacuna in the text after this enunciation ;
for the next words are " And we have again to divide 2 as before"
1 Cf. note on v. 15. In this case, if one of the cubes is chosen arbitrarily and m 3
is another, we have only to put ($k*  m) for the side of the third cube in order that, in the
expression to be made a square, the term in rtfi may vanish, and the term in w 2 may be a
square.
216 THE ARITHMETICA
whereas there is nothing in our text to which they can refer, and
the lines which follow are clearly no part of the solution of V. 19.
Bachet first noticed the probability that three problems inter
vened between v. 19 and v. 20, and he gave solutions of them.
But he seems to have failed to observe that the eight lines or so in
the text between the enunciation of v. 19 and the enunciation of
v. 20 belonged to the solution of the last of the three missing
problems. The first of the missing problems is connected with
V. 1 8 and 19, making a natural trio with them, while the second and
third similarly make with V. 20 a set of three. The enunciations
were doubtless somewhat as follows.
19 a. To find three numbers such that their sum is a square
and any one of them minus the cube of their sum gives a square.
igb. To find three numbers such that their sum is a given
number and the cube of their sum plus any one of them gives a
square.
igc. To find three numbers such that their sum is a given
number and the cube of their sum minus any one of them gives a
square.
The words then in the text after the enunciation of v. 19
evidently belong to this last problem.]
The given sum is 2, the cube of which is 8.
We have to subtract each of the numbers from 8 and
thereby make a square.
Therefore we have to divide 22 into three squares, each
of which is greater than 6 ;
after which, by subtracting each of the squares from 8, we
find the required numbers.
But we have already shown [cf. V. n] how to divide 22
into three squares, each of which is greater than 6
and less than 8, Diophantus should have added.
[The above is explained by the fact that, by addition,
three times the cube of the sum minus the sum itself
is the sum of three squares, and three times the
cube of the sum minus the sum = 3.82 = 22.] *
1 Wertheim adds a solution in Diophantus 1 manner. We have to find what small
fraction of the form ^ we have to add to or , and therefore to 66, in order to
3 9
make a square. In order that 66 +~ may be a square, we put
s square  ( r + 8*) 2 , say,
which gives #=8 and 0^=64.
BOOK V 217
20. To divide a given fraction into three parts such that any
one of them minus the cube of their sum gives a square.
Given fraction J.
Therefore each part = ^ ? 4 a square.
Therefore the sum of the three =  = the sum of three
squares f $%.
Hence we have to divide f into three squares, " which is
easy 1 ."
21. To find three squares such that their continued product
added to any one of them gives a square.
Let the " solid content " = x\
We want now three squares, each of which increased by I
gives a square.
They can be got from rightangled triangles 2 by dividing
the square of one of the sides about the right angle
by the square of the other.
Let the squares then be
The continued product = ^tfife* 6 = # 2 , by hypothesis.
Therefore ff# 2 = i ; and, if ff were a square, the problem
would be solved.
We have therefore to increase 66 by g , and therefore 7^ by > , in order to make a
square. And in fact 7$ + ^ = ( /"
Next, since 12 = 3 2 f 3 3 + 2 2 , and 65  48 = 1 7, while 72  65 = 7, we put
22 = (3  7*) 2 + (3  7*) 3 + (2 + 1 7.r) 2 .
16
and x = 5
387
Therefore the sides of the squares are * , jr= ~o7T
, , , 1100401 1100401 1094116
the squares themselves ~ ,  ~ , *~
149769 149769 149/69
and the required parts of, are 4ZZS' JH2* , 124596
H97<>9 X 497^9 r 497^9
1 As Wertheim observes, i= = ~ + + , and the required fractions into which
64 64 25 400
i . j. ., , 250 89 61
 is divided are ~ , ~ , ^ .
4 1600 1600 1600
2 If a, b be the perpendiculars, c the hypotenuse in a rightangled triangle,
+ i =  2 =a square.
Diophantus uses the triangles (3, 4, 5), (5, 12, 13), (8, 15, tf).
2i8 THE ARITHMETIC A
As it is not, we must find three rightangled triangles such
that, if b's are their bases, and /'s are their perpen
diculars, A AA WA = a square ;
and, if we assume one triangle arbitrarily (3, 4, 5), we
have to make 12/j^A a square, or 3/A/#A a square.
"This is easy 1 ," and the three triangles are (3, 4, 5),
(9, 40, 41), (8, 15, 17) or similar to them.
1 Diophantus does not give the work here, but only the result. Bachet obtains it
in this way. Suppose it required to find three rational rightangled triangles (h^ fa j),
(^2J*s ^2) and (^3,/s, 3) such that p\fafa\b\b$z is the ratio of a square to a square.
One triangle (^ , p l , J being chosen arbitrarily, form two others by putting
and we have
f Pi \ 2
( ^ ] =a square.
Wi/
If now ^==5,^ = 4, ^=3, the triangles (/*j,/2, &>) and (/& 3> ^3> a) are ( 4 i, 9, 40)
and (34, 1 6, 36) respectively. Dividing the sides of the latter throughout by 2 (which
does not alter the ratio), we have Diophantus' second and third triangles (9, 40, 41) and
& I5> 17).
Ferxnat, in his note on the problem, gives the following general rule for finding two
rightangled triangles the areas of which are in the ratio m : n (m>n}.
Form (i) the greater triangle from im + n, m  #, and the lesser from m + in> m  ,
or (2) the greater from tin  n, m + n, the lesser from in  m, m + n,
or (3) the greater from 6m, im  n, the lesser from 4#j + n, yn  2,
or (4) the greater from m i 4, im  4, the lesser from 6w, min.
The alternative (2) gives Diophantus' solution if we put #3 = 3, n=i and substitute
m 2 for a  m. ,
Fermat continues as follows : We can deduce a method of finding three rightangled
triangles the areas of which are in the ratio of three given numbers, provided that the
sum of two of these numbers is equal to four times the third. Suppose e.g. that m, , q
are three numbers such that m + q 4** (m > q} Now form the following triangles :
(1) from m + 4, im  4,
(2) from 6n, m 2,
(3) from 4+f, 42f.
[If AI , AZ, A 3 be the areas, we have, as a matter of fact,
We can derive, says Fermat, a method of finding three rightangled triangles the areas
of which themselves form a rightangled triangle. For we have only to find a .triangle
such that the sum of the base and hypotenuse is four times the perpendicular. This is
easy, and the triangle will be similar to (17, 15, 8) ; the three triangles will then be formed
(1) from 17 + 4.8, 2. 174.8 or 49, 2,
(2) from 6.8, 172*8 or 48, i,
(3) from4.8+i5, 4.82.15 or 47, 2.
[The areas of the three rightangled triangles are in fact 234906, 110544 arid 207270,
and these numbers form the sides of a rightangled triangle.]
Hence also we can derive a method of finding three rightangled triangles the areas
of which are in the ratio of three given squares such that the sum of two of them is equal to
BOOK V 219
Starting again, we put for the squares
Equating the product of these to x* t we find x to be
rational \x=^ and the squares are ~, ^, 2 ~].
22. To find three squares such that their continued product
minus any one of them gives a square.
Let the solid content be # 2 , and let the numbers be
obtained from rightangled triangles, being
Therefore the continued product [   jjy 6 ;
If then   were a fourth power. i.e. if  ' were
1221025 ^ ' 5.13.17
a square, the problem would be solved.
We have therefore to find three rightangled triangles
with hypotenuses 7t l9 7/ 2 , // 3 respectively, and with
A> A> A as one of the perpendiculars in each re
spectively, such that
1 VfeAAA = a square.
Assuming one of the triangles to be (3, 4, 5), so that eg.
Ihp* = 5 . 4 = 20, we must have
S^iA^aA = a square.
This is satisfied if ^A = S^A
With a view to this we have first (cf. the last proposition)
to find two rightangled triangles such that, if x^ y^
are the two perpendiculars in one and x^ jr t the two
perpendiculars in the other, x^ = 5*y a . From such
a pair of triangles we can form two more right
angled triangles such that the product of the
hypotenuse and one perpendicular in one is five times
the product of the hypotenuse and one perpendicular in
the other 1 .
four times the third^ and we can also in the same way find three rightangled triangles of
the same area; we can also construct^ in an infinite number of ways, two rightangled
triangles the areas of which are in a given ratio^ by multiplying one of the terms of the
ratio or the two terms by given squares, etc.
1 Diophantus* procedure is only obscurely indicated in the Greek text. It was
explained by Schulz in his edition {cf. Tannery in Oewvres de Fermat^ I. p. 323, note).
220 THE ARITHMETICA
Since the triangles found satisfying the relation ^^=
are (5, 12, 13) and (3, 4, 5) respectively 1 , we have in
fact to find two new rightangled triangles from them,
namely the triangles (7^, p lt b^) and (A 2 , A> ^a)> such that
&ifi = 30 and hzpt = 6,
the numbers 30 and 6 being the areas of the two
triangles mentioned.
These triangles are (6J, ft, [^]) and (2*, ^, [&]) re
spectively.
Starting again, we take for the numbers
[Y divided by 2\ gives ff , and fg divided by 6 gives ff]
The product = jt 8 :
therefore, taking the square root, we have
4.24.120
5.25.169*''
so that #=ff , and the required squares are found.
23. To find three squares such that each minus the product of
the three gives a square.
Having given a rational rightangled triangle (s, x, y)> Diophantus knows how to find a
rational rightangled triangle (,/, b) such that hpxy. We have in fact to put
Thus, having found two triangles (5, 12, 13) and (3, 4, 5) with areas in the ratio of 5
to i (see next paragraph of text with note thereon), Diophantus takes
,i , . 5.12 60 , . .. , , i . , 3.4 12
fii= . 13 =6i,/ 1 =^ = ; and similarly /& 2 =  . 5=2}, /s = ~~ ~ =  .
* *33 55
Cossali (after Bachet) gives a formula for three rightangled triangles such that the
solid content of the three hypotenuses has to the solid content of three perpendiculars
(one in each triangle) the ratio of a square to a square ; his triangles are
(i)
and, in fact,
If z'=5, ^=4, /=3, we can get from this triangle the triangles (13, 5, 12) and
(65, 63, 16), and our equation is '^ x? = i.
1 These triangles can be obtained by putting m = $,n=i in Fermat*s^r^ formula
(note on last proposition). By that formula the triangles are formed from (9, 6) and
(6, 3) respectively ; and, dividing out by 3, we form the triangles from (3, 2) and (2, i)
respectively.
BOOK V 221
Let the "solid content" be x*, and let the squares be
formed from rightangled triangles, as before.
If we take the same triangles as those found in the last
problem and put for the three squares
***. m*. fffs** 9 .
each of these minus the continued product (x*) gives a
square.
It remains that their product ~x*\
this gives # = $, and the problem is solved.
24. To find three squares such that the product of any two
increased by r gives a square.
Product of first and second 4 1 = a square, and the third
is a square ; therefore " solid content " h each = a
square.
The problem therefore reduces to V. 21 above 1 .
1 De Billy in the Invention Novum, Part n. paragraph 18 (Oeuvres de F&rmatt III.
pp. 3734), extends this problem, showing how to findyfrw numbers, three of which (only)
are squares, having the given property, t.e. to solve the equations
First seek three square numbers satisfying the conditions of Diophantus' problem
v. 24, say ( % , , ^ j, the solution of v. 2 1 given in Bachet's edition. We have then
to find a fourth number (JT, say) such that
x+ i N are all squares.
Substitute y* + y for x, so as to make the first expression a square. We have
then to solve the doubleequation
100 9
r 2
9 '
4296 2 +
729 ^ 729
which can be solved by the ordinary method.
De Billy does not give the solution, but it may be easily supplied thus.
THedi^ence .(2 + g) ( ? _ g)
154 (16 2.26N
^vl y\  J.
7^W 2 7 /
222 THE ARITHMETICA
25. To find three squares such that the product of any two
minus i .gives a square.
This reduces, similarly, to V. 22 above.
26. To find three squares such that, if we subtract the product
of any two of them from unity, the result is a square.
This again reduces to an earlier problem, V. 23.
27. Given a number, to find three squares such that the sum of
any two added to the given number makes a square.
Given number 15.
Let one of the required squares be 9 ;
I have then to find two other squares such that each
4 24 = a square, and their sum + 1 5 = a square. .
To find two squares, each of which + 24 = a square, take
two pairs of numbers which have 24 for their pro
, duct 1 .
Let one pair of factors be 4/#, 6x, and let the side of one
2
square be half their difference or jar.
Let the other pair of factors be 3/ar, &r, and let the
side of the other square be half their difference or
ii*,
X ^
Therefore each of the squares 4 24 gives a square.
It remains that their sum + 15 =a square;
therefore
( 4#J 4 ( 3#) + 1 5 = a square,
Equating the square of half the sum of the factors to the larger expression, we have
Therefore *= 0^ + 2^)= X V A" , which satisfies the equations. In fact
9 /ii467\ 2 25 /3275\ 2 , 256
^rjff i = ( * ' ) ,~x+ i =(* ) , and o
16 V5/ 4 \3456/ 81
But even here, as the value of x which we have found is negative, we ought, strictly
speaking, to deduce a further value by substituting y  "^~r for x in the equations
and solving again, which would of course lead to very large numbers.
1 The text adds the words " and [let us take] sides about the right angle in a right
angled triangle." I think these words must be a careless interpolation : they are not
wanted and give no sense; nor do t.hey occur in the corresponding place in the next
problem.
BOOK V 223
or ~ + 2$x* 9 =s a square = 25# 2 , say.
x
Therefore ^=, and the problem is solved 1 .
28. Given a number, to find three squares such that the sum of
any two minus the given number makes a square.
Given number 13.
Let one of the squares be 25 ;
I have then to find two other squares such that each
+ 12 = a square, and (sum of both) 13 = a square.
Divide 12 into factors in two ways, and let the factors be
(3*, 4/x) and (4*, 3/3).
Take as the sides of the squares half the differences of the
factors, i.e. let the squares be
Each of these 4 12 gives a square.
It remains that the sum of the squares 13 = a square,
or f f 6J 3? 25 = a square = f , say.
Therefore x = 2, and the problem is solved 2 .
Diophantus has found values of , 97, f satisfying the equations
a= 2 \
*=*4.
a w 2 )
Fermat shows how to find four numbers (not squares) satisfying the corresponding
conditions, namely that the sum of any two added to a shall give a square. Suppose #= 15,
Take three numbers satisfying the conditions of Diophantus' problem, say 9, , ^2 1
Assume a?  15 as the first of the four required numbers; and let the second be &rf 9
(because 9 is one of the square numbers taken and 6 is twice its side) ; for the same
reason let the third number be  x + and the fourth ~ x 4 ^^ .
5 loo 15 225
Three of the conditions are now fulfilled since each of the last three numbers added to
the first (x*  15) plus 15 gives a square. The three remaining conditions give the triple
equation
1^6 520 136
2*+ 9+ 5 ^+i5=^jr
15 225 * 15
15* ioo 225 15
2 Fermat observes that four numbers (not squares) with the property indicated can
be found by the same procedure as that shown in the note to the preceding problem.
If a is the given number, put x* +a for the first of the four required numbers.
224 THE ARITHMETICA
29. To find three squares such that the sum of their squares is
a square.
Let the squares be jr 2 , 4, 9 respectively 1 .
Therefore x* f 97 = a square = (x io) 2 , say ;
whence x* = ^.
If the ratio of 3 to 20 were the ratio of a square to a
square, the problem would be solved ; but it is not.
Therefore / have to find two squares (/ 2 , q, say) and a
number (m, say) such tJiat m / 4 (f has to 2m the
ratio of a square to a square.
Therefore m 2  ^  0* = (0 2 f 4) 2  &  1 6 = 8# 2 .
Hence S* 2 /^ 2 + 8), or 48 8 /( < s 8 + 4), must be the ratio of a
square to a square.
Put .z 2 + 4 = 0+i) 2 , say;
therefore z i^, and the squares are / 2 =2j, 2 = 4, while
i6J;
or, if we take 4 times each, p* = 9, # 2 = 16, #2 = 25.
Starting again, we put for the squares ;tr 2 , 9, 16;
then the sum of the squares = x* + 337 = (;r 3 25) 2 , and
*~*
The required squares are ^, 9, 16.
30. [The enunciation of this problem is in the form of an
epigram, the meaning of which is as follows.]
A man buys a certain number of measures (%oe9) of wine, some
at 8 drachmas, some at 5 drachmas each. He pays for them a
square number of drachmas ; and if we add 60 to this number, the
result is a square, the side of which is equal to the whole number
of measures. Find how many he bought at each price.
Let x the whole number of measures ; therefore ^ 60
was the price paid, which is a square = (x trif, say.
If now 2, /2, * 2 represent three numbers satisfying the conditions of the present
problem of Diophantus, put for the second of the required numbers 2/fo. +/& 2 , for the third
2/j:t/ 2 , and for the fourth 3r+* 2 . These satisfy three conditions, since each of the
last three numbers added to the first (x* + a) less the number a gives a square. The
remaining three conditions give a tripleequation.
1 "Why," says Fermat, "does not Diophantus seek two fourth powers such 'that
their sum is a square ? This problem is in fact impossible, as by my method I am in
a position to prove with all rigour." It is probable that Diophantus knew the fact
without being able to prove it generally. That neither the sum nor the difference of
two fourth powers can be a square was proved by Euler (Commentationes arithmeticae, I.
pp. 24sqq., and Algebra^ Part II. c. xni.).
BOOK V 225
Now of the price of the fivedrachma measures + of
the price of the eightdrachma measures =^r;
so that x* 60, the total price, has to be divided into
two parts such that of one + of the other =x.
We cannot have a real solution of this unless
Therefore 5* < x*  60 < &r.
( I ) Since x* > $x + 60,
^ 2 = 5#+ a number greater than 60,
whence x is 1 not less than 1 1.
(2) x* < 8* + 60
or ^r a = &r f some number less than 60,
whence # is 1 not greater than 12.
Therefore 1 1 < x < 12.
Now (from above) ;r= (w a + 6o)/2*#;
therefore 22 w < ;// 2 f 6b < 24;;^.
Thus (i) 22wz = w 2 f (some number less than 60),
and therefore m is 2 not less than 19.
(2) 24771 = m + (some number greater than 60),
and therefore ;;/ is 2 less t/ian 21.
Hence we put m = 20, and
.a:* 60 =5 (x  20) 2 ,
so that x*= i ij, AT 2 132^, and x z 60 = 72^.
Thus we have to divide 72^ into two parts such that
of one part p fas % of the other = 1 1^.
Let the first part be 5^.
Therefore J (second part) = ii ^,
or second part = 92 8# ;
therefore 5^ f 92 8^r = 72 ,
.
12
Therefore the number of fivedrachma %oe<? = 25 .
eightdrachma g.
1 For an explanation of these limits see p. 60, ante,
3 See p. 62,
H.
226 THE ARITHMETIC A
BOOK VI
i. To find a (rational) rightangled triangle such that the
hypotenuse minus each of the sides gives a cube 1 .
Let the required triangle be formed from x, 3.
Therefore hypotenuse = x + 9, perpendicular = 6x, base
= #9.
Thus x* + 9 (# 2 9)= 1 8 should be a cube, but it is not.
Now 18 = 2 . 3 8 ; therefore we must replace 3 by m, where
2 . m is a cube ; and m = 2.
We form, therefore, a rightangled triangle from x, 2,
namely (^ 2 + 4, 4*, # 4)', and one condition is
satisfied.
The other gives x~ 42* f 4 = a cube ;
therefore (x 2f is a cube, or JIT  2 is a cube = 8, say.
and the triangle is (40, 96, 104).
2. To find a rightangled triangle such that the hypotenuse
added to each side gives a cube.
Form a triangle, as before, from two numbers ; and, as
before, one of them must be such that twice its
square is a cube, i.e. mu'st be 2.
We form a triangle from x, 2, namely ^ 2 f4, 4^, 4 x*\
therefore ^ + 4^ + 4 must be a cube, while x 2 must
be less than 4, or x< 2.
Thus ;tr + 2 = a cube which must be < 4 and > 2 = ^, say.
Therefore ;tr = J^,
and the triangle is (S3 , 5 J , fZ) ,
or, if we multiply by the common denominator, (135,
352, 377)
3. To find a rightangled triangle such that its area added to
a given number makes a square.
Let 5 be the given number, (3^, 4^, $x) the required
triangle.
1 Diophantus' expressions are 6 fr r# #7romvoi50fl, * the (number) in (or represent
ing) the hypotenuse," 6 fr tKarfyg, ruv 6p6uv, "the (number) in (or representing) each
of the perpendicular sides," 6 & T< evfiadf, "the (number) in (or representing) the area,"
etc. It will be convenient to say "the hypotenuse," etc. simply. It will be observed
that, as between the numbers representing sides and area, all idea of dimension is ignored.
BOOK VI 227
Therefore 6x* + 5 = a square = gx 9 say,
or 3* 2 =5.
But 3 should have to 5 the ratio of a square to a square.
Therefore we must find a rightangled triangle and a
number such that the difference between the square
of the number and the area of the triangle has to 5 the
ratio of a square to a square, i.e. = % of a square.
Form a rightangled triangle from (m, ) ;
thus the area is m z  , .
m
2 c
Let the number be m f , so that we must have
m
therefore 4.254 *f = a square,
or zoo;;/ 2 4 505 = a square = (iom 4 5) 2 , say,
and ?#  ^t.
The auxiliary triangle must therefore be formed from % t
j 5 ? , and the auxiliary number sought is *$$.
Put now for the original triangle (Jix,px, bx\ where (//,/, b
is the rightangled triangle formed from ^, ^ ;
this gives i/^ 2 + 5 = HSM^^
and we have the solution.
[The perpendicular sides of the rightangled triangle are
7s T^2 ) X ~ 14400 ^ anc * 2;ir '
.5 2 4/
whence ^W^ 2 + S = :
and the triangle is
( s m&, it.
4. To find a rightangled triangle such that its area minus a
given number makes a square.
Given number 6, triangle (3^, 4^, 5ar), say.
Therefore 6^ 2 6 square = 4ar 3 , say.
Thus, in this case, we must find a rightangled triangle
and a number such that
(area of triangle) (number) 2 = of a square.
Form a triangle from ;;/, .
s m
152
228 THE ARITHMETICA
Its area is m s   , and let the number be m f . .
*
m
Therefore 6   i (a square),
or 36m 60 = a square = (6m 2) 2 , say.
Therefore m = f , and the auxiliary triangle is formed from
(f , ), the auxiliary number being f.
We start again, substituting for 3, 4, 5 in the original
hypothesis the sides of the auxiliary triangle just
found, and putting (fJ) 3 J^ in place of 4^; and the
solution is obvious.
[The auxiliary triangle is (*$, 2,  4 ^ 7 ), whence
^j* 6 = (#)*, and *=f,
so that the required triangle is (^tf, \*, ^jtff).]
5. To find a rightangled triangle such that, if its area be
subtracted from a given number, the remainder is a square.
Given number 10, triangle (3^', 4*, 5^), say.
Thus 10 6r 2 = a square; and we have to find a right
angled triangle and a number such that
(area of triangle) h (number) 2 = ^ of a square.
Form a triangle from m, , the area being #z 2  ^ ,
s ' m s m*
and let the number be h $m.
m
Therefore 26;;z' 2 f 10 = fa of a square,
or 26o;# 2 f ioo = a square,
or again 65 ;;z 2 f 25 = a square = (8?# + S) 2 , say,
whence #z = 80.
The rest is obvious.
The required triangle is *MMB a . T f , *jiHHHM
6. To find a rightangled triangle such that the area added
to one of the perpendiculars makes a given number.
Given number 7, triangle (3^r, 4#, 5^).
Therefore 6^ + 3^ = 7.
In order that this might be solved, it would be necessary that
(half coefficient of xf \product of coefficient of 3? and
absolute term should be a square ;
but (i) 2 f 6.7 is not a square.
Hence we must find, to replace (3, 4, 5), a rightangled
triangle such that
(^ one perpendicular) 2 f 7 times area = a square.
Let one perpendicular be m, the other I.
BOOK VI 229
Therefore 3J//2 + i= a square, or 14^ f i = a square.)
Also, since the triangle is rational, ;;; 2 4 I = a square.}
The difference m* i^m = ; (;# 14) ;
and putting, as usual, 7 2 = 14;;^ f i,
we have ?;/ = .
The auxiliary triangle is therefore (^, i, *) or (24, 7, 25).
Starting afresh, we take as the triangle (24*, 7^, 25^).
Therefore 84*r 2 + 7^ = 7,
and #=rj.
We have then (e, ^ > f) as the solution 1 .
7. To find a rightangled triangle such that its area minus one
of the perpendiculars is a given number.
Given number 7.
As before, we have to find a rightangled triangle such that
( one perpendicular) 2 + 7 times area a square ;
this triangle is (7, 24, 25).
Let then the triangle of the problem be (7^, 24^, 25^).
Therefore 84^  7^ = 7,
* = i,
and the problem is solved 2 .
1 Fermat observes that this problem and the next can be solved by another method.
"Form in this case,'* he says, "a triangle from the given number and i, and divide
the sides by the sum of the given number and i ; the quotients will give the required
triangle."
In fact, if we take as the sides of the required triangle
where a is the given number, we have
one root of which is *= 5 +
~ = ,
and the sides of the required triangle are therefore
a+i
The solution is really the same as that of Diophantus.
3 Similarly in this case we may, with Fermat, form the triangle from the given number
and i, and divide the sides by the difference between the given number and i, and we
shall have the required triangle.
In vi. 6, 7, Diophantus has found triangles , , t\ (^ being the hypotenuse), such that
(i) i7+;=tf,
and W &?{=
Fermat enunciates the third case
(3) 7"*
230 ' THE ARITHMETICA
8. To find a rightangled triangle such that the area added to
the sum of the perpendiculars makes a given number.
Given number 6.
Again I have to find a rightangled triangle such that
(i sum of perpendiculars) 2 f 6 times area = a square.
Let m, i be the perpendicular sides of this triangle ;
therefore J (m + i) 2 4 yn = \m* + 3i/# + i = a square,
while m + i must also be a square.
Therefore **+ '^ jj are both squares.
The difference is 2m . 7, and we put
whence ; = f ,
and the auxiliary triangle is (ff , i, ff), or (45> 2 ^ S3)
Assume now for the triangle of the problem
(45*, 28*, 53*).
Therefore 63O* 2 + 73* = 6 ;
* is rational [= y^], and the solution follows.
9. To find a rightangled triangle such that the area minus the
sum of the perpendiculars is a given number.
Given number 6.
As before, we find a subsidiary rightangled triangle such
that(sum of perpendiculars ) 3 + 6 times area= a square.
This is found to be (28, 45, 53) as before.
Taking (28*, 45*, 53*) for the required triangle,
630* 3 73* = 6;
x~w&> and the problem is solved 1 .
10. To find a rightangled triangle such that the sum of its
area, the hypotenuse, and one of the perpendiculars is a given
number.
observing that Diophantus and Bachet appear not to have known the solution, but that
it can be solved "by our method." He does not actually give the solution ; but we may
compare his solutions of similar problems in the Inventum Novum, e.g. those given in
the notes to vi. 1 1 and vi. 15 below and in the Supplement. The essence of the method
is that, if the 6rst value of x found in the ordinary course is such as to give a negative
value for one of the sides, we can derive from it a fresh value which will make all the
sides positive.
1 Here likewise, Diophantus having solved the problem
Fermat enunciates, as to be solved by his method, the corresponding problem
&?=a.
BOOK VI 231
Given number 4.
If we assumed as the triangle (//,r, px, bx\ we should have
and, in order that the solution may be rational, we must
find a rightangled triangle such that
J (hyp. f one perp.) 2 4 4 times area = a square.
Form a rightangled triangle from i, ;;/ f i.
Then  (hyp. + one perp.) 3 = i (m* + 2; + 2 + ;;/ 2 4 2;#) 2
= v + 4#2 3 f 6;;/ 2 + 4m j i ,
and 4 times area =4 (m 4 i) (; s 4
Therefore
;/2 4 + 8?/2 3 + 1 8;;2 2 h 12m 4 1 = a square = (6;/2 + 1 ^ 3 ) 2 , say,
whence #/ = f , and the auxiliary triangle is formed from
(i, f) or (5, 9). This triangle is (56, 90, 106) or
(28, 45, 53)
We assume therefore 28;r, 45^, 53^ for the original triangle,
and we have 630^ + 8i^r = 4.
Therefore ^ = 7^, and the problem is solved.
n. To find a rightangled triangle such that its area mimes
the sum of the hypotenuse and one of the perpendiculars is a given
number.
Given number 4.
We have then to find an auxiliary triangle with the same
property as in the last problem ;
therefore (28, 45, 53) will serve the purpose.
We put for the triangle of the problem (28^,45^, 5 jar), and
we have 63or 2 8 \x = 4 ;
# = , and the problem is solved 1 .
1 Diophantus has in VI. io 3 u shown us how to find a rational rightangled triangle
> fj (f being the hypotenuse) such that
(i) &
Fermat, in the Iwventwn Navum* Part in. paragraph 33 (Qcwvres dt Fermat, in.
p 389), propounds and solves the corresponding problem
(3)
In the particular case taken by Fermat a =4. He proceeds thus :
First find a rational rightangled triangle in which (since a =4)
(f +*)}* 4 .^ = a square.
232 THE ARITHMETICA
Lemma I to tke following problem.
To find a rightangled triangle such that the difference of the
perpendiculars is a square, the greater alone is a square, and further
the area added to the lesser perpendicular gives a square.
Let the triangle be formed from two numbers, the greater
perpendicular being twice their product.
Hence I must find two numbers such that (i) twice their
product is a square and (2) twice their product exceeds
the difference of their squares by a square.
This is true of any two numbers the greater of which
= twice the lesser.
Form then the triangle from x, 2x, and two conditions are
satisfied.
The third gives 6x* + 3r 2 = a square, or d^H 3 = a square.
I have therefore to find a number such that 6 times its
square f 3 = a square ;
one such number is i, and there are an infinite number of
others 1 .
If #= i, the triangle is formed from I, 2.
Suppose it formed from jr+ i, x, the sides then are
f
Thus +*) 4~
= a square
say.
Therefore 6^=6^4^, *=,and *+i=.
o 3
The triangle formed from * , ^ is { ^ ^ ,  ) . Thus we may take as the auxiliary
..,. . 33V99 9/
triangle (17, 15, 8).
Take now 17^, 15^:, 8* for the sides of the triangle originally required to be found.
We have then
whence *=:, and the required triangle is ( , ,  ).
3 \ 3 3 3/
[The auxiliary rightangled triangle was of course necessary to be found in order to
make the final quadratic give a rational result.]
Bachet adds after vi. n a solution of the problem represented by
fr?f=a,
to which Fermat adds the enunciation of the corresponding problem
*;&>.
1 Though there are an infinite number of values of x for which 6>* + 3 becomes a square,
the resulting triangles are all similar. For, if x be any one of the values, the triangle is
BOOK VI 233
Lemma II to the following problem.
Given two numbers the sum of which is a square, an infinite
number of squares can be found such that, when the square is multi
plied by one of the given numbers and the product is added to
the other, the result is a square.
Given numbers 3, 6.
Let x* \~2x\i be the required square which, say, when
multiplied by 3 and then increased by 6, gives a square.
We have ^ + 6x + 9 = a square ;
and, since the absolute term is a square, an infinite number
of solutions can be found.
Suppose, e.g. yp + 6x + 9 = (3  yc)\
and x = 4.
The side of the required square is 5, and an infinite
number of other solutions can be found.
1 2. To find a rightangled triangle such that the area added
to either of the perpendiculars gives a square.
Let the triangle be (5,r, ixr, 134).
Therefore ( i) 30^ f 1 2x = a square = 36^, say,
and x = 2.
But (2) we must also have
3Or 3 + 5# = a square.
This is however not a square when x = 2.
Therefore I must find a square m*x*> to replace 36^, such
that i2/(iw 2 30), the value of x obtained from the
first equation, is real and satisfies the condition
301^ f $x = a square.
This gives, by substitution,
(6o;# 2 + 2S2o)/(;;^  6om + 900) = a square,
or 6o;# 3 + 2520 == a square.
This could be solved [by the preceding Lemma II] if
.60+2520 were equal to a square.
Now 60 arises from 5. 12, i.e. from the product of the
perpendicular sides of (5, 12, 13);
 2520 is 30. 1 2. (12 5), i.e. the continued product of the
area, the greater perpendicular, and the difference
between the perpendiculars.
formed from JT, ur, and its sides are therefore s* 2 , 4# 2 , $x 2 ; that is, the triangles are all
similar to (3, 4, 5). Fermat shows in his note on the following problem, vi. 12, how to
find any number of triangles satisfying the conditions of this Lemma and not similar to
(3, 4, 5). See p. 235, note.
234 THE ARITHMETICA
Hence we must find an auxiliary triangle such that
(product of perps.) + (continued product of area,
greater perp. and difference of perps.) = a square.
Or, if we make the greater perpendicular a square and
divide out by it, we must have
(lesser perp.) + (product of area and diff. of perps.)
= a square.
Then, assuming that we have found two numbers, (i) the
product of the area and the difference of the perpen
diculars and (2) the lesser perpendicular, satisfying
these conditions, we have to find a square (m 2 ) such
that the product of this square into the second of
the numbers, when added to the first number, gives
a square 1 .
1 The text of this sentence is unsatisfactory. Bachet altered the reading of the MSS.
So did Tannery, but more by way of filling out. The version above follows Tannery's text,
which is as follows : airdyerai eZs TO Stio dpt,&fjt,ovs ebp6vTas [for ovras of MSS.] < TOP re wrd>
roG ^/ijffaSov Kcd rrjs inrepoxijs riav dp^wr, < Kal rbv tv rrj \d<r<rovi r&v 6p6Qv > , avQts [for
avTTfc of MSS.] frr&v QOV TW a, 6s 7roXXcnrXa<rtaa0eis irl &a TOP 5o^vra, < /cat TrpocrXajSwi/
rbv %Tpov> t TOLCI rerpdywov.
The argument would then be this. If (k, p t b) be the triangle (>/), we have to make
bp+ bp(bp)ba, square,
or, if b is a square, f\ bp (bp) must be a square.
The ultimate equation to be solved (corresponding to 6o/// 2 + 252o=a square) is
bpnfi+ bp (b p] b=a. square,
or, if b is a square, pm* + l>p(b /) = a square ;
and therefore, according to Tannery's text, "the problem is reduced to this: Having found
two numbers  bp (b p) and p [satisfying the conditions, namely that their sum is a
square, while b is also a square], to find after that a square such that the product of it
and the latter number added to the former number gives a square."
The difficulty is that, with the above readings, there is nothing to conespond exactly to
the phraseology of the enunciation of Lemma I, which speaks, not of making p +  bp (b  p)
a square when b is a square, but of making b p, b and/H bp all simultaneously iquares.
But fat particular solution of the Lemma is really equivalent to making b and/ f  bp (b p)
simultaneously squares. For the triangle is formed from #, ia\ this method of making
b a square (=4<z 2 ) incidentally makes bp a square ( = a 2 ), and/ +  bp becomes 3<z 2 + 6"a 4 ,
while/h3/(J/) becomes 3a?+6a*. Since the solution actually used is a = i, the
effect is the same whichever way the problem is stated. And in any case, whether the
expression to be made a square is s^wHfo* or 3^ 2 >w 2 i6 6 , the problem equally reduces
to that of making 3/tf /a + 6 a square.
BOOK VI 235
How to solve these problems is shown in the Lemmas,
The auxiliary triangle is (3, 4, 5). [Lemma L]
Accordingly, putting for the original triangle (3^, 4*, $x),
, 6jtr 2 r4r) , .
we nave ^ > both squares.
OJT 4 3^,J ^
4
Let .# = ^  be the solution of the first equation ;
then ^ = ^ .
up I2m 2 + 36
The second equation therefore gives
96 12 ^ a s uare
^ 4 1 2m* 4 3^ 7 ^ 2 ~ 6
whence 1 2;^ 2 + 24 = a square,
and we have therefore to find a square (m z ) such that
twelve times it + 24 is a square; this is possible, since
12 + 24. is a square [Lemma II].
A solution is n? = 25,
whence x = A,
and 5?, ?9 is the required triangle 1 .
13. To find a rightangled triangle such that its area minus
either perpendicular gives a square.
We have to find an auxiliary triangle exactly as in the
last problem ;
Bachet's reading is dirrfrerai s r6 &5o apiB^fav foBtrrw roO re e/ij9a5oO, mil r^y
t\d<7<TOJfos rwv irepi rijv dpQiljv, ai5ro?s frrrew rtTp&ywbv riva, 6s iro\XairXa<rta(r06ts tirl
Hva, T&V 8o0frT(aj>i Kal Trpoa^CLp&v TOV Zrepov, TrotJ Terpdycwov.
1 Fermat observes that Diophantus gives only one species of triangle satisfying the
condition, namely triangles similar to (3, 4, 5), but that by his (Fermat's) method an infinite
number of triangles of different species can be found to satisfy the conditions, the first
being derived from Diophantus' triangle, the second from the new triangle, and so on.
Suppose that the triangle (3, 4, 5) has been found satisfying the condition that
 1)  ft = a square,
where , rj are the perpendicular sides and
To derive a second such triangle from the first (3, 4, 5), assume the greater of the two
perpendicular sides to be 4 and the lesser 3+jr.
Then '7 + ('7) .  ^=36  i2r8jf a = a square.
Also f 2 =3+i7 2 =25f 6jrl.* 2 ==asquare.
We have therefore simply to solve the doubleequation
3612^8^=2^1
25+ 6x+ # 2 r=z;2j '
which is a matter of no difficulty. As a matter of fact, the usual method gives
20667 j i_  i / 20667 2 373Qi65\
3=   , and the triangle is (  ' , 4, D V 1 .
' s *' 593^289;
236 THE ARITHMETIC A
this triangle is (3, 4, 5), and accordingly we assume for
the triangle of the problem (3r, 4^, 5.tr).
One condition then gives 6^4^= a square = ra 2 ;^, say
and Jtr= .
6 m*
The second condition gives 6.r 2 3^= a square; and, by
substitution,
96 12 __
"
or 2411 2m 2 = a square.
This is satisfied by m I,
whence x=$> and the required triangle is (~ , ~ , 4 V
Or, if we do not wish to use the value i for m,
let m = ^f i, and (dividing by 4) we have
3?;/ 2 + 6 = 32* f 6,0 + 9 =r a square ;
# must be found to be not greater than ^ (in order that
m* may be less than 6), and m will not be greater than
^f. The solution is then rational 1 .
14. To find a rightangled triangle such that its area minus the
hypotenuse or minus one of the perpendiculars gives a square.
Let the triangle be (jr, 4^, SJF).
_
Therefore ., . \ are both squares.
6x*  $x\ ^
Making the latter a square (= m*x*\ we have
1 Diophantus having solved the problem of finding a rightangled triangle
( being the hypotenuse) such that
> are both squares,
Fermat enunciates, as susceptible of solution by his method, but otherwise very difficult,
the corresponding problem of making
>]
> both squares.
,^J
This problem was solved by Euler (Novi Commentarii Acad. PetropoL 1749, JI 
pp. 49 sqq. = Commentationes artihmeticae t i. pp. 6272).
BOOK VI 237
The first equation then gives
54 IS
 i  _ TL_ = a square,
m* 1 2m* + 36 6 m n
or i s?;/ 2 36 = a square.
TXif equation we cannot solve because 1 5 is not the sum of
two squares^. Therefore we must change the assumed
triangle.
Now (with reference to the triangle 3, 4, 5) i5;;/ 2 = the
continued product of a square less than the area, the
hypotenuse, and one perpendicular ;
while 36 = the continued product of the area, the perpen
dicular, and the difference between the hypotenuse
and the perpendicular.
Therefore we have to find a rightangled triangle (/t, /, b,
say) and a square (m*) less than 6 such that
m*hp \pb ,p (Ji /) is a square.
If we form the triangle from two numbers A' n X and
suppose that p zX^X^ and if we then divide
throughout by (X l A r a ) 2 which is equal to h p, we
must find a square * [= m*l(Xi X,^f\ such that
zhp \pb ./ is a square.
The problem can be solved if X^ A" 2 are "similar plane
numbers*?
Form the auxiliary triangle from similar plane numbers
accordingly, say 4, i. [The conditions are then
satisfied 8 .]
[The equation for m then becomes
8 . 17#z 2 4 . 1 5 . 8 . 9 = a square,
or 1 367;^ 4320 = a square.]
Let 4 ? 2 = 36. [This satisfies the equation, and 36 < area
of triangle.]
1 See p. 70 above.
2 Diophantus states this without proof. [A '* plane number " being of the form a . b,
... . . f ,  mm, m , _
a plane number similar to it is of the form a. b or 5 ao.~\
r n n n*
The fact stated may be verified thus. We have
s* (X*+X) *XiXiXiXt (Xi*X) 2X^2=3. square.
The condition is satisfied if z 2 = Jfi X Zt for the expression then redu
In that case XiX% is a square, or X\\X% is a square.
3 Since ATi = 4) ^2=i we ^a.ve h = 17,^=8, ^=15, s 2 = A"i^Y" 2 =4, and
z L hppb.p^.+, 17, 84. 15. 8 = 2. 32 = 64, a square.
* The reason for this assumption is that, by hypothesis, ^fn\
and w 2 = g6.
238 THE ARITHMETICA
The triangle formed from 4, I being (8, 15, 17), we assume
S;tr, i$x t ijx for the original triangle.
We now put 6ox*  8^' = 36^,
and x = J.
The required triangle is therefore (, 5, ^ j .
Lemma to the following problem.
Given two numbers, if, when some square is multiplied into
one of the numbers and the other number is subtracted from
the product, the result is a square, another square larger than
the aforesaid square can always be found which has the same
property.
Given numbers 3, II, side of square 5, say, so that
3 . 25 1 1 = 64, a square.
Let the required square be (# + 5) 2 .
Therefore
3 (*+ S) 2  1 1 = 3^+ 30* + 64 = a square
= (8  2*) 2 , say,
and .#=62.
The side of the new square is 67, and the square itself
4489.
15. To find a rightangled triangle such that the area added
to either the hypotenuse or one of the perpendiculars gives a
square.
In order to guide us to a proper assumption for the
required triangle, we have, in this case, to seek a
triangle (//, /, b, say) and a square (in?) such that
m* > \pb> the area, and
w&hp \pb .p (ft p) is a square.
Let the triangle be formed from 4, i, the square (m*)
being 36, as before ;
but, the triangle being (8, 15, 17), the square is not
greater than the area.
We must therefore, as in the preceding Lemma, replace
36 by a greater square.
Now hp = 136, and \pb .p (Ji /) = 60 . 8 . 9 = 4320,
so that 136^  4320 = a square,
which is satisfied by m* = 36 ; and we have to find a larger
square (z*) such that
 4320 = a square.
BOOK VI 239
Put z = m f 6, and we have
(m* + 1 2m H 36) 1 36  4320 s= a square,
or 136;;^ f 1632;^ 4 576 = a square = (;# 24), say.
This equation has any number of solutions ; e.g. t putting
k = 1 6, we have
m = 20, = 26, and z = 676.
We therefore put (&tr, 15^, 17^) for the original triangle,
and then assume
6o^ 2 f 8^ = 676^,
whence x = fa> and the problem is solved 1 .
1 In vi. 14, 15 Diophantus has shown how to find a rational rightangled triangle
& 17, (where $ is the hypotenuse) such that
(0 Jfcr]
> are both squares,
;*)
are both squares.
In the Inuentitm Novum, Part I. paragraphs 26, 40 (Oetivres de Fermat t ill. pp. 341
2, 34950) is given Fermat's solution of a third case in which
fifcl
> are both squares.
*H
This depends on the Lemma : To find a rational rightangled triangle in which
m+ j?)  \&=* square.
Form a rightangled triangle from x+ 1, i ; the sides are then
We must therefore have
(x~ + ix + 2) (x* + $x + 1)  (x + 1) (x* + ix) = a square,
or **+ 4J^+6* 2 + 6jrh 2 =a square
= (^+2^+ 1) 2 , say.
Therefore *= , and the triangle has one of its sides x + ix negative. Instead
2
therefore of forming the triangle from , i or from i, 7, we form it from x+i t i and
repeat the operation. The sides are then
and we have
x+ 5) (o: 2 f 4^ x) ~ (aj:+ 2) (jr 2 + 2^ s)=a square,
x* + qx? + 6^ + 20^*+ 1 =a square
2 4 o THE ARITHMETICA
16. To find a rightangled triangle such that the number
representing the (portion intercepted within the triangle of the)
bisector of an acute angle is rational 1 .
A
4*
33* D** B
Suppose the bisector AD $x, and one segment of the
base (DB) = $x ; therefore the perpendicular = 44.
Let the whole base CB be some multiple of 3, say 3 ; then
0) = 33*.
But, since AD bisects the angle CAB,
AC:CD = AB:BD]
therefore the hypotenuse A C = f (3 $x) = 4 ^x.
whence r = ~, and the required auxiliary triangle is formed from ,2 or from 29, 12,
the sides being accordir.gly 985, 697, 696.
(Fermat observes that the same result is obtained by putting y   for x in the
expression Jt 4 + ^ + 6.r 2 4 6jc + 2 ; for we must have
+y +  a square =f 5^ y , say,
whence j>=~ 3 so that x = y~ = ~, and the triangle is formed from , i or from
29, 12, as before)
We now return to the original problem of solving
We assume for the required triangle (985,2;, 697*, 696:*:) and we have 17 = 2
so that
985^242556^) A , , ,
^ ., ! must both, be squares.
697^242556^]
Assume that 697^  242556^ =: (697A) 2 ,
and we have x  348^ =697^,
whence A~=  , and the required triangle is ( ", ~, 2 ).
1045 ^ a \i045 7 1045' 1045;
[The 985^242556^ is a square by virtue of the sides 985, 697, 696 satisfying the
conditions of the Lemma; for 98 5* 2 425 56^ =~r   f^ ' ^ > which is a square
if 985 . 1045    697 . 696 is a square, and 1045 = 697 +  . 696.]
1 "Why did not Diophantus propound the analogous problem " To find a rightangled
triangle such that the sides are rational and the bisector of the right angle is also rational"?
Evidently because he knew it to be impossible, as is clear when (a, c being the perpen
diculars) the bisector is expressed as  */2. (Loria, op. cit. p. 148 n.)
BOOK VI 241
Therefore [by EucL I. 47]
i6or 2 32^4 16 = 16^ + 9,
and x h
If we multiply throughout by 32, the perpendicular = 28,
the base = 96, the hypotenuse = 100, and the bisector
= 35
17. To find a rightangled triangle such that the area added
to the hypotenuse gives a square, while the perimeter is a cube.
Let the area be x and the hypotenuse some square
minus x, say 16 x.
The product of the perpendiculars = 2x ;
therefore, if one of them be 2, the other is x, and the
perimeter = 18, which is not a cube.
Therefore we must find some square which, when 2 is
added to it, becomes a cube 1 .
1 "Did Diophantus know that the equation tt*+'2=zP only admits of otie solution
= 5, z/=3? Probably not" (Loria, op. cit. p. 155). The fact was noted by Fermat
(on the present proposition) and proved by Euler.
Euler's proof (Algebra^ Part II. Arts. [88, 193) is, I think, not too long to be given
here. Art. 188 shows how to find x, y such that ax* + cy~ may be a cube. Separate
ax* + cy* into its factors x >Ja +y \/( ^)> x *Ja ~y *J (  0> an d assume
the product (ajP+cff being a cube and equal
To find values for x and y, we write out the expansions of the cubes in fall, and
whence
y
For example, suppose it is required to make jp+}P a cube. Here ai and c=i 9
so that
. If now/ = 2 and y=i, we find xi and^=n, whence
Now (Art, 193) let it be required to find, if possible, in integral numbers, other squares
besides 25 which, when added to 2, give cubes.
Since ^ + 2 has to be made a cube, and i is double of a square, let us first determine
the cases in which jc 3 + 27 s becomes a cube. Here a=i 9 c=i,so that
x=p6p&, y=3pQ*Pi
therefore, since y= i, we must have
3/V*^ 8 or ^(3/ >2 ~ 2 ^ 2 )= I J
consequently q must be a divisor of i .
Let, then, q~ i, and we shall have 3^2= r.
With the upper sign we have 3/^=3 and, taking p  i, we find x 5 ; with the lower
sign we get an irrational value of p which is of no use.
H. D. l6
242 THE ARITHMETICA
Let the side of the square be m+ i, and that of the cube
m i.
Therefore *  $m* + ^m  i = ;;* 2 + 2m + 3>
from which aw is found 1 to be 4.
Hence the side of the square = 5, and that of the cube 3.
Assuming now x for the area of the original triangle,
25  x for its hypotenuse, and 2, x for the perpen
diculars, we find that the perimeter is a cube.
But (hypotenuse) 2 = sum of squares of perpendiculars ;
therefore x* SOT + 62 5 = x* + 4 ;
#= fisi^ and the problem is solved.
1 8. To find a rightangled triangle such that the area added
to the hypotenuse gives a cube, while the perimeter is a square.
Area JF, hypotenuse some cube minus x, perpendiculars #, 2.
Therefore we have to find a cube which, when 2 is added
to it, becomes a square.
Let the side of the cube be m I.
Therefore m s $m? + $m + i = a square = (ijw + i) 2 , say.
Thus * = Jy, and the cube = (Jf ) 8 = i.
Put now x for the area, x, 2 for the perpendiculars, and
4 li^ ~ x f r ^ e hypotenuse;
and x is found from the equation (*ffi xf = x* + 4.
[*4Him and the trian s le is (2, ^IMI 11 ^ 1 *HiW a )]
19. To find a rightangled triangle such that its area added to
one of the perpendiculars gives a square, while the perimeter is
a cube.
Make a rightangled triangle from some indeterminate odd
number*^ say 2x+ I ;
then the altitude =2#+ i, the base = 2x* + sar, and the
hypotenuse = 2# 2 + 2x+ i.
It follows that there is no square except 25 which has the required property.
Fermat says ("Relation des nouvelles de"couvertes en la science des nombres,'*
Oeiwres, n. pp. 4334) that it was by a special application of his method of descents,
such as that by which he proved that a cube cannot be the sum of two cubes, that he proved
(1) that there is only one integral square which when increased by i gives a cube, and
(2) that there are only two squares in integers which, when added to 4, give cubes. The
latter squares are 4, 121 (as proved by Euler, Algebra, Part II. Art. 192).
1 See pp. 66, 67 above.
2 This is the method of formation of rightangled triangles attributed to Pythagoras.
If m is any odd number, the sides of the rightangled triangle formed therefrom are m,
i (m*  1), i (m z f i ), for m* + Ji (m*  i)l * = fe (m 2 + i) i *. Cf. Proclus, Comment.
on Eucl. i. (ed. Friedlein), p. 428, 7 sqq., etc. etc.
BOOK VI 243
Since the perimeter = a cube,
) = a cube;
and, if we divide all the sides by x+ r, we have to make
43: + 2 a cube.
Again, the area + one perpendicular = a square.
~ u r  * + x
1 herefore 7    1  = a square :
(x+if *
. , . . 
that is,  ^  3  = 2^r + I == a square.
*  2.2: + I ^
But 4*rf 2 a cube;
therefore we must find a cube which is double of a
square ; this is of course 8.
Therefore 4^ + 28, and #= ij.
The required triangle is (? , S 3 S\
20. To find a rightangled triangle such that the sum of its
area and one perpendicular is a cube, while its perimeter is a
square.
Proceeding as in the last problem, we have to make
AtX h 2 a square)
2;tr+ I a cube J
We have therefore to seek a square which is double of a
cube ; this is 16, which is double of 8.
Therefore 4^+2=16, and x^\.
The triangle is ,, .
21. To find a rightangled triangle such that its perimeter is
a square, while its perimeter added to its area gives a cube.
Form a rightangled triangle from x, i.
The perpendiculars are then 2x, x* i, and the hypotenuse
Hence 2x* \ 2x should be a square,
and x? 4 2.3? \x a cube.
It is easy to make 2x* + 2x3 square ; let 22? + 2x =
therefore x 2/(m*  2).
By the second condition,
8 8 2 A , ,
7 s ^> + ;^ s TO H s must be a cube,
(w 2 2) 3 (;^ 2 2) 2 m* 2
2m* .
2.^. ,  ^rr = a cube.
(m 2)
1 6 2
244 THE ARITHMETICA
Therefore 2?;z 4 = a cube, or 2m = a cube = 8, say.
Thus w = 4, *=& = !, ^ 2 = ^.
But one of the perpendiculars of the triangle is x*  i, and
we cannot subtract I from ^.
Therefore we must find another value for x greater than i ;
hence 2<m*< 4.
And we have therefore to find a cube such that \ of the
square of it is greater than 2, but less than 4.
If z 3 be this cube,
2<z*< 4,
or 8 < z*< 16.
This is satisfied by z* = ^, or * = ^.
Therefore *0 = fJ, ^ 2 = iff and * = fT7> the square of
which is > i.
Thus the triangle is known
22. To find a rightangled triangle such that its perimeter is
a cube, while the perimeter added to the area gives a square.
(1) We must first see how, given two numbers, a triangle
may be formed such that its perimeter one of
the numbers and its area = the other.
Let 12, 7 be the numbers, 12 being the perimeter, 7 the
area.
Therefore the product of the two perpendiculars
= 14 = . I4#.
^ x ^
If then ,14^; are the perpendiculars,
hypotenuse = perimeter sum of perps. = 12  14^.
x
Therefore [by Eucl. I. 47]
1+196^ + 172^336^^ + 196^;
that is, 172 = 3364;+ ^ ,
or 172^=336^+24.
This equation gives no rational solution, because 86 2 24 . 3 36
is not a square.
Now 172 = (perimeter) 2 + 4 times area,
24. 336 = 8 times area multiplied by (perimeter) 2 .
(2) Let now the area = m, and the perimeter = any
number which is both a square and a cube, say 64.
BOOK VI 245
Therefore { (64 3 4 4^)} 2 8 . 64* . 7/2 must be a square,
or 4//2 2 245767/2 +4194304 = a square.
Therefore m 61447/2 4 1048576 = a square]
Also m + 64 = a square/ *
To solve this doubleequation, multiply the second by
such a number as will make the absolute term the
same as the absolute term in the first.
Then, if we take the difference and the factors as usual,
the equations are solved.
[After the second equation is multiplied by 16384, the
doubleequation becomes
7*2 2 6144272 4 1048576 a square]
163847/2 4 1048576 == a square] "
The difference is m z 225287/2.
If 7/2, 7/2 22528 are taken as the factors, we find m = 7680,
which is an impossible value for the area of a right
angled triangle of perimeter 64.
We therefore take as the factors 1 17/2, ^7/2 2048 ; then,
when the square of half the difference is equated to
the smaller of the two expressions to be made squares,
we have
(eo m + io24) a = 163842/2 4 1048576,
and m =
Returning now to the original problem, we put , 2,mx
for the perpendicular sides of the required triangle,
and we have
(64  I  2^) 2 = ~ + 4^^
which leads, when the value of m is substituted, to
the equation
78848^ 8432^4225 = o.
The solution of this equation is rational, namely
527 23_ 25 9
9856 "448 176"
Diophantus would of course use the first value, which
would give (^^, J ^ & , J %^) as the required right
angled triangle. The second value of x clearly gives
the sarne triangle.]
246 THE ARITHMETICA
23. To find a rightangled triangle such that the square of its
hypotenuse is also the sum of a different square and the side of
that square, while the quotient obtained by dividing the square
of the hypotenuse by one of the perpendiculars of the triangle is
the sum of a cube and the side of the cube.
Let one of the perpendiculars be #, the other # 2 .
Therefore (hypotenuse) 2 = the sum of a square and its
^4 i yfi
side; also  ==^ 8 +^r= the sum of a cube and its
side.
It remains that & \x* must be a square.
Therefore tf + I = a square = (# 2) 2 , say.
Therefore x f , and the triangle is found [f , T 9 ^, {].
24. To find a rightangled triangle such that one perpendicular
is a cube, the other is the difference between a cube and its side,
and the hypotenuse is the sum of a cube and its side.
Let the hypotenuse be x*+x, and one perpendicular
jp x.
Therefore the other perpendicular = 2^ = a cube = X s , say.
Thus x 2, and the triangle is (6, 8, 10).
It is on Bachet's note to vi. 22 that Fermat explains his method of solving
tripleequations, as to which see the Supplement, Section v.
[No. 20 of the problems on rightangled triangles which Bachet
appended to Book vi. ("To find a rightangled triangle such that its area
is equal to a given number ") is the occasion of Fermat's remarkable note
upon the theorem discovered by him to the effect that The area of a right
angled triangle the sides of which are rational numbers cannot be a square
number.
This note will be given in full, with other information on the same
subject, in the Supplement]
ON POLYGONAL NUMBERS
All numbers from 3 upwards in order are polygonal, containing
as many angles as they have units, e.g. 3, 4, 5, etc.
" As with regard to squares it is obvious that they are such
because they arise from the multiplication of a number into
itself, so it was found that any polygonal multiplied into a
certain number depending on the number of its angles, with
the addition to the product of a certain square also depending
on the number of the angles, turned out to be a square. This
I shall prove, first showing how any assigned polygonal
number may be found from a given side, and the side from
a given polygonal number. I shall begin by proving the pre
liminary propositions which are required for the purpose."
1. If there are three numbers with a common difference, then
8 times the product of the greatest and middle + the square of the
least = a square, the side of which is the sum of the greatest and
twice the middle number.
Let the numbers be AB, BC, BD in the figure, and we
have to prove SAJ3 . BC + BD* = (AB + zBC)\
E A c D B
1 j 1 2.
By hypothesis A C= CD, AB=BC+ CD, BD= BC CD.
Now 8AB.BC=4A.jBC+(4.C*+4BC. CD).
Therefore ZAB.BC+BD*
. C+ 4C* + (4BC . CD + BD*)
C+4BC* + AB Z , [Eucl. II. 8]
and we have to see how AB* 4 ^AB . BC+ 4#C 2 ,.can
be made a square.
[Diophantus does this by producing BA to E, so that
AE = BC, and then proving that
AB* h 4AB . BC+ A<BC* = (BE + EA ) 2 .]
It is indeed obvious that
AB* H AB . BC+4BC* = (AB + zBC)*.
2. If there are any numbers, as many as we please, in A.P.,
the difference between the greatest and the least is equal to the
common difference multiplied by the number of terms less one.
248 ON POLYGONAL NUMBERS
[That is, if in an A.P. the first term is a, the common
difference b and the greatest term /, n being the
number of terms, then
/ * = (!)&]
Let AB, BC, BD, BE have a common difference.
Now AC, CD, DE are all equal.
Therefore EA = ACx (number of terms AC, CD, DE)
(number of terms in series i).
3. If there are as many numbers as we please in A. P., then
(greatest + least) x number of terms = double the sum of the
terms.
[That is, with the usual notation, 2s = n (/+ #).]
(i) Let the numbers be A, B, C, D, E, F, the number of
them being even.
A . B ,
L M
Let GH contain as many units as there are numbers,
and let GH, being even, be bisected at K. Divide
GK into units at L, M.
Since F D=CA,
But F + A =(F + A).GL;
therefore C+D = (F + A) . LM.
Similarly E + B = (F + A ) . MK.
Therefore, by addition,
Therefore 2(A ++ ...)= 2 (F+A). GK
(2) Let the number of terms be odd, the terms being
A, B y C, D, E.
.' D , E
F H L K
Let there be as many units in FG as there are terms,
so that there is an odd number of units.
Let FH be one of them ; bisect HG at K, and divide HK
into units, at Z.
ON POLYGONAL NUMBERS
249
Since
Similarly B + D = 2 C . LH.
Therefore A +E + B + D = iC
= C. HG.
Also C=C.HF;
therefore, by addition,
and, since 2C=A+E,
4. If there are as many numbers as we please beginning with
i and increasing by a common difference, then the sum of all
x 8 times the common difference + the square of (common
difference  2) = a square, the side of which diminished by 2
= the common difference multiplied by a number which when
increased by i is double of the number of terms.
[The A,P. being i, 1+5, ... i+(n \}b, and s the sum,
we have to prove that
S . S& + ( 2) 2 = {b (20  I) + 2J 2 ,
z>. 8t>s = 4<5V  4 (b  2) nb,
or 2s = br? (b 2)n
= n{2 + (n 1)6}.
The proof being cumbrous, I shall add the generalised
algebraic equivalent in a column parallel to the
text]
Let A, CD, EF be the terms in
A.P. after i.
1+6,1+26,1 +3*,....
P A
K N
M
Let GH contain as many units
as there are terms including i.
Difference between EF and I
= (diff. between Atfand i) x (GH 1). / I
[Prop. 2]
25
ON POLYGONAL NUMBERS
Put AK, EL, GM each equal
to unity.
Therefore LF=KB .MH.
Make KN=2, and we have to
inquire whether
(sum of terms) x 8KB f NB*
[Prop. 3]
= i (KB . MH. GH + 2GH\
since LF= KB.MH [above].
Bisecting MH at 0, \ve have
(sum of terms)
Now sum of terms
We have therefore to inquire
whether
is a square.
Now KB.GH.HO.ZKB
Is then
a square ?
Now ZGH.KB
Also
and, adding NB* 3 the righthand side
becomes KB* + KN\ [Eucl. n. 7]
Is then ^GH.HM.KB*
a square ?
Again, ^T^ 2 + 4GH. HM . KB*
= GM* . .ST^ 2 + 4G/T. HM. KB*
= (Gff+ffMy. KB\ [Eucl. n. 8]
Is then (GJf+ HM}* . .fiT.5 2
a square?
Make the number NO' equal to
Call the expression on
the lefthand side X.
+ 4 { + (n  i)} 3 + b* + 2*
thus
will be proved later 1 .
Also
ON POLYGONAL NUMBERS
as
Therefore
251
.NK, since
Is then NO* + NK* f 2NO . NK
a square ?
Yes ; it is the square on KO r .
And
OK  2 = NO = KB (GH + HM\
while GH+HM+ i = (twice number
of terms).
Thus the proposition is proved.
" The above being premised, I say that,
[5] If there be as many terms as we please in A.P. beginning
from i, the sum of the terms is polygonal; for it has as many
angles as the common difference increased by 2 contains units, and
its side is the number of the terms set out including i."
The numbers being as set out in the figure of Prop. 4, we
have, by that proposition,
(sum of terms) . %KB 4 NB* = KO~\
Taking another unit AP, we have KP = 2> while KN=2\
therefore PB, BK> BN are in A.P., so that
&PJ3 . BK + NB* = (PB 4 2KBJ ; [Prop, i]
and PB 4 2KB 2 = PB + 2KB PK = $KB 3
while 3 + i = 2 . 2, or 3 is one less than the double of 2.
Now, since the sum of the terms of the progression
1 Deferred lemma.
To prove that (GH+HM}*. KB*
Place DE (equal to a) and F (equal to ]3) in a
straight line.
Describe squares DH, EL on Z>#, EF and com
plete the figure.
Then DE : EF=Dffi HF,
and HE : EK= HF : EL.
Therefore HF is a mean proportional between the two squares,
that is
M
252 ON POLYGONAL NUMBERS
including unity satisfies the same formula 1 [literally
" does the same problem "] as PB does,
while PB is any number and is also always a polygonal,
the first after unity (for AP is a unit and AB is the
term next after it), and has 2 for its side,
it follows that the sum of all the terms of the progression
is a polygonal with the same number of angles as PB,
the number of its angles being the same as the
number of units in the number which is greater by 2,
or PK, than the common difference KB, and that its
side is GH which is equal to the number of terms
including I.
And thus is demonstrated what is stated by Hypsicles in
his definition, namely, that,
"If there are as many numbers as we please beginning
from i and increasing by the same common difference,
then, when the common difference is i, the sum of all
the terms is a triangular number ; when 2, a square ;
when 3, a pentagonal number [and so on]. And the
number of the angles is called after the number
exceeding the common difference by 2, and the side
after the number of terms including I."
[In other words, if there be an arithmetical progression
the sum of the n terms, or \n {2 f (# i) }, is the
nth polygonal number which has (b + 2) angles.]
Hence, since we have triangles when the common dif
ference is i, the sides of the triangles will be the
greatest term in each case, and the product of the
greatest term and the greatest term increased by I
is double the triangle.
1 Nesselmann (pp. 4756), exhibits this result thus.
Take the A.P. i , i + b, 1 + 2^, ... r + ( i) b.
If j is the sum, &r*+ (b  2) 2 = {b (in  i) + 2} 2 .
If now we take the three terms 2, b t b+z, also in A.P.,
Now 5+2 is the sum of the first two terms of the first series, and corresponds there
fore to s when n = 2 ; and 3 = 2 . i  i , so that 3 corresponds to in  j.
Hence j and b + 2 are subject to the same law ; and therefore, as b + 2 is a polygonal
number with 3 + 2 angles, s is also a polygonal number (the nth) with b + i angles.
ON POLYGONAL NUMBERS 253
And, since PB is a polygonal with as many angles as
there are units in it,
and SP . (PB  2) + (/Iff 4)* = a square (from above,
BK being equal to PB  2, and NB to P~4),
the definition of polygonal numbers will be as follows :
Every polygonal multiplied by 8 times (number of angles
2) f square of (number of angles 4) = a square 1 .
The Hypsiclean definition and the new one being thus
simultaneously proved, it remains to show how, when
the side is given, the prescribed polygonal is found.
For, having given the side GH and the number of angles,
we know KB.
Therefore (GH+ HM} KB, which is equal to NO\ is also
given ; therefore KO'(=NO'+NK or N0'+ 2) is given.
Therefore K0' z is given ;
and, subtracting from it the given square on NB, we
obtain the remaining term which is equal to the
required polygonal multiplied by %KB. Thus the
required polygonal can be found.
Similarly, given the polygonal number, we can find its
side GH. Q. E.D.
Rules for practical use.
(i) To find the number from the side.
Take the side, double it, subtract I, and multiply the
remainder by (number of angles 2). Add 2 to the
product ; and from the square of the sum subtract
the square of (number of angles 4). Dividing the
remainder by 8 times (number of angles  2), we
have the required number*
1 Hultsch points out (art. Diophantos in PaulyWissowa's RealEncyclopadie der
dassischen Altertumswissenschaften) that this formula
SP (a  2) + (a  4) 2 = a square
shows that Diophantus intended it to be applied not only to cases where a is greater than
4 but also where 0=4 or less. For 36, as Diophantus must have known, besides being
the second s6gon, is also a triangle, a square, and a 13gon, inasmuch as
And indeed it is evident from Def. 9 of the Arithmetica that (3 ~4) 2 = i, while.it is
equally obvious that (4 4)2=0.
254 ON POLYGONAL NUMBERS
[If P be the th agona.\ number,
p _
(2)
To find the side from the number.
Multiply the number by 8 times (number of angles 2) ;
add to the product the square of (number of angles 4).
We thus get a square. Subtract 2 from the side of
this square and divide the remainder by (number of
angles 2). Add i to the quotient, and half the
result gives the side required 1 ,
2
02
Given a number, to find in how many ways it can be polygonal
Let AB be the given number, BC \ [Algebraical equivalent^
the number of angles, and in BC take
CD = 2, C=4
Number A B = P.
Number of angles SC a.
M
Since the polygonal AB has BC
angles,
(i) 8 AS. BD + BE*= a square =
say.
Cut off AH equal to i ;
therefore ZAB.BD
But
say.
2)
(^2). 2+4(2^1X02).
Make DK equal to 4 (AB + BH\
and for $AE '. BD put 2BD.DE.
1 Fermat has the following note. " A. very beautiful and wonderful proposition which
I have discovered shall be set down here without proof. If, in the series of natural
numbers beginning with i, any number n be multiplied into the next following, n+ i,
the product is twice the nth triangular number; if n be multiplied into the (n+i)th
triangular number, the product is three times the nth tetrahedral member; if n be
multiplied into the (n+i)th tetrahedral number, the product is four times the nth triangulo
triangular number \Jigured number of tfh order}; and so on, ad injinitum. I do not
think there can be, in the theory of numbers, any theorem more beautiful or more
general. The margin is too small, and I am not at liberty, to give the proof." (Cf.
Letter to Roberval of 4 November 1636, Oeuvres de Fermat, ir. pp. 84, 85.) For a proof,
see Wertheim's Diophantus, pp. 31820.
ON POLYGONAL NUMBERS
2 55
Therefore
(2)
(3)
[Eucl. II. 7]
(4) = KB.BD+DE*. [EucL II. i]
But, since
and DC half 4 or 2 ;
therefore AT > C>.
Therefore, if DK be bisected at
L, L falls between C and 7lT.
And, since DK is bisected at Z,
whence KB .BD = LB 2  LD\
Therefore, by (4) above,
(5)
or
(6) or
Again, since ED^DC, and
is produced to Z,
therefore EL .LC= DL* DC*
(7)
Put FM=BL (for BL>FG,
since
while
Therefore FM*  FG* = EL.LC.
Now } DK being bisected at L and
being equal to
And DC=2AH.
Therefore CL = $BH,
or BH=ICL.
But
therefore
while BH=\CL.
Therefore
or EL.LC=i6AB.BH.
{2(2/ ? l)} 2 2 2
256 ON POLYGONAL NUMBERS
(8) Therefore
(9)
Therefore GM is even.
Let GM be bisected at N
i6P(Pi)
[={2(2.P2)2(;jl)02)j 2
[Here the fragment ends, and the question of course arises whether
Diophantus ever actually solved the problem of finding in how many
different ways a given number can be a polygonal. Tannery went so far
as to call the whole of the fragment, from and including the enunciation
of the problem, the "vain attempt of a commentator ;j to solve it 1 .
Wertheirn 2 has however shown grounds for thinking that Diophantus did
solve the problem and that the fragment is a genuine part of his argument
leading to that result The equation
easily reduces (by algebra) to
SP(a  2} = 4 (a  2) { 2 + (  1) (a  2)},
or 2P=n{2 + (n i) (0 2)}.
Wertheim has shown how this result can be obtained by a continuation
of the work, from the point where the fragment leaves off, in the same
geometrical form which is used up to that point 3 , and how, when the
1 Dioph. I. pp. 4767, notes.
2 Zeitschriftfilr Math. u. Physik, hist. litt. Abtheilung, 1897, pp. 1216.
3 The only thing, so far as I can see, tending to raise doubt as to the correctness of
this restoration is the fact that, supposing it to be required to prove geometrically, from
the geometrical equivalent of
W > (a)+(*4) a ={a+(a* i)(a)},
that iP=n{i + (ni)(a>2)\,
it can be done much more easily than it is in Diophantus' proposition as extended by
Wertheim.
For let FG a + ( in  i ) (a  2). Cut off FR equal to 2, and produce F to 5 so that
T
We have now 8P. SR=FG*  SF*
Bisect at 7 1 , and divide out by 4 ;
therefore 2^. S=ST*  ST. SF
=ST(STSF)
Now ST=n. S, and FJ?=i, while J?7"=( i) . SR=(n i) (a 2).
It follows that iP=n {2 + (n  i) (a  2)}.
ON POLYGONAL NUMBERS
25?
formula is thus obtained, it can be used for the purpose of finding the
number of ways in which P can be a polygonal number. The portion of
the geometrical argument which has to be supplied is, it is true, somewhat
long, and its length and difficulty may, as Wertheim suggests, account for
the copyist having failed, as it were, to see his way through it and having
stopped through discouragement when he had lost his bearings.
I shall now reproduce Wertheim's suggested restoration of the rest of
the problem. The figure requires some extension, and I accordingly give
a new one after Wertheim.
B
H 1
S R Q N
The last step in the above fragment is
(9) 2FG . GM + GM* = i6AB . BH.
Bisect GM'mN,
so that GN= NM.
Therefore, if we divide by 4,
(10) FG. GN+
M
or
(u)
FN.NG^^AB.BH.
Put now FR=2AB, and
so that GSRN, and we have
GN,
(12)
Substituting in (n), we have
GN=NM
2 (/>!
 i)(a 2)
!)( 2)}
(a 2)}
{2 (/>!) (!) (**)}
= 2 + (n L) (a 2}
2P+n(a'2\ from above
=2(Pi) (ni)(a 2)
= FM
H. D,
ON POLYGONAL NUMBERS
or
(i3)
 RN. FS =
. AH.
Therefore
(14)
or
(15)
~2P{(ni)(a2)+2n(a2)\
n(a 2) {(n  i) (a  2) + 2}
= 4^4^
2P{2(a2}}
+ n (a  2) {(n  i) (a  2) + 2}
n(a2){(n i)(#2) + 2}
Now RN= FN FR = FM NM FR = FM\ GM FR
= BL\ GM 2AB = D + \DK \ GM~ 2AB
= BD + 2AB + 2BH
and FS=FRRS
= 2AB\
Therefore RN FS = BD+ 2BH *AB
and RN FS+ 2 AH = BD.
Again, we have
= BD + 2BH \ GM= BD + 2BH~
2BH \BD \DL + \FG
2BH \DL + \FG
= BD + 2BH (AB + BH]
But, from the rule just preceding this proposition,
(ini) + 2\
= 2n . BD + 2,
therefore
or
therefore RN= n .
Accordingly the equation (15) above becomes
(16) 2AB.BD = n.BD.FS, 2P(a2]
or
(17)
= n {(n i)(a  2) + 2}
2}
Thus the double of any polygonal number must be divisible by its
side, and the quotient is the number arrived at by adding 2 to the product
of (side  r) and (number of angles 2).
For a triangular number the quotient is n + i, and is therefore greater
ON POLYGONAL NUMBERS
2 S9
than the side; and, as the quotient increases by ;/ i for every increase
of i in the number of angles (a), it is always greater than the side.
We can therefore use the above formula (17) to find the number of
ways in which a given number P can be a polygonal number. Separate 2P
into two factors in all possible ways, excluding i . zP. Take the smaller
factor as the side (n). Then take the other factor, subtract 2 from it,
and divide the remainder by (n  i). If (n  i) divides it without a
remainder, the particular factors taken answer the purpose, and the quotient
increased by 2 gives the number of angles (a). If the second factor
diminished by 2 is not divisible by (n i) without a remainder, the
particular division into factors is useless for the purpose. The number of
ways in which P can be a polygonal is the number of pairs of factors
which answer the purpose. There is always one pair of factors which will
serve, namely 2 and P itself.
The process of finding pairs of factors is shortened by the 1 following
considerations.
zPn {(n  i) (a 2) + 2} ;
therefore 2Pfn = 4 + an a 271, :
2 (/>)
and = 2 + , ';
(!)'
therefore not only zP\n but also . ^ must be a whole number and,
as a is not less than 3,
or=
and consequently
n < or =
Thus in choosing values for the factor n we need not go beyond that
shown in the righthand expression.
Example i. In what ways is 325 a polygonal number?
Here  i + ^/(i + 8/>) =  i + ^(260 1) = 50. Therefore n cannot be
greater than 25. Now 2 . 325 = 2.5.5.13, and the only possible values
for n are therefore 2, 5, 10, 13, 25. The corresponding values for a are
shown in the following table.
n
2
5
34
10
13
25
a
325
9
6
3
Example 2. P 120.
n
2
3
4
5
6
8
10
12
15
a
120
41
_
6
3
172
CONSPECTUS OF THE ARITHMETICA
Equations of the first degree with one unknown.
i. 7. x a = m (x V).
i. 8. x + a = m (x + b\
I. 9. ax = m(bx).
i. 10. x + & = m(a x).
I. ii. x + b m (x a).
I. 39. (a + re) <5 + (b + #) = 2 (a + <5) #, 1
or (a + ^) x + (^ + #) a = 2 (a + x) 6, (a> ff).
or (a + b)x + (a + x)b=2(b+x)a,,
Determinate systems of equations of the first degree.
(I. 2. x+y = a, x = my.
li. 4. tf j/~a, x = my.
i. 3. a? +jy = 2, ^ = my + b.
m n
16  JL _I A
m n
. 12. ^ + a: 2 = y l +1*2 = a, ^ =
i. 15. x + a m (y a), y + b n (x b).
Ji. 16. y + s = a,z + x
/i. 18. y + zx=a, z
I. 19. y + z + w x = a, z
x + y + z  ze' = d.
I, 20. ic + y + z = a, x +y = #20, j; + = nx.
i i i
I. 21* #
77*
Determinate systems of equations reducible to the first degree
L 26. ## = a 2 , bx = oi.
1.29. ic + JK = a, x* y 1 *  b.
* Probably spurious.
CONSPECTUS OF ARITHMET1CA 261
i. 31. x = my, x 2 fjy 2 = n (x +y).
i. 33. x = my, x*y* =
i. 34. x = my, x*y 2 =
i. 34. Cor. T. x = my, xy = n(x
Cor. 2. x = wjy, #y = # (# j').
:i. 35. x = my 9 jP=nx.
i. 36. x = my, y^ ny,
i. 37. # = #zy, j/ 2 = n (x +y).
i. 38. x= my, y = ;/ (x y).
1.38. Cor. x = my 9 x*=ny.
X = OTX, x* = nx.
a; = #y/, tf 2 = (x y).
n. 6*. # j/ = a, x? ~y = jc j/ f ^.
iv. 36. yz = m (y + z), zx = n (z + #), ^y =/ (x +y).
Determinate systems reducible to equations of second degree.
I. 28. x+y = a, y?+y* = b,
iv. i. x*+y* = a, x \yb.
iv. 2. x 3 y* = a, xy = b.
iv. 15. (y t z) x = <2, (2 + 3;)j; = ^, (ie+j;)2; = ^
IV  34 yz + (j> + z) = a*ij zx + (z + x) Fi, xy + (x +y) = c* i.
IV  37 yzm(x +y + z), zx = n (x + y + s), xy =p (x +y + 2).
Lemma to v. 8. yz = a 2 , zx = b*, xyc*.
Systems of equations apparently indeterminate but really reduced, by
arbitrary assumptions, to determinate equations of the first degree.
1.14. xy = m(x+y) [value of y arbitrarily assumed].
/ ii. 3*. xy = m (x +y) and xy = m (x y)
I n. t*. (cf. i. 31.) x*+f = m(x+y)
j n. 2*. (cf. I. 34.) *f = m(X'y)
n. 4*. (cf. I. 32.) x*+y* = m (xy)
' n. 5*. (cf. I. 33.) x*f = m (x +y)
ii. 7*. x 2 y z = m(x y) + a [Diophantus assumes x y = 2].
ii ii ii r .. ,_
1.22. x x + z=y y +  # = ** + y I value of y assumed \.
M p n m j) H L J
i ii
+ w=y y f x
q * n j m
ii i i r , r ,,
= ^~  +y = w w + z value of v assumed).
p n q p L ^ J
* Probably spurious.
[x assumed = zy\.
262 CONSPECTUS OF ARITHMETIC A
T
r
[value of y 4 z assumed].
1.24. x+
i. 25. x + ^ 0' + z + a/) =J +  (
(w + ac+j;)
[value ofy + 2 + w assumed].
= s +  (ze; + # + j/) = w f  (x +y + z)
11.17*. (cf. I. 22.) tfftf+fl J + \I
x+a ) =
[ratio of # to y assumed]
IV. 33.
Diophantus assumes j> = i .
L z J
Indeterminate equations of the first degree.
Lemma to iv. 34. xy + (x +y) = a} [Solutions lv &opurr<p.
ii IV  35 ^7 ~ (# + J') = fl r J 7 practically found in
iv. 36. xy = mj(x+y) J terms of x.]
Indeterminate analysis of the second degree.
(u. 8.
n. 9.
I II. 10.
n. n.
n. 12. ax=u*, bx = tf.
n. 13. xa = u* 9 xb = &.
II. 14 = 111.21. #+jp = #, jc i s 2 = w 2 , j; 4 =
II. 15 = 111. 20. # +j/ = cz, &*x = u?, zPy^
ll. 16. ^ = wy, a 2 + x = w 2 , a 2 +^ = ^.
11.19. ^c 2 y = ^(jV 2  2 2 )
II. 2O. ^C 2 +J^ = W 2 , J^ + X = Z* 3 .
II. 21. x z y = u*, f~x = tf.
f II. 22. ^+(^+j)=W 2 , /+(^4J/) = ^.
II. 23. # 2  (x + j;) = w 2 , y 5  (* + y) = zr 2 .
II. 24. (x +y) z + x = u\ (x +y)* +yv*.
( II. 25. (x +j/) 2 x=u\ (x + j;) 2 j; = ^.
II. 26.
n. 27.
t n. 28.
I n. 29.
Probably spurious.
CONSPECTUS OF ARITHMETICA
263
II. 31.
( II. 32.
t 11.33.
n. 34.
n. 35.
in. i*.
im. 2*.
in. 3*.
nr. 4*.
in. 5.
in. 6.
in. 7.
pii. 8.
in. 9.
in. 10.
in. u.
fm. 12.
I in. 13.
in. 14.
fill. 15.
lin. 16.
C in. 17.
(in. 18.
(x +y) = u\ xy (x +y) = v\ x+y = w\
y*z = u\ z~x = v\ x*y = w.
s? 4 (x + y + 2) = &? 2 , y 2 + (x +y + 2) = zr 2 , z 2 + (# + y +
tf 2  (# +jv + s) = 2* 2 , y*(x +y + z} = v\  (x
)  # 2 = w 2 , (^ +j/ + s) j; s = # 2 , (^ + y +
) 2 + re  a 2 , (jc +JK 4 ^) 2 +y = ??, (x +y +
(x +y + z)* x = u*, (x +y + zf y = tf, (x +y
x(x +y + 0) a = 2 , 7 
#+_y + 2 =/ 2 , j/45 ^c
x +y + z ^ y + z = u\ z + x = v* s x +y = y
# jv =j^ ~ z, y + z = # 2 , 2? + ^ = z; 2 ,
x+y + z + <2 = / 2 , y + z + a =
# H jy + s a P, y {z a u 2 , z + x a v^^ x + y 
= 2/ 2 , 5^p +y = v\ xy + z w
x = u z , zxy v*, xy~z=w
v 2 , xy +z* =
+ z = 2^ 2 , 25C +
j/2 ( y + z) = u*, zx
xy + (# + y) = u* 9 xy + x = v*, xy +y
xy  (x + y) = u\ xyx = v 2 , xyy
in. 19. (x 1 + x. 2 + x* + xtf Xi= j J
: 4 ) 2 ^ 2
# 3 =
riv. 4.
1 iv. 5.
iv. 13,
iv. 14.
riv. 16.
[iv. 17.
iv, 19.
IV. 20.
iv. 21.
flV. 22.
Uv. 23.
= u,
^2^3 + I =
I = ^ XiX 2 + I =
Probably spurious,
264 CONSPECTUS OF ARITHMETICA
f IV. 29. X* +y' + Z~ + Ur + # + y + Z 4 / = <2.
t iv. 30. x f y + 2 + w 1  (x +y + s + w) = a.
iv. 31. x +y = i, (x + #) (_r f />) = z* 2 .
IV. 32. .V 4 J + 2 = (7, #V + S = 2 , A^'  = Z' 2 .
iv. 39. x v = m (y  s), y + z = u*, s + x = v*
iv. 40. x r = m (y  5), y + s = ?/, s + x =
C v. i. .rs^y 3 , x a = u\ y a = v*, z a~
I v. 2. jus =jt 2 , a* + a = w 2 , J J + a = v~, s + a =
v. 3. ^ + a = r 2 , j + = J 2 , 5 + ^z^/ 3 ,
j^'s + a = z/ 2 , s^; + a = y 2 , ^y t a = w 2 .
v. 4. # a = r\ y  a = J 2 , s  a P.
yz a = M\ sx a = zP, xy  a = 2# 2 .
v. 5.
V. 6. X 2 = r 3 , y 2 = J 2 , Z 2 = /J 2 ,
j/s  j; z u?, zx zx^v 2 , xy x y = ze/ 2 ,
J/JST x = z/ 2 , 2:^? y =v' 2 , xy z = w'\
Lemma i to v. 7. xy + #* + y 2 = w 2 .
v. 7.
v. 8. yg(x+j> + z) = 2 , ^ (* f y + g) =
V. 9. (cf. ii. ii.)
v. ii. ^+j' + ^=i
V. 10. x+y=i, x + a = u\ y + b = v\
v. 12. a; + y + z = i, x + a = 2 , y + 3 = zr 1 ,
V. 13.
V. 14.
a: + y + 5 = j 2 , y + z + a' = / 2 , 2; + ^ + x = ^ 2 , a/ + x +y
v. 21.
v. 22.
v. 23.
v. 24.
v. 25.
V. 26.
v. 27.
v. 28.
v. 30.
Lemma 2 to vi. 1 2. ax* + b = u* (where a + ^ = r>).
Lemma to VL 1 5 . a# s  <5 = ^ ^ w h er e o^ 3  b = ^ is known).
CONSPECTUS OF ARITHMETICA
([111.15]. xy + x + y = u 2 , x f 1 =  9 (y + i).
[iv. 32]. o?+ i =(*!).
Indeterminate analysis of the third degree.
fiv. 6.
I iv. 7.
fiv. 8. a
liv. 9. a;
fiv. 10. x s +y s = x+y
iv. IT. a^j^tf ~_y]
iv. 12. a^Jj; =y s + x)
iv. 1 8. x 3 +y = rf, y* + x =
iv. 24.
iv. 25.
fiv. 26.
265
3 = u\ x + y  u
, , .
the same problem
r
Really reducible to second
degree.
(x >y > z).
liv. 27.
iv. 28.
[iv. 28].
iv. 38. (x +y + z)x = \u (u + i), (x +y + z)y = p 2 ,
v. 15. (x +y + zf + x = w 3 ,
v. 16. (#H
v. 17. 
v. 1 8.
2? = z
= ^ 2 , (a? +y + jsr) 3
(x +y + z)* +z =
v. 19.
v.
v.
v. 20.
[iv. 8]. x
[iv. 9, 10].
[iv. u]. *
= u\ y(x+y + z) s = iP,
,
(x +y + zf 4 z
y = tf,
(x +y + z)* z
0) 3 = ?; 2 ,
 (x +y + ^) 3
266 CONSPECTUS OF ARITHMETICA
[v. 15].
[v. 16]. 3 (
[V. 17].
Indeterminate analysis of the fourth degree,
v. 29.
[v. 18].
Problems of constructing rightangled triangles with sides in rational
numbers and satisfying various other conditions.
[N.B. I shall use x, y for the perpendicular sides and z for the
hypotenuse in all cases, so that the condition x* +y z = z* must be under
stood to apply in every case in addition to the other conditions specified.]
Lemma to v. 7. xy =
fvi. i. zx = tf, zy=
tvi. 2.
[vi. 3.
\ vi. 4.
(vi. 5.
Jvi. 6,
Ivi. 7. ^y x = a.
fvi. 8. Jo;_y + (x +y) =
Ivi. 9.
[vi. 10.
tvi. ii.
Lemma i to vi. 12.
fvi, 12.
tvi. 13.
(vi. 14.
vi. 15.
vi. 16. + TI = X,
fvi. 17.
\vi. 18.
fvi. 19.
Ivi. 20.
fvi. 21. #+^ + Jsr = # 2 , lay + (^c +j; + z) = V s .
Ivi. 22. x +y f js = ^ 3 , J^y + (^ +y + 5f) = ^.
vi. 23.
vi. 24,
[vi. 6, 7].
[vi. 8,9]. $(x+y)}*+ \rnxy = u\
[vi. 10, 1 1]. { (z + jc)} a + \rnxy =
[vi. 12], y + (xy).%xy = u\ x =
[vi. 14, 1 5]. u*zx ^xy.x(zx) = ^ (u* < or >
JU11 LlilVlliiN 1
ADDITIONAL NOTES, THEOREMS AND PROBLEMS BY FERMAT,
TO WHICH ARE ADDED SOME SOLUTIONS BY EULER
I HAVE generally referred to the notes of Fermat, and allied propositions
of his, on the particular problems of Diophantus which were the occasion
of such notes, illustrations or extensions ; but there are some cases where
the notes would have been of disproportionate length to give in the places
where they occur. Again, some further explanations and additional
theorems and problems given by Fermat are not in the notes to Diophantus
but elsewhere, namely in his correspondence or in the Doctrinae Anafyticae
Inventum Novum of Jacques de Billy "based on various letters sent to
him from time to time by Pierre de Fermat" and originally included at the
beginning of the 2nd (1670) edition of Sachet's Diophantus (the Inventum
Novum is also published, in a free French translation by Tannery, in
Oeuwes de Fermat, Vol. in, pp. 323398), Some of these theorems and
problems are not so closely connected with particular problems in Dio
phantus as to suggest that they should be given as notes in one place
rather than another. In these circumstances it seemed best to collect the
additional matter at the end of the book by way of Supplement.
In the chapter on the Porisms and other assumptions in Diophantus
(pp. 106110 above) I quoted some famous propositions of Fermat on the
subject of numbers which are the sums of two, three or four square numbers
respectively. The first section of this Supplement shall be devoted to
completing, so far as possible, the story of Fermat's connexion with these
theorems.
SECTION I.
ON NUMBERS SEPARABLE INTO INTEGRAL SQUARES.
As already noted, Fermat enunciated, on Diophantus iv. 29, a very
general theorem of which one part states that Every number is either a
square or the sum of two, three or four squares. We shall return to this
later, and shall begin with the case of numbers which are the sum of
two squares.
266 CONSPECTUS OF ARITHMETICA
[v. 15]. X s +y* f z 3 3 = u\
[v. 1 6]. 3(
[v. 17],
Indeterminate analysis of the fourth degree,
v. 29.
[v. 18].
Problems of constructing rightangled triangles with sides in rational
numbers and satisfying various other conditions.
[N.B. I shall use #, y for the perpendicular sides and z for the
hypotenuse in all cases, so that the condition x* +y* = z 2 must be under
stood to apply in every case in addition to the other conditions specified.]
Lemma to v. 7. xy = x^ = x 2 y z .
(vi. i. z x = u 3 , z~y = v*.
(.VI. 2.
I vi. 3.
vi. 4.
vi. 5.
Jvi. 6.
Ivi. 7.
Jvi. 8.
Ivi. 9.
Jvi. 10.
Ivi. ii. ^ny; (x + z) = a.
Lemma i to vi. 1 2. x = u*, x y =
(vi. 12.
Ivi. 13.
vi. 14.
15.
vi. 16.
1. 17.
.vi. 18.
r
Iv:
c
<s
19.
I vi. 20. ^xy + x =
VI. 21.
22.
vi. 23.
VI. 24,
[vi. 6, 7].
[vi. 8, 9], { (# +j/)} 2 + \rnxy = u\
[vi. 10, n], {^ (z { x)} z + ^mxy ~ u\
[vi. 12]. y f (x y) . \xy  u 2 , x = v* (x>y)
[vi. 14, 15]. uhx \xy.x(zx)^v^ (i/ < or >
SUPPLEMENT
ADDITIONAL NOTES, THEOREMS AND PROBLEMS BY FERMAT,
TO WHICH ARE ADDED SOME SOLUTIONS BY EULER
I HAVE generally referred to the notes of Fermat, and allied propositions
of his, on the particular problems of Diophantus which were the occasion
of such notes, illustrations or extensions ; but there are some cases where
the notes would have been of disproportionate length to give in the places
where they occur. Again, some further explanations and additional
theorems and problems given by Fermat are not in the notes to Diophantus
but elsewhere, namely in his correspondence or in the Doctrinae Analyticae
Inventum Novum of Jacques de Billy " based on various letters sent to
him from time to time by Pierre de Fermat " and originally included at the
beginning of the 2nd (1670) edition of Sachet's Diophantus (the Inventum
Novum is also published, in a free French translation by Tannery, in
Oeuvres de Fermat, Vol. in. pp. 323398). Some of these theorems and
problems are not so closely connected with particular problems in Dio
phantus as to suggest that they should be given as notes in one place
rather than another. In these circumstances it seemed best to collect the
additional matter at the end of the book by way of Supplement
In the chapter on the Porisms and other assumptions in Diophantus
(pp. 106110 above) I quoted some famous propositions of Fermat on the
subject of numbers which are the sums of two, three or four square numbers
respectively. The first section of this Supplement shall be devoted to
completing, so far as possible, the story of Fermat's connexion with these
theorems.
SECTION I.
ON NUMBERS SEPARABLE INTO INTEGRAL SQUARES.
As already noted, Fermat enunciated, on Diophantus iv. 29, a very
general theorem of which one part states that Every number is either a
square or the sum of two> three or four squares. We shall return to this
later, and shall begin with the case of numbers which are the sum of
two squares,
2 68 SUPPLEMENT
i. On* numbers which are the sum of two squares.
I may repeat the beginning of the note on m. 79 already quoted (p. 106).
" A prime number of the for^ ^i +i is the hypotenuse of a rightangled
triangle in one way only, its square i^^sp in two ways, its cube in three, its
biquadrate in four ways, and so on ad injfnitum.
"The same prime number 472+1 and its square are the sum of two
squares in one way only, its cube and its biquadrate in two ways, its fifth
and sixth powers in three ways, and so on ad infinitum.
" If a prime number which is the sum of two squares be multiplied into
another prime number which is also the sum of two squares, the product
will be the sum of two squares in two ways ; if the first prime be multiplied
into the square of the second prime, the product will be the sum of two
squares in three ways ; if the first prime be multiplied into the cube of the
second, the product will be the sum of two squares in four ways, and so on
ad infinitum"
Before proceeding further with this remarkable note, it is natural to
ask how Fermat could possibly have proved the general proposition that
(a) Every prime number of the form. 47111 is the sum of two square
numbers, which was actually proved by Euler 1 . Fortunately we have
in this case a clear statement bj Fermat himself of the line which his
argument took. In his " Relation des nouvelles decouvertes en la science
des nombres" sent by Fermat to Carcavi and shortly after (14 August,
1659) communicated by the latter to Huygens, Fermat begins by a refer
ence to his method of proof by indefinite diminution (descente infinie or
indefinie) and proceeds 2 thus: "I was a long time before I was able to
apply my method to affirmative questions because the way and manner
of getting at them is much more difficult than that which I employ with
negative theorems. So much so that, when I had to prove that every
prime number of the form 4^+1 is made up of two squares^ I found myself
in a pretty fix. But at last a certain reflection many times repeated gave
me the necessary light, and affirmative questions yielded to my method,
with the aid of some new principles by which sheer necessity compelled me
to supplement it. This development of my argument in the case of these
affirmative questions takes the following line : if a prime number of the
form 473 + i selected at random is not made up of two squares, there will
exist another prime number of the same sort but less than the given
number, and again a third still smaller and so on, descending ad infinitum^
until you arrive at the number 5 which is the smallest of all numbers of
1 Novi Commentarii Academiae Petropolitanae 1752 and 1753, Vol. iv. (1758),
PP 3~4 1754 and 1755, Vol. v. (1760), pp. $$=Cowmentations arithmetics
collectae, 1849, I. pp. 155173 and pp. 210233.
2 Oeuvres de Format, II. p. 432.
THEOREMS AND PROBLEMS BY FERMA'jr *w
the kind in question and which the argument would require not to be made
up of two squares, although, in fact, it is so made up. From which we
are obliged to infer, by reductio ad absurdum, that all numbers of the kind
in question are in consequence made up of two squares."
The rest of the note to Diophantus in. 19 is as follows.
" From this consideration it is easy to deduce a solution of the problem
" To find in how many ways a given number can be the hypotenuse of
a rightangled triangle.
" Take all the primes of the form 472+ i, e.g. 5, 13, 17, which measure
the given number.
" If powers of these primes measure the given number, set out the
exponents of the powers ; e.g. let the given number be measured by the
cube of 5, the square of 13, and by 17 itself but no other power of 17 \
and set out the exponents in order, as 3, 2, i.
" Take now the product of the first of these and twice the second, and
add to the product the sum of the first and second : this gives 1 7. Multiply
this by twice the third exponent and add to the product the sum of 1 7 and
the third exponent: this gives 52, which is the number of the different right
angled triangles which have the given number for hypotenuse. [If a, &, c be
the exponents, the number of the triangles is ^abc + 2(bc + ca + ab) + a + b + c.~\
We proceed similarly whatever the number of divisors and exponents.
"Other prime factors which are not of the form 4^+1, and their
powers, do not increase or diminish the number of the rightangled triangles
which have the given hypotenuse.
" PROBLEM i. To find a number which is a hypotenuse in any assigned
number of ways.
"Let the given number of times be 7. We double 7 : this gives 14.
Add i, which makes 15. Then seek all the prime numbers which measure
it, Le. 3 and 5. Next subtract i from each and bisect the remainders.
This gives i and 2. [In explanation of the process it is only necessary to
observe that, for example, 2 {^abc+ 2 (&c + ca + ab) + a + b + \ + i is equal
to (204 i)(2b + 1) (2c+ i), and so on.] Now choose as many prime
numbers of the form 472 t 1 as there are numbers in the result just arrived
at, i.e. in this case two. Give to these primes the exponents i, 2 re
spectively and multiply the results, Le. take one of the primes and multiply
it into the square of the other.
" It is clear from this that it is easy to find the smallest number which
is the hypotenuse of a rightangled triangle in a given number of ways."
[Fermat illustrates the above further in a letter of 25 December 1640
to Mersenne 1 '.
To find a number which is the hypotenuse of 367 different rightangled
triangles and no more.
1 Oeuvres de Fermat ', II. pp. 214 sq.
2 68
SUPPLEMENT
Double the number and add i ; this gives 735. Take all the divisors
which are prime numbers : these are 3, 5, 7, 7. Subtract i from each and
then divide by 2 ; this gives i, 2, 3, 3. We have then to take four prime
numbers of the form 4 i 1 and give them i, 2, 3, 3 respectively as ex
ponents. The product of these powers is the number required.
To find the least such number^ we must take the four least primes of the
form 4 + i, i.e. 5, 13, 17, 29, and we must give the smallest of them,
in order, the largest exponent; I.e. we must take 5 s , 13*, if and 29 in this
case, and the product of these four numbers is the least number which is
the hypotenuse of 367 rightangled triangles and no more.
If the double of the given number + i is a prime number, then there is
only one possible divisor. Suppose the given number is 20; the double
i it plus i is 41. Subtracting unity and bisecting, we have 20, so that the
number to be taken is some prime number of the form 4/4 + 1 to the power
of 20,]
" PROBLEM 2. To find a number which shall be the sum of two squares
in any assigned number of ways.
" Let the given number be 10. Its double is 20, which, when separated
into its prime factors, is 2.2.5. Subtract i from each, leaving .1, i, 4.
Take three different prime numbers of the form 473+ i, say 5, 13, 17, and
multiply the biquadrate of one (the exponent being 4) by the product
of the other two. The result is the required number.
" By means of this it is easy to find the smallest number which is the
sum of two squares in a given number of ways.
" In order to solve the converse problem,
" To find in how many ways a given number is the sum of two squares,
"let the given number be 325. The prime factors of the form 4;; + i
contained in this number are 5, 13, the latter being so contained once only,
the former to the second power. Set out the exponents 2, i. Multiply
them and add to the product the sum of the two : this gives 5. Add i,
making 6, and take the half of this, namely 3. This is the number of ways
in which 325 is the sum of two squares.
"If there were three exponents, as 2, 2, i, we should proceed thus.
Take the product of the first two and add it to their sum : this gives 8.
Multiply 8 into the third and add the product to the sum of 8 and the
third: this gives 17. Add r, making 18, and take half of this or 9. This
is the number of ways in which the number taken in this second case is
the sum of two squares. [If #, , c be the three exponents, the number
of ways is \ {afc+ (fc+a + aS) + (a + 6 + c)+i} provided that the number
represented by this expression is an integer.]
"If the last number which has to be bisected should be odd, we
must subtract i and take half the remainder.
27*
THEOREMS AND PROBLEMS BY FERMAT
" But suppose we are next given the following problem to solve :
"To find a whole number which, when a given number is added to zV,
becomes a square, and which is the hypotenuse of any assigned number of
rightangled triangles.
" This is difficult. Suppose e.g. that a number has to be found which
is a hypotenuse in two ways and which, when 2 is added to it, becomes
a square.
" The required number will be 2023, and there are an infinite number
of others with the same property, as 3362 etc."
2. On numbers which cannot be the sum of two squares.
In his note on Diophantus v. 9 Fermat took up a remark of Bachet's
to the effect that he believes it to be impossible to divide 21 into two
squares because " it is neither a square nor by its nature made up of two
squares." Fermat's note was: "The number 21 cannot be divided into
two squares (even) in fractions. That I can easily prove. And generally
a number divisible by 3 which is not also divisible by 9 cannot be divided
into two squares either integral or fractional."
He discusses the matter more generally in a letter of August 1640
to Roberval 1 .
"I have made a discovery a propos of the i2th [gth] proposition of
the fifth Book of Diophantus (that on which I have supplied what Bachet
confesses that he did not know and at the same time restored the corrupted
text, a story too long to develop here). I need only enunciate to you my
theorem, while reminding you that I proved some time ago that
"A number of the form 4^1 is neither a square nor the sum of two
squares, either in integers or. fractions"
[This proposition was sent by Mersenne to Descartes, on 22 March
1638, as having been proved by Fermat.]
" For the time I rested there, although there are many numbers of the
form 472 + i which are not squares or the sums of squares either, e.g. 21,
33> 77? etc ' a f act which made Bachet say on the proposed division of 21
into two squares 'It is, I believe, impossible since 21 is neither a square
nor by its nature made up of two squares/ where the word rear (I think)
clearly shows that he was not aware of the proof of the impossibility.
This 1 have at last discovered and comprehended in the following general
proposition.
" If a given number is divided by the greatest square which measures it,
and the quotient is measured by a prime number of the form 40 1, the given
number is neither a square nor the sum of two squares either integral or
fractional.
1 Oeuores de Fermat, II. pp. 2034.
^
SUPPLEMENT
" EXAMPLE. Let the given number be 84. The greatest square which
measures it is 4, and the quotient is 21 which is measured by 3 or by
7, both 3 and 7 being of the form 4^ i. I say that 84 is neither a square
nor the sum of two squares either integral or fractional.
"Let the given number be 77. The greatest square which measures it
is i, and the quotient is 77 which is here the same as the given number
and is measured by n or by 7, each of these numbers being of the form
4~i. I say that 77 is neither a square nor the sum of two squares,
either in integers or fractions.
" I confess to you frankly that I have found nothing in the theory of
numbers which has pleased me so much as the proof of this proposition,
and I shall be glad if you will try to discover it, if only for the purpose
of showing me whether I think more of my discovery than it deserves.
"Following on this I have proved the following proposition, which
is of assistance in the finding of prime numbers.
" If a number is the sum of two squares prime to one another, I say
thai it cannot be divided by any prime number of the form 4721.
"For example, add i, if you will, to an even square, say the square
10000000000, making 10000000001. I say that 10000000001 cannot
be divided by any prime number of the form ^n i, and accordingly,
when you would try whether it is a prime number, you need not divide by
3, 7, ii etc."
(The theorem that Numbers which are the sum of two squares prime to
one another have no divisors except such as are likewise the sum of two squares
was proved by Euler 1 .)
3. Numbers (i) which are always, (2) which can never be, the sum
of three squares.
(i) The number which is double of any prime number of the form
8n i is the sum of three squares (Letter to Ken elm Digby of June 1658)*.
E.g. the numbers 7, 23, 31, 47 etc. are primes of the form 8# i ; the
doubles are 14, 46, 62, 94 etc.; and the latter numbers are the sums of
three squares.
Fermat adds " I assert that this proposition is true, though I do so in
the manner of Conon, an Archimedes not having yet arisen to assert it
or prove it."
Lagrange 3 remarks that he has not yet been able to prove the pro
position completely. The form &n i reduces to one or other of the three
1 Novi Commentarii Acad. PetropoL 1752 and 1753, Vol. IV. (1758), pp. 340=
Commentations* arithmeticae^ I. pp. 155173.
2 Oettvres de Fermat, II. pp. 402 sqq.
3 "Recherches d'Arithm&ique" in Berlin Mtmoires 1773 and itf^Oeuvres de
Lagrange, III. p. 795.
THEOREMS AND PROBLEMS BY FERMAT 273
forms 24^  i, 24^ + 7, 2472 + 15, of which the first two only are primes.
Lagrange had previously proved that every prime number of the form
2472 + 7 is of the form x* + 6y\ The double of this is 2x* + i2jj/ 2 , and
2X* + 1 2J/ 2 = (X + 2yf + (X tyf + (2J/) 2 ,
that is, 2x*+ i2jy 2 is the sum of three squares.
The theorem was thus proved for prime numbers of the form 8^ i,
wherever n is not a multiple of 3, but not for prime numbers of the form
2473 i.
Legendre 1 , however, has the theorem that Every number which is the
double of an odd number is the sum of three squares.
(2) No number of the form 24/2 + 7 or 4 (2472 + 7) can be the sum
of three squares.
This theorem is substantially stated in Fermat's note on Dioph. v. n.
We may, as a matter of fact, substitute for the forms which he gives the
forms B?i + 7 and 4 (8^ + 7) respectively.
Legendre 3 proved that numbers of the form 8^ + 7 are the only odd
numbers which are not the sum of three squares.
4. Every number is either a square or the sum of two, three or
four squares.
This theorem is also mentioned in the "Relation des nouvelles de
couvertes en la science des nombres " already quoted, as a case to which
Fermat ultimately found himself able to apply the method of proof by
descente. He says 3 that there are some other problems which require new
principles in order to enable the method of descente to be applied, and the
discovery of such new principles is sometimes so difficult that they cannot
be arrived at except after very great trouble.
" Such is the following question which Bachet on Diophantus admits
that he could never prove, and as to which Descartes in one of his letters
makes the same statement, going so far as to admit that he regards it as
so difficult that he does not see any means of solving it.
"Every number is a square or the sum of two, three or four squares.
" I have at last brought this under my method, and I prove that, if
a given number were not of this nature, there would exist a number smaller
than it which would not be so either, and again a third number smaller
than the second, etc. ad infinitum ; whence we infer that all numbers are
of the nature indicated."
In another place (letter to Pascal of 25 September, 1654)*, after quoting
the more general proposition, including the above, that every number is
1 Legendre, Zahlentheorie, tr. Maser, i. p. 387.
2 Ibid. p. 386.
3 Oeuvres de Permed, II. p. 433.
4 Ibid. p. 313.
H. D. l8
SUPPLEMENT
made up (i) of one, two, or three triangles, (2) of one, two, three or four
squares, (3) of one, two, three, four or five pentagons, and so on adinfinitum,
Fermat adds that " to arrive at this it is necessary
(1) To prove that every prime number of the form 4^+ i is the sum
of two squares, e.g. 5, 13, 17, 29, 37, etc. ;
(2) Given a prime number of the form 4^+1, as 53, to find, by a
general rule, the two squares of which it is the sum.
(3) Every prime number of the form 3*2+1 is of the form x? 4 y 2 ,
*# 7> J 3> Z 9> 3*1 37> *&
(4) Every prime number of the form Sn + i or Sn f 3 is of the form
*? + 2}? 9 e.g. ii, 17, 19* 4i. 43> **'
(5) There is no rational rightangled triangle in whole numbers the
area of which is a square.
"This will lead to the discovery of many propositions which Bachet
admits to have been unknown to him and which are wanting in Diophantus.
"I am persuaded that, when you have become acquainted with my
method of proof in this kind of proposition, you will think it beautiful, and
it will enable you to make many new discoveries, for it is necessary, as you
know, that multi pertr'anseant ut augeatur sciejitia [Bacon]."
Propositions (3) and (4) will be mentioned again, and a full account
will be given in Section in. of this Supplement of Fermat's method, or
methods, of proving (5).
The main theorem now in question that every integral number is the
sum of four or fewer squares was attacked by Euler in the paper 1 (1754
1755) m which he finally proved the proposition (i) above about primes
of the form 4^+1; but, though he obtained important results, he did not
then succeed in completing the proof. Lagrange followed up Euler's
results and finally established the proposition in i77o 2 . Euler returned
to the subject in 1772 ; he found Lagrange's proof long and difficult, and
set himself to simplify it 3 .
(The rest of the more general theorem of Fermat quoted above, the
portion of it, that is, which relates to numbers as the sum of ;/ or fewer
ngonal numbers^ was proved by Cauchy 4 .)
1 Novi Commentarii Acad. PetropoL for 17545, Vol. v. (1760), pp. 358 =Com
mentationes arithmeticae collcctae, 1849, I. pp. 210233.
2 Nouveavx Mtmoires de VAcad. Roy. des Sciences de Berlin, annee 1770, Berlin 1772,
pp. 123133= Oeuvres de Lagrange, III. pp. 187201: cf. Wertheim's account in his
Diophantus, pp. 324330.
3 " Novae demonstrationes circa resolutionem numeromm in quadrata," Acta Erudit.
Lips. I773>P J 93; Acta Petrop* r. n. 1775, p. 48; Comment, arithm. I. pp. 538548.
4 Cauchy, "Demonstration du theoreme ge"n<ral de Fermat sur les nombres polygones,"
O&uvres, n e Serie, Vol. vi. pp. 320353. See also Legendre, Zahlentheorie, tr. Maser,
" PP 332343
THEOREMS AND PROBLEMS BY FERMAT 275
Under this heading may be added the further proposition that
" Any number whatever of the form 8;z i can only be represe?ited as the
sum of four squares, not only in integers (as others may have seen) but in
fractions also, as I promise that I will prove 1 ."
5. On numbers of the forms x? + 2y, o<? + $y\ x* + $y* respectively*
(1) Every prime number of the form %n + i or 8 + 3 is of the form
y? + 2y\
This is one of the theorems enunciated in the letter of 25 Sept., 1654,
to Pascal 2 and also in the letter of June, 1658, to Kenelm Digby 8 .
[In a paper of 1754 Euler says that he does not yet see his way to
prove either part of the theorem 4 . In 1759 he says 5 he can prove the
truth of the theorem for a prime number of the form 8n + i, but not for
a prime of the form 8^ + 3. Later, however, he proved it for prime
numbers of both forms 6 . Lagrange 7 also proved it for primes of the form
87243.]
(2) Every prime number of the form 3*7 + 1 is of the for?n x? + 3^.
The theorem is stated in the same two letters to Pascal and Digby
respectively.
Lagrange naturally quotes it as "All prime numbers of the form 6 + i
are of the form x* + $y\" for of course 3^ + i is not a prime number unless
n is even.
The proposition was proved by Euler 8 . Lagrange proved 9 (a) that all
prime numbers of the form 12^5 are of the form x*+ sy*, (8) that all
prime numbers of the form 12/2 i are of the form 3# 2 y 2 , and (c) that
all prime numbers of the form 12^ + 1 are of both the forms x? 4 3jv 2 and
*&
(3) No number of the form 3^1 can be of the form of + 3J/ 2 .
In the "Relation des nouvelles de'couvertes en la science des nombres 10 "
Fermat says that this was one of the negative propositions which he proved
by his method of descente.
1 Letter to Mersenne of Sept. or Oct. 1636, Oeuvres de Fermat, II. p. 66.
2 Oeuvres de Fermat, u. p. 313.
3 Ibid. n. p. 403.
4 u Specimen de usu observationum in mathesi pura (De numeris formae iaa + bb} " in
Novi Commentarii Acad. Petrop. 17567, Vol. VI, (1761), pp. 185230= Comment.
arithm. I. pp, 174192.
fi Novi Commentarii Acad. Petrop. 17601, Vol. VIII. (1763), pp. 1268 = Comment.
arithm. I. p. 296.
6 Commentationes arithmeticae^ n. p. 607.
7 "Recherches d' Arithme'tique " in Oeuvres de Lagrange, in. pp. 776, 784.
8 " Supplementum quorundam theorematum arithmeticorum, quae in nonnullis de
monstrationibus supponuntur (De numeris formae aa + ^bb}" in Nwi Comment. Acad.
Pttrop. 17601, Vol. VIII. (1763), pp. 105128 = Comment, arithm. I. pp. 287296.
9 Op. cit., Oeuvres de Lagrange,, ill. pp. 784, 791.
lu Oeuvres de Fermat^ n. p. 431.
18 2
^76 SUPPLEMENT
(4) If two prime numbers ending in either 3 or 7 which are also of the
form 4*3+3 are multiplied together^ the product is of the form x* + 5jv 2 .
This theorem also is enunciated in the letter of June, 1658, to Ken elm
Digby. Fermat instances 3, 7, 23, 43, 47, 67 etc. as numbers of the kind
indicated. Take, he says, two of these; e.g. 7 and 23. The product 161
will be the sum of a square and 5 times another square, namely 81 + 5 . 16.
He admits, however, that he has not yet proved the theorem generally :
" I assert that this theorem is true generally, and I am only waiting for
a proof of it. Moreover the square of each of the said numbers is the sum of
a square and 5 times another square : this, too, I should like to see proved."
Lagrange proved this theorem also 1 . He observes that the numbers
described are either of the form 20/2 + 3 or of the form 2072 + 7, and he
proves that all prime numbers of these forms are necessarily of the form
2^2^j + 3/. He has then only to prove that the product of two
numbers of the latter form is of the form x* + 5/.
This is easy, for
2xy + 3/) (2x'' 2 + 2x'y + 3/ 2 ) = (2x3? + xy'+yx' + ^yyj + 5 (xy f yx')*.
6. Numbers of the forms y?*f and 2x?y\
* Fermat's way of expressing the fact that a number is of one of these
forms is to say that it is the sum of, or the difference between, the two smaller
sides, .e. the perpendicular sides, of a rightangled triangle. Like Diophantus,
he " forms" a rational rightangled triangle from two numbers x, y, taking
as the three sides the numbers x 2 + y>, x 2 ^ 2xy respectively. The sum
therefore of the perpendicular sides is x 2 + 2xyjP or (x+yf 2y\ and
their difference is either x 2 2xy}? * 2xy(x*y*), that is, either
(x y)*  2^ or 2JV 2  (x jy) 3 .
The main theorem on the subject of numbers of these forms is, as
a matter of fact, contained, not in a letter of Fermat's, but in two letters
of Fre'nicle to Fermat dated 2nd August and 6th Sept., 1641, respectively 2 .
It is, however, clear (cf. the letter in which Fermat had on i5th June, 1641,
propounded to Fre'nicle a problem on such numbers) that the theorem was
at any rate common property between the two.
Frenicle's two statements of the theorem are as follows :
" Every prime number of the form 8# i is the sum of the two smaller
sides of a (rightangled) triangle, and every number which is the sum of the
two smaller sides of a (rightangled) triangle with sides prime to one another
is of the form Sn i."
"Every prime number of the form 8i, or which is the product
of such prime numbers exclusively, is the difference between the two
smaller sides of an infinite number of primitive rightangled triangles."
1 Op. dt., Oeuvres de Lagrange, n. pp. 784, 7889.
2 Oeztvres de Fermat, ] I. pp. 231, 235.
THEOREMS AND PROBLEMS BY FERMAT 277
Lagrange 1 quotes the theorem in the form
All prime numbers of the form Sn i are of the form y 2  2t\
Lagrange himself proves 2 that all prime numbers of the form Bn i are
of both the forms x?  2y 2 and 23? y 2 , and observes 3 that this theorem is more
general than that of Fermat so far as prime numbers of the form 8#  i are
concerned. This, however, seems scarcely correct if the further explanations
given by Fre'nicle are taken into account. For Frenicle shows clearly,
in the second of the two letters referred to 4 , that he was fully alive to
the fact that numbers which are of the form x? 2j/ 2 are also of the form
2 # 2 jy 2 ; and indeed it is obvious that he was aware that
tf 2 2j/ 2 = 2 (x +y) 2 (x + zyf.
Lagrange proved in addition 5 that
Every prime number of the form Bn + i is at the same time of the three
forms y? + 2jy 2 ) x~ 2jy 2 , 2X? y*.
This is, I think, really included in Frgnicle's statements when combined
with Fermafs theorem (i) above to the effect that every prime number
of the form 8n + i is of the form x* + 2y\
The problem propounded by Fermat to Frenicle in connexion with the
numbers now under consideration was:
Given a number , to find in how many ways it can be the sum of the two
smaller sides of a rightangled triangle.
Fre'nicle replied that this involved also the problem of finding a number
which will be the sum of the two smaller sides of a rightangled triangle in
an assigned number of ways and no more, and tried, but unsuccessfully 6 ,
to bring these problems under a rule corresponding to that by which
Fermat found the number of ways in which a prime number of the form
4+i can be the hypotenuse of a rightangled triangle (see p. 269 above),
but with a prime number of the form 8 i substituted for the prime
number of the form 4^ + 1. I cannot find that Fermat ever communicated
his own solution, at all events in the correspondence which we possess.
SECTION II.
EQUATION X*Ay*=l.
History of the equation up to Fermat' s time.
Fermat was not the first to propound, or even to discover a general
method of solving, the problem of finding any number of integral values of
x, y satisfying the above equation, .wherein A is any integral number not
a square. But Fermat rediscovered the problem and was perhaps the first
1 Op* tit., Oeuvres de Lagrange, ill. p. 775. 2 Ibid. p. 784. 3 Ibid. p. 788,
4 Oeuvres de Fermat^ II. pp. 235240.
5 Op. dt. t Oeuvres de Logrange, III. p. 790.
6 See Oeuvres de format, n. pp. 331, 238 sqcj.
278 SUPPLEMENT
to assert that the general solution is always possible whatever be the
(nonsquare) value of A. The equation has a history of over 2000 years,
and that history, even in outline, requires, as it has now obtained, a book
to itself 1 . This note will therefore be confined, practically, to recalling, in
the briefest possible way, the recorded stages anterior to Fermat, and then
to setting out somewhat fully the passages in Fermat's writings which throw
the most light on his connexion with the subject.
The Pythagoreans.
We have seen (p. 117 above) that the Pythagoreans had already
discovered a general solution of a particular equation of this type, namely
2# 2 y = + i,
by which all the successive values of x 9 y satisfying the equation were
ascertained. If x =/, y = q satisfies the equation 2X 2 y = + i, they proved
that the equation 23? y 1 = + i is satisfied by
A=^ + ?
the equation 2X 2 ~y*  i again by
=
and so on. As/= r, ^=i satisfies 2^ 2 ~y = +i, we have all the suc
cessive solutions of 2^ 2 / = r by forming (j) 1} ft), (/ 2 , ft ) etc. in accord
ance with the law.
Archimedes,
The solution of the above equation by the Pythagoreans was evidently
used in order to obtain successive approximations to ^2.
Consequently, when we find Archimedes giving, without explanation, the
fractions f and 2fijfr as being approximately equal to ^3, the hypothesis of
Zeuthen and Tannery that he arrived at these approximations by obtaining
successive solutions of equations of a similar form, but with 3 substituted
for 2, is one of the most natural that have been suggested 2 . The equations
are in this case
Tannery shows how the law for forming successive solutions of such
simple cases as these can easily be found when we have found by trial
(which is not difficult) the three simplest solutions. If we take the more
general equation
y? ay* r,
1 H. Konen, Geschichted&r Gleickungt* Difi=i, Leipzig (S. Hirzel), 1901.
3 Zeuthen, " Nogle hypotheser om Arkhimedes kvadratrodsberegning," Tidsskrift for
Mathematik, vi. Raekke, 3. Aargang, pp. 150 sqq.; P. Tannery, "Sur la mesure du cercle
d'Archimede" in Mtmoires de la soc. des sciences phys. et not. de Bordeaux, lie Se"r. iv.
1882, p. 303; see Gunther, "Die quadratischen Irrationalitaten der Alten und deren
Entwickelungsmethoden" in Abhanflwgen zvr Gach. der Mathematik, Heft iv. 1882
pp. 8791; Konen, of, cit. p. 13,
THEOREMS AND PROBLEMS BY FERMAT 279
of which x p^ y q is a known solution, and put
A = <# + ? ffi = yp + fy,
it is sufficient to know the three simplest solutions in order to find a, /?, y,
S for, substituting the values of (p, q\ (p^ ft) and (/ a , ^ 3 ) where (/ 3 , $r 2 )
are formed from (/ 1? ft) by the same law as (p^ ft) are formed from (/, q\
we have four simultaneous equations in four unknown quantities. Taking
the particular equation
^3/=i,
we easily find the first three solutions, namely (/ = i, q o), (p l  2, ft = i)
and (/ 2 = 7, ^ 3 = 4), whence
2 = a, i=y,
7 = 2a + j8, 4
and a = 2, /? = 3, y = i, 8 = 2, so that
But there is evidence that Archimedes dealt with much more difficult
equations of the type, for (as stated above, p. 123) the Cattle Problem
attributed to him requires us to solve in positive integers the equation
3 s  47294947*= I.
There is this difference between this equation and the simpler ones,,
above that, while the first solutions of the latter can be found by trial,
the simplest solution of this equation cannot, so that some general method,
e.g. that of continued fractions, is necessary to find even the least solution
in integers. Whether Archimedes was actually able to solve this particular
equation is a question on which there is difference of opinion j Tannery
thought it not impossible, but, as the smallest values of #, y satisfying the
equation have 46 and 41. digits respectively, we may, with Giinther, feel
doubt on the subject 1 . There is, however, nothing impossible in the
supposition that Archimedes was in possession of a general method of
solving such equations where the numbers involved were not too great for
manipulation in the Greek numeral notation.
Diophantus.
Tannery 2 was of opinion that Diophantus dealt with the equation
tfAj? = j
somewhere in the lost Books of the Arithmetica. Diophantus does indeed
say (Lemma to vi. 1 5) that, if a, b are any numbers and ay? b is a square
when x is given a certain value/, then other values of x greater than p can
also be found which have the same property ; and Tannery points out that
1 Giinther, op. dt., pp. 9293 note. Cf. Konen, op. at., p. 14.
2 Tannery, " L'Arithm&ique des Grecs dans Pappus" in Mtmoires de la soc. des
sciences phys. et nat. de Bordeaux, n e Sr. ill., 1880, pp. 370 sq.
2 8o SUPPLEMENT
we can, by making suppositions of the same kind as Diophantus makes,
deduce a more general solution of the equation
x?  Af = i
when one solution (/, q) is known.
Put pi = mxft qi = x + q,
and suppose
p?  Aq? = mW  2mpx f/ 2  Acs?  zAqx  Af = i ;
therefore (since/ 2  Atf = i)
mp + Aq
X 2~  ~ ,
m*A
and, by substitution in the expressions for / 19 ft, we have
_ (m* + A)p + zAmq __ imp + (m* + A) q
A ~ ^^ ' ft ~ w 8 ^
and in fact/! 2  ^ 2 = i.
If an integral solution is wanted, one way of obtaining it is to substitute
?(/vfor m where u*Av=i, i.e. where u, v is another solution of the
original equation, and we then have
/ L = (fcj 2 + Av*)p + zAuvq^ qL ipuv + (u* + Av*) q.
But this is all that we can get out of Diophantus as we have him, and
it will be observed that here too we must have ascertained two solutions of
the one equation, or one solution of it and a solution of an auxiliary equation,
before we can apply the method 1 .
1 It may be observed that, in the particular case of the equation x 2 ~3y 2 i, the
assumption, of u, v satisfying the equation will not enable us to obtain from the formula
Pi = (#2 + ^^,2) p _f_ 2 ^^, ^ = ipuv + (z/ 2 + A v 2 ) q
above given the simpler formula otherwise obtained by Tannery (p. 279 above), namely
for, if (/i, sft) is to be a different solution from (/, q\ we cannot make u=i t v = o, but
must take u = i 9 v = i, whence, putting A^ we obtain
which is the same as/ 2 , ?2, the next solution to/i = 2
In order to get the latter we have to take u, v satisfying, not x 2  $y 2 = i, but
#2_ 3 yj = _ 2 .
The values u= i, v i satisfy x 2  $y*  a, and
and of course /i = +(^+3^), ^= h(/ + 2^) can be taken, since they equally satisfy
THEOREMS AND PROBLEMS BY FERMAT 281
The Indian Solution.
If the Greeks did not accomplish the general solution of our equation,
it is all the more extraordinary that we should have such a general solution
in practical use among the Indians as early as the time of Brahmagupta
(born 598 A.D.) under the name of the "cyclic method." Whether this
method was evolved by the Indians themselves, or was due to Greek
influence and inspiration, is disputed. Hankel held the former view 1 ;
Tannery held the latter and showed how, from the Greek manner of
deducing from one approximation to a surd a nearer approximation, it is
possible, by simple steps, to pass to the Indian method 2 . The question
presumably cannot be finally decided unless by the discovery of fresh
documents; but, so far as the other cases of solution of indeterminate
equations by the Indians help to suggest a presumption on the subject,
they are, I think, rather in favour of the hypothesis of ultimate Greek
origin. Thus the solution of the equation axt>y = c given by Aryabhata
(born 476 A.D.) as well as by Brahmagupta and Bhaskara, though it
anticipated Sachet's solution which is really equivalent to our method of
solution by continued fractions, is an easy development from Euclid's
method of finding the greatest common measure or proving by that process
that two numbers have no common factor (Eucl. vn. i, 2, x. 2, 3), and
it would be strange if the Greeks had not taken this step. The Indian
solution of the equation xy~ax + by + 1, by the geometrical form in which
it was clothed, suggests Greek origin 4 .
The "cyclic method" of solving the equation
is found in Brahmagupta and Bhaskara 15 (bora 1114 A.D.) and is well
described by Hankel, Cantor and Konen*.
The method is given in the form of dogmatic rules, without any proof
of the assumptions made, but is equivalent to a preliminary lemma followed
by the solution proper.
1 Hankel, Zur Geschichte der Math, wi Alterthitm und Mittelalter, pp. 2034.
2 Tannery, "Sur la mesure du cercle d'Archimede" in Mini, de fa soc. des sciences
phys. et nat. de Bordeaux, ii e Se"r, IV., 1882, p. 325; cf. Konen, pp. 2728; Zeuthen,
" L'Oeuvre de Paul Tannery comme historien des mathematiques " in Bibliotheca Mathe
matica, vi 3 , 19056, pp. 271273.
3 G. R. Kaye, "Notes on Indian mathematics, No. 2, Aryabhata" in Joitrnal of the
Asiatic Society of Bengal, Vol. IV. No. 3, 1908, pp. 135138.
4 Cf. the description of the solution in Hankel, p. 199; Cantor, Gesch. d. Math. I 3 ,
p. 631.
5 The mathematical chapters in the works of these writers containing the solution in
question are contained in H. T. Colebrooke's Algebra 'with arithmetic and mensuration
from the Sanskrit of Brahmegupta and Bhaskara^ London, 1817.
5 Hankel, pp. 200203; Cantor, I 8 , pp. 632633; Konen, op. ci.> pp. 1926,
282 SUPPLEMENT
Lemma,
If xp^ yq be a solution of the equation
Ay* + s = x?,
and x =/', y = q a solution of the equation
then, say the Indians, #=//' ^/, _y =// / ^ a solution of the equation
In other words, if
then A (pq p'qf f / = (//
This is easily verified 1 .
In particular, taking s = s f , we find, from any solution #=/, jv^ of
the equation
a solution x p^ + ^f 2 , jy = 2/^ of the equation
Again, particular use of the lemma can be made when s = i or 5 = + 2.
(0) If 5 1 = + i, and x =/, j = ^ is a solution of
4/4 I =*,
then jc =/ 3 + ^^, jv = 2/^ is another solution of the same equation.
If s  i, and x =/, y  q is a solution of
^yi=a*.
then # =/ 2 + ^^, j; = 2/^ is a solution of
^J/ 3 f I = tf 3 .
(^) If ^ = + 2, and x = ^, ji/ = ^ is a solution of
^/2=^,
then re =/ 2 + AqP, y = 2^ is a solution of
^ + 4 = ^.
In this case, since zpq is even, the whole result when the values of
re, y are substituted must be divisible by 4, and we have x \ (p* +
y =pq as a solution of the equation
Af + i = x*.
1 For, since s=p  ^^ 2 , /=/ 2  Aq'\
THEOREMS AND PROBLEMS BY FERMAT 283
Solution proper of the equation o^ Ay 1  i.
We take two numbers prime to one another, / 3 ^, and a third number s
with no square factor, such that
the numbers being also chosen (in order to abbreviate the solution) such
that s is as small as possible, though this is not absolutely necessary.
(This is a purely empirical matter; we have only to take a rough ap
proximation to <JA in the form of a fraction pjq.)
[It follows that J, q can have no common factor ; for, if 8 were a
common factor of s, q, it would also be a factor of/ 5 , and/ 2 , f would have
a common factor. But/, q are prime to one another.]
Now find a number r such that
^ =  is a whole number.
s
[This would be done by the Indian method called cuttaca (" pulveriser "),
corresponding to our method by continued fractions.]
Of the possible values of r a value is taken which will make r  A
as small as possible.
Now, say the Indians, we shall have :
is an integral number,
"
and
(Again the proofs are not given; they are however supplied by Hankel 1 .)
1 Since ^=1 j s an integral number, all the letters in qs=p + qr represent
integers.
Further, s=pAq*\
therefore, eliminating s, we have
or
Since p^ q have no' common factor, q must divide J>q\ i ; that is,
pq\  l
 =an integer.
We have next to prove that ^(r 2  A]\s is an integer.
since
therefore . j ? 1I is an integer.
and, since j, q have no common factor, it follows that
y#A ...
= j is an integer.
Also ji = ** ~ = V l * S " 2 ^ + T = rf ( ^ ~ A ^ ~ *&* + ' = /^M " i V a
S ^ q* \ q ) y j '
284 SUPPLEMENT
We have therefore satisfied a new equation of the same form as that
originally taken 1 .
We proceed in this way, obtaining fresh results of this kind, until we
arrive at one in which s = + i or + 2 or + 4, when, by means of the lemma,
we obtain a solution of
Ay 2 +i=x\
Example. To solve the equation 677* + i = tf 3 .
Since 8 2 is the nearest square to 67, we take as our first auxiliary equation
67. i*3 = 8 2 , so
Thus ft =   . We put r = 7, which makes ft an integer and at the
7 2 67
same time makes s 1 =    = 6 as small as possible.
o
Thus ft =  5, ft = (/ft  i)fq =  41,
and we have satisfied the new equation
Next we take ? 8 = , ? and we put 73 = 5, giving ft=n; thus
6
r* 67
3 =  7 and fa = (A&  i)/ft = 9, and
Next f 3 =    , and we put r% = 9, giving ^ =  27 ; therefore
90.271
=   = 
,
1, and
As we have now brought our j down to 2, we can use the lemma, and
67(2. 27 . 22i) 2 + 4 = (22i 2 f 67 . 27 2 ) 2 ,
or 67(ii934) 2 + 4 = (97684) 2 ;
therefore, dividing by 4, we have
67 (5967)'+ i =(48842)'.
Of this Indian method Hankel says, "It is above all praise; it is
certainly the finest thing which was achieved in the theory of numbers
1 Hankel conjectures that the Indian method may have been evolved somewhat in
this way.
si=p is given, and if we put ^ /2 + j'=/ 2 , then
Now suppose/', q' to be determined as whole numbers from the equation// p'q^=. i,
and let the resulting integral value of pp'  Aqq' be r.
Then ^ + j > / = r 2 , and accordingly r*A must be divisible by j, or s'=(Ar 2 )ls is 1
a whole number.
Eliminating/' from the two equations in/', /, we obtain
and, as stated in the rule, r has therefore to be 50 chosen that (/+^r)/f is an integer.
THEOREMS AND PROBLEMS BY FERMAT 285
before Lagrange " ; and, although this may seem an exaggeration when we
think of the extraordinary achievements of a Fermat, it is true that the
Indian method is, remarkably enough, the same as that which was redis
covered and expounded by Lagrange in his memoir of ij6B l . Nothing is
wanting to the cyclic method except the proof that it will in every case
lead to the desired result whenever A is a number which is not a square ;
and it was this proof which Lagrange first supplied.
Fermat.
As we have already said, Fermat rediscovered our problem and was
the first to assert that the equation
where A is any integer not a square, always has an unlimited number
of solutions in integers.
His statement was made in a letter to Frenicle of February, r657 2 .
Fermat asks Frenicle for a general rule for finding, when any number not a
square is given, squares which, when they are respectively multiplied by the
given number and unity is added to the product, give squares. If, says
Fermat, Frenicle cannot give a general rule, will he give the smallest value
of y which will satisfy the equations 6ijv 2 + i = x? and 109^ + i = # 2 ? 3
At the same time Fermat issued a challenge to the same effect to
mathematicians in general, prefacing it by some remarks which are worth
quoting in full 4 .
" There is hardly any one who propounds purely arithmetical questions,
hardly any one who understands them. Is this due to the fact that up to
now arithmetic has been treated geometrically rather than arithmetically?
This has indeed generally been the case both in ancient and modern
works; even Diophantus is an instance. For, although he has freed
himself from geometry a little more than others have in that he confines
his analysis to the consideration of rational numbers, yet even there
geometry is not entirely absent, as is sufficiently proved by the Zetetica
of Vieta, where the method of Diophantus is extended to continuous
magnitude and therefore to geometry.
" Now arithmetic has, so to speak, a special domain of its own, the
theory of integral numbers. This was only lightly touched upon by Euclid
in his Elements, and was not sufficiently studied by those who followed
him (unless, perchance, it is contained in those Books of Diophantus of
1 "Sur la solution des problemes indeterminds du second degre"" in Mcmoires de
VAcad. Royale des Sciences et BellesLettres de Berlin, t. xxm. 1769 ( Oeuvres de
Lagrange, n. pp. 377 sqq.). The comparison between Lagrange's procedure and the
Indian is given by Konen, pp. 7577.
2 Oeuvres de Fermat, n. pp. 3334.
3 Fermat evidently chose these cases for their difficulty ; the smallest values satisfying
the first equation are ^=226153980, #=1766319049, and the smallest values satisfying
the second are y 15 140424455100, #=158070671986249.
**Oeuvres de Fermat, n. pp. 3345.
286 SUPPLEMENT
which the ravages of time have robbed us); arithmeticians have therefore
now to develop it or restore it.
" To arithmeticians therefore, by way of lighting up the road to be
followed, I propose the following theorem to be proved or problem to
be solved. If they succeed in discovering the proof or solution, they will
admit that questions of this kind are not inferior to the more celebrated
questions in geometry in respect of beauty, difficulty or method of proof.
" Given any number whatever which is not a square, there are also given
an infinite number of squares such that, if the square is multiplied into the
given number and unity is added to the product^ the result is a square.
" Example. Let 3, which is not a square, be the given number; when
it is multiplied into the square i, and i is added to the product, the result
is 4, being a square.
"The same 3 multiplied by the square 16 gives a product which, if
increased by i, becomes 49, a square.
"And an infinite number of squares besides i and 16 can be found
which have the same property.
" But I ask for a general rule of solution when any number not a square
is given.
" E.g. let it be required to find a square such that, if the product of the
square and the number 149, or 109, or 433 etc. be increased by i, the
result is a square."
The challenge was taken up in England by William, Viscount Brouncker,
first President of the Royal Society, and Wallis 1 . At first, owing apparently
to some misunderstanding, they thought that only rational, and not neces
sarily integral, solutions were wanted, and found of course no difficulty in
solving this easy problem. Fermat was, naturally, not satisfied with this
solution, and Brouncker, attacking the problem again, finally succeeded in
solving it. The method is set out in letters of Wallis 2 of i;th December,
1657, and 3oth January, 1658, and in Chapter xcvm. of Wallis' Algebra ;
Euler also explains it fully in his Algebra*, wrongly attributing it to Pell 4 .
1 An excellent summary of the whole story is given in Wertheim's paper " Pierre
Fermat's Streit mit John Wallis" in Abhandlnngen zur Gesch. der Math. IX. Heft
(Cantor Festschrift), 1899, pp. 557576. See also Konen, pp. 2943.
2 Oeuwes de Fermat^ m. pp. 457480, 490503. Wallis gives the solution of each
of the three difficult cases last mentioned.
3 Euler, Algebra^ Part n. chap. vii.
4 This was the origin of the erroneous description of our equation as the " Pellian"
equation. Hankel (p. 203) supposed that the equation was so called because the solution
was reproduced by Pell in an English translation (1668) by Thomas Brancker of Rahn's
Algebra*, but this is a misapprehension, as the socalled "Pellian" equation is not so
much as mentioned in Pell's additions (Wertheim in Bibliotheca Mathematics in s ,
1902, pp. 1246; Konen, pp. 334 note). The attribution of the solution to Pell was a
pure mistake of Euler's, probably due to a cursory reading by him of the second volume
of Wallis' Opera where the solution 'of the equation ax 2 + i => 2 is given as well as informa
tion as to Pell's work in indeterminate analysis. But Pell is not mentioned in connexion
with the equation at all (Enestrom in Bibliotheca Mathematica, m 3 , 1902, p. 206).
THEOREMS AND PROBLEMS BY FERMAT 287
Fermat appears to have been satisfied with the actual sohttioif) but
later he points out that, although Frenicle and Wallis had given many
particular solutions, they had not supplied a general /ra?/" 2 (i.e. presumably
that the solution is always possible and that the method will always lead
to the solution sought for). He says, " I prove it by the method of
descente applied in a quite special manner. ...The general demonstration
will be found by means of the descente duly and appropriately applied."
Further on, Fermat says he has discovered " general rules for solving
the simple and double equations of Diophantus."
" Suppose, for example, that we have to make
23? + 7967 equal to a square.
" / have a general rule for solving this equation^ if it is possible, or
discovering its impossibility ', and similarly in all cases and for all values
of the coefficient of y? and of the absolute term.
"Suppose we have to solve the doubleequation
2% + 3 = square j
2x + 5 = square } '
"Bachet boasts, in his commentary on Diophantus 3 , of having dis
covered a rule for solving in two particular cases ; I make it general for
all kinds of cases and can determine, by rule, whether it is possible or not 4 ."
Thus Fermat asserts that he can solve, when it is possible to solve
it, and can determine, by a general method, whether it is possible or
impossible to solve, for any particular values of the constants, the more
general equation
a*Af = JB.
This more general equation was of course solved by Lagrange. How
Fermat solved it we do not know. It is true that he has sometimes been
1 Letter of June, 1658, to Kenelm Digby, Oeuvres de Fermat, II. p. 402.
2 " Relation des nouvelles decouvertes en la science des nombres," Oeuvres, 11. p. 433.
3 See on Diophantus iv. 39, and above, pp. 8082.
4 With this should be compared Fermat's note on Dioph. iv. 39, where he says,
similarly :
" Suppose, if you will, that the doubleequation to be solved is
2jr+5=square(
6jr+3=square j "
"The first square must be made equal to 16 and the second to 36; and others will be
found ad injinitu?ti satisfying the question. Nor is it difficult to propound a general rule
for the solution of this kind of question."
No doubt the doubleequation in this case, as in the others referred to in the "Relation,"
would be transformed into the single equation
t*Au*=B
by eliminating x. I think this shows how Fermat 'was led to investigate our equation :
a question which seems to have puzzled Konen (p. 29), in view of the fact that the actual
equation is not mentioned in the notes to Diophantus. The comparison of the two places
seems to make the matter clear. For example, the two equations mentioned above in
this note lead to the equation * 2  3^2= 12, and the solution *=6, = 4 is easily
obtained.
288 SUPPLEMENT
credited with the very same solution of the equation x* Ay 1 = i as that
given by Brouncker and Waltis ; but this idea seems to be based on a
misapprehension of a sentence in Ozanam's Algebra (1702). Ozanam
gives the Brouncker Wallis solution as "une regie generate pour rdsoudre
cette question, qui est de M. de Fermat"; and possibly the ambiguity
of the reference of "qui" may have misled Lagrange and others into
supposing that the "regie" was due to Fermat.
For the history of the equation after Fermat's time I must refer to
other works and particularly that of Konen 1 . Euler, Lagrange, Gauss,
Jacobi, Dirichlet, Kronecker are the great names associated with it. I
will only add a few particulars with regard to Euler 2 as coming nearest
to Fermat.
In a letter to Goldbach 3 of loth August, 1730, Euler mentions that he
requires the solution of the equation x?  Af  i in order to make
ay? + bx + c a complete square. He goes on to observe that the problem
of solving tf 2  Af  i in integers was discussed between Wallis and
Fermat and that the solution (which he already attributes to Pell) was
set out in Wallis' Opera. There is an indication in this very passage that
Euler had then only read the BrounckerWallis correspondence cursorily,
for he speaks of the equation iogy* + i op as being the most difficult
case solved by them, whereas the most difficult examples actually solved
were 433jy 2 + i = x? and 31 3j/ 2 f i = x?.
A paper of a year or two later 4 contained the proof that the evolution
of successive solutions of at& + bx + c=y* when one is known requires that
one solution of at? + i = if must also be known. Similarly, in his Algebra*,
he shows that the solution of the latter equation is necessary for finding all
the possible solutions of the equation ay? + b =y 2 , the importance of which
remark is emphasised by Lagrange 6 .
In the paper quoted in the last paragraph Euler finds any number
of successive solutions of ax? + &x + c=;y*, and the law for forming them,
when we are given one value n of x which will make ax* + bx + c a
complete square and one value p of $ which will make at? 4 1 a complete
square, or, in other words, when an*+bn + c=m !> and a/ 1 + i = ^. He
then takes the particular case ax? + bx + d z = jr where (since x ~ o, y = d
satisfies the equation) we can substitute o for n and d for m in the
expressions representing the successive solutions of ax? + bx + <:=/. Then
again, putting <$ = o and d=i, he is in a position to write down any
1 Konen, op. /.; cf. Cantor's Geschichte der Mathematik, iv. Abschnitt xx., as
regards Euler and Lagrange.
3 Cf. Konen, op. cit. pp. 4758.
3 Correspondance mattihnaHque et physique de quelques cllttres gtomttres du xvm#wa
siSe/e, publie'e par P. H. Fuss,Petersbourg, 1843, I. p. 37.
4 "De solutione problematum Diophanteorum per numeros integros" in Commentarii
Acad. PetropoL 17323, vi. (1738), pp. 175 sqq. = Commentationes arithm. I. pp. 410.
5 Algebra, Part II. ch. VI.
6 Additions to Euler's Algebra, ch. vm.
THEOREMS AND PROBLEMS BY FERMAT 289
number of successive solutions of ag* + i = rf when one solution =/,
t\ q is known. The successive values of are
o, p, 2fq, 4p?fr
and the corresponding values of 77 are
i, q, 2?> i, i? 3 M,... ' . :; 
the law of formation being in each case that, if A> B be consecutive values
in either series, the next following is 2qB A.
The question then arises how to find the first values /, q which will
satisfy the equation. Euler first points out that, when a has one of many
particular forms, values of/, q can at once be written down which satisfy
the equation. The following are such cases with the obvious values of
/ and q,
a = aV & 2a^ >  1 ' 9 p = e, q = ae*+ l i
(where a may even be fractional provided ae^" 1 is an integer),
a = (a* 6 + $<?)* + 2ae b ~ 1 + 2^~ l ; f = e, q = aJ> +l + /3^ +1
a = aW & a**' 1 ; ^ke.^ atf^ i.
But, if a cannot be put into such forms as the above, then the method
explained by Wallis must be used. Euler illustrates by finding the least
values /, q which will satisfy the equation 3i 2 i 1 =?/*, and then adds a
table of the least solutions of the equation at? + i = if for all values of
a (which are not squares) from 2 to 68.
The important remark follows (g 18) that the above procedure at once
gives a very easy way of finding closer and closer approximations to the
value of any surd *Ja. For, since ap z + i = q\ we have *Ja = *J(gP  i)fe,
and, if q (and therefore p also) is large, q\p is a close approximation to *ja ;
the error is not greater than i/(2/V^) Euler illustrates by taking ^/6.
The first solution of 6^ + i = rf (after f = o, vj = i ) is p  2, q = 5. Taking
then the series of values" above given for a 2 + i = i? 3 , namely
=, A a/^ WA 4 B, zqBA,
17=1, q, 2^1, 4^3^,...^, F, 2qFE,
and substituting p  2, q = 5, the successive corresponding values jP } Q
of , 77 respectively become
P~o, 2, 20, 198, 1960, 19402, 192060, 1901198, ...
<2=i, 5, 49, 485, 4801, 475 2 5> 47449 4656965,
and the successive values QjP are closer and closer approximations to *J6.
It will be observed that the method of obtaining successive approximations
H. D. 19
2 9 o SUPPLEMENT
to ^a from successive solutions of a? + i = if is the same as that which,
according to the hypothesis of Zeuthen and Tannery, Archimedes used in
order to find his approximations to ^3.
The converse process of finding successive solutions of ag* + i = rf by
developing >Ja as a continued fraction did not apparently occur to Euler
till later. In two letters 1 to Goldbach of 4th Sept. 1753 and 23rd Sept.
1755 he speaks of a "certain" method and of improvements which he had
made in the " Pellian " method but gives no details. His next paper on
the same subject 3 returns to the problem of finding all the solutions of
a3? + bx + c=y* or ax? + b=y* when one is known, and in the course of
his discussion of the latter he arrives at " the following remarkable theorem
which contains within it the foundation of higher solutions.
"If xa,yb satisfies az?+p =y 2 ,
and x = c, y^d satisfies aa,* 2 4 q =y\
then x = bc ad> y~bd aac satisfies ax? + pq =jA"
That is to say, Euler rediscovers and recognises the importance of the
lemma to the Indian solution, as Lagrange did later.
More important is the paper of about three years later 3 in which Euler
obtained the solution of the equation x?  Ay* i by the process of con
verting tJA into a continued fraction, this course being the reverse of that
which was s according to the hypothesis of Tannery and Zeuthen, followed
by Archimedes, and to the feasibility of which Euler had called attention
in 17323. He begins by stating, without proof, that, if ^ 3 = ^ 2 i 1, then
q\p is an approximation to ^7, and q\p is " such a fraction as expresses
the value of >Jl so nearly or exceeds it so little that a closer approximation
cannot be made except by bringing in greater numbers." Next he develops
certain particular surds, namely ^(13), V(6i) and J(6?\ after which he
states the process generally thus. If \Jz be the given surd and v the root
of the greatest integral square which is less than 0, the process will give
/ *
Jz = v +  i
v a+ 7 i
b+ i
' + 5+ etc.
the successive quotients a, b, ^ d, being found by means of the process
shown in the following table :
1 Corre$$ondance etc., ed. Fuss, pp. 614 sq., 629 sq.
2 "De resolutione formularum qu^draticarum indeterminatarum per numerosintegros"
in Novi Comuuntarii Acad. PetropoL 1 7623, IX. ( r 764), pp. 3 sqq. = Commentat. aritkm.
I. pp. 497315.
3 "De usu novi algorithm! in problemate Pelliano solvendo" in Novi Commentarii
Acad. Petropol. 1765, XI. (1767), pp. 2866= Commentat. arithm* I. pp. 316336. The
paper seems to have been read as early as 15 Oct. 1759.
THEOREMS AND PROBLEMS BY FERMAT
291
Take
I. A=v .
II. fi=a.aA
III. C=pbJB
IV. D=yc~C
V. E=UD
and
7=
0^_
a
T~ =
~T" "
~T~ =
etc.
It follows that
,*+
etc.
(This is of course exactly the process given in text books of Algebra,
e.g. Todhunter's.)
Euler now remarks as follows.
1. The numbers A, B^ C, D ... cannot exceed #; the first, A, is equal
to #; since a (v + A)/a, 9 aa A = J3 ^LV, and so on.
2. Unless where one of the numbers a, , y, 8 ... is equal to unity,
none of the corresponding quotients a, b, c, d ... can exceed v.
3. When we arrive at a quotient equal to 2?;, the next quotients will be
a, b) c, d . . . in the same order.
4. Similar periods occur with the letters a, /?, y, S... and the term
of this series corresponding to a quotient 2V is always i.
The successive convergents to the continued fraction are then investi
gated and it is shown that, for successive convergents q\p beginning
with z?/i,
f  zf =  a, + fa  y, +8,  c etc. in order.
It follows that the problem is solved whenever one of the terms with a
positive sign, fa 8, etc., becomes i.
Since unity for one of the terms a, /?, y, 8 corresponds to the quotient
2i)\ and each fresh period begins with 227, the first period will produce
a convergent qlp such that f zfi* = i ; and the negative sign will apply
if the number of quotients constituting the period is odd, while the positive
sign will apply if the number of quotients is even. In the latter case we
have a solution of our equation at once; if, however, f zjP = i, we
must go on to the end of the second period in order to get an even number
of quotients and so satisfy the equation ^ 2 / J = + i. Or, says Euler,
instead of going on and completing the second period, we can satisfy
the latter equation more easily thus.
Suppose fqp = i, and assume
Then
g" 2  zf 1 = 4^ +
+ i 
19 2
292 SUPPLEMENT
[This last derivation of a solution of y*zx*  i from a known solution
of y  zx? =  i is of course the same as the Indian method of doing the
same thing, for they assumed/ = zpg, <?' = <? + zp*, and ^ + zp* = <f+ (<f + i).]
We thus see that in Euler's method there is everything necessary to the
complete solution of our equation except the proof that it must always lead
to the desired result. Unless it is proved that the quotient zv will actually
occur in the development of the continued fraction in every case, we cannot
be sure that the equation has any solution except x = o, y = i.
I cannot, I think, do better than conclude by a quotation from
H. J. S. Smith 1 , the first part of which is well known 2 . " Euler observed
that [if T* DU* = i] T\U is itself necessarily a convergent to the value
of JD, so that to obtain the numbers T and U it suffices to develop >JD
as a continued fraction. It is singular, however, that it never seems to
have occurred to him that, to complete the theory of the problem, it was
necessary to demonstrate that the equation is always resoluble and that
all its solutions are given by the development of sJD. His memoir
contains all the elements necessary to the demonstration, but here, as
in some other instances, Euler is satisfied with an induction which does
not amount, to a rigorous proof. The first admissible proof of the re
solubility of the equation was given by Lagrange in the Melanges de la
Sodefe de Turin, Vol. iv. p. 4i 3 . He there shows that in the development
of ,JD we shall obtain an infinite number of solutions of some equation of
the form T^DU^ = A and that, by multiplying together a sufficient
number of these equations, we can deduce solutions of the equation
T^DU^i. But the simpler demonstration of its solubility which
is now to be found in most books on algebra, and which depends on
the completion of the theory (left unfinished by Euler) of the development
of a quadratic surd as a continued fraction, was first given by Lagrange
in the Hist, de F Academic de Berlin for 1767 and 1768, Vol. xxm. p. 272,
and VoL xxiv. p. 236*, and, in a simpler form, in the Additions to Euler's
Algebra*, Art. 37."
1 "Report on the Theory of Numbers, Part \\i." British Association Reports for 1861,
London, 1862, p. $i$?= Collected Works, Vol. i., Oxford, 1894, p. 192.
3 It is given in Cantor, Gesck. d* Math. iv. 1908, p. 159, and referred to by Konen,
op. af. p. 51.
3 "Solution d'un probleme d'Arithmetique," finished at Berlin on soth Sept. 1768
and published in Miscellanea Taurinensia, iv. 17661 769 =.0euvres de Lagrange^ \.
pp. 671731.
4 The references are: " Sur la solution des problemes indetermines du second degre,"
read 34th Nov. 1768 and published in the MSmoires de FAcadtmie Royale des Sciences
et Belleslettres de Berlin, Vol. xxm., 1769, pp. i6$3io=0euvres de Lagrange, II.
PP 377535 5 " Nouvelle mithode pour resoudre les problemes indetermines en nombres
entiers," read 2ist June, 1770, and published in Mhnoires de PAcademie Royale des
Sciences et Belleslettres de Berlin^ VoL xxiv., 1770, pp. 181156 Oeuvres de Lagrange,
II. pp. 655726.
5 The Additions of Lagrange were first printed as an appendix to tiviens tVAlgebre
par M. L. uler traduits de Vallemand^ Vol. u., Lyons, 1774; second edition, Paris,
1798; they were thence incorporated in Oeuvres de, Lagrange^ vn. pp 158 sqq.
293
SECTION III.
THEOREMS AND PROBLEMS ON RATIONAL RIGHTANGLED TRIANGLES.
i. On No. 20 of the problems about rightangled triangles added
by Bachet to Book vi. ("To find a rightangled triangle such that its
area is equal to a given number") Fermat has a note which shall be
quoted in full, not only for the sake of the famous theorem enunciated
in it, but because, exceptionally, it indicates the lines on which his proof
of the theorem proceeded.
"The area of a rightangled triangle the sides of which
are rational numbers cannot be a square number.
" This proposition, which is ray own discovery, I have at length
succeeded in proving, though not without much labour and hard thinking.
I give the proof here, as this method will enable extraordinary develop
ments to be made in the theory of numbers.
" If the area of a rightangled triangle were a square, there would exist
two biquadrates the difference of which would be a square number. Con
sequently there would exist two square numbers the sum and difference of
which would both be squares. Therefore we should have a square number
which would be equal to the sum of a square and the double of another
square, while the squares of which this sum is made up would themselves
[*.*. taken once each] have a square number for their sum. But if a square
is made up of a square and the double of another square, its side, as I can
very easily prove, is also similarly made up of a square and the double of
another square. From this we conclude that the said side is the sum of the
sides about the right angle in a rightangled triangle, and that the simple
square contained in the sum is the base and the double of the other square
the perpendicular.
"This rightangled triangle will thus be formed from two squares,
the sum and the difference of which will be squares. But both these
squares can be shown to be smaller than the squares originally assumed
to be such that both their sum and their difference are squares. Thus,
if there exist two squares such that their sum and difference are both
squares, there will also exist two other integer squares which have the same
property but have a smaller sum. By the same reasoning we .find a sum
still smaller than that last found, and we can go on ad infinitum finding
integer square numbers smaller and smaller which have the same property.
This is, however, impossible because there cannot be an infinite series
of numbers smaller than any given integer we please. The margin is too
small to enable me to give the proof completely and with all detail
" By means of these considerations I have also discovered and proved
that no triangular number except i can be a biquadrate"
294 SUPPLEMENT
As Wertheim says, it may have been by following out the indications
thus given by Fermat that Euler succeeded in proving the propositions
that x 4 y^ and # 4 +y i cannot be squares, as well as a number of other
theorems connected therewith (Co?nmentationes arithmeticae collectae^ i.
pp. 24 sqq. j Algebra, Part n. Chapter xni.).
Zeuthen 1 suggests a method of filling out Fermat's argument, thus.
The sides of a rational rightangled triangle can be expressed as
As a common factor in the sides would appear as a square in the
number representing the area, we can neglect such a factor, and assume
that sP y* and therefore also x+y and x y are odd numbers and that
x, y are prime to one another, so that x, y, x +y t x y are all prime to
one another.
We have now to test the assumption that the area of the triangle
*y(xy)(x+y)
is a square. If so, the separate factors must be squares, or
(" There would exist two biquadrates the difference of which f// 4  z/ 4 ] would
be a square, and conseque?itly there would exist two squares the sum and differ
ence of which [? + # 2 , 2  z/ would both be squares? Fermat.)
From the last two equations we obtain
(" We should have a square number which would be equal to the sum
of a square and the double of another square \_p^ 21? + ^ 2 ]," Fermat.)
Now / + q and p q are both even numbers because, on the above
assumptions, f* and q* are both odd; but they cannot have any other
common factor except 2, since u* and ^ are prime to one another. It
follows therefore from the last equation that
where n is an even number.
We obtain, therefore,
The whole numbers m* and are therefore sides of a new rightangled
triangle with the square area ^^ .
4
1 Zeuthen, Geschichte der Mathematik im XVI. itnd xvn. Jahrhundert> 1903, p. 163.
THEOREMS AND PROBLEMS BY FERMAT 295
(" If a square is made up of a square and the double of another square
[/ 2 = 2 ?; 2 + ^ 2 ], its side is, as I can very easily prove, also made up of a
square and the double of another square # = m z + z( \ . From this we
conclude that the said side is the sum of the sides about the right angle
in a rightangled triangle, the square [m*\ being the base and the double of
the other square 2(j the perpendicular," Fermat.)
That the sides of the new triangle are less than those of the original
triangle is ^ clear from the fact that the square on its hypotenuse u* or
x is a factor of one of the perpendicular sides of the original triangle 1 .
As now an infinite series of diminishing positive whole numbers is
impossible, the original assumption from which we started is also impossible.
It will be observed, as Zeuthen says, that the proof includes also the
proof of the fact that i tf cannot be a square and therefore cannot be
a fourth power, from which it follows that the equation w 4 = #* + a/* cannot
be solved in whole numbers, and consequently cannot be solved in rational
numbers either.
The history of this theorem would not be complete without an account
of a " proof originating with Fermat " which Wertheim has reproduced 2 .
In the small paper of Fermat's entitled " Relation des nouvelles decouvertes
en la science des nombres 3 " containing a statement of his method
of *' diminution without limit" (descente infinie or indefinie) and of a
number of theorems which he proved by means of it, there is a remark
that he had sent to Carcavi and Frenicle some proofs based on this
method. And, sure enough, Frenicle gives a proof by this method of
the theorem now. in question in his "Traite des triangles rectangles en
nombres 4 ." Wertheim accordingly concludes that we have here a proof
of Fermat's. A short explanation is necessary before we come to Frenicle's
proof.
We obtain a rightangled triangle 0, x, y in rational numbers (# 2 =
if, a, b being any integers and a>b, we put
If a is prime to b and one of these numbers is even, the other odd, then
it is easily shown that the greatest common measure of x, y, z is i.
In the rightangled triangle # 2 b* and zab are the perpendicular sides,
1 Zeuthen's inference at this point diverges slightly in form from what we actually find
in Fermat's own statement of his argument. Fermat does not actually say that the new
rightangled triangle is a triangle in smaller numbers than the original triangle and with
the same assumed property, but that its formation gives us two new square numbers the
sum and difference of which are squares, and which are smaller than the two squares
originally assumed to have this property.
2 Zeitschriftfiir Math. it. Physik, hist, litt Abtheilung XLIV. 1899, pp. 46.
3 Oeuvres de Permed, Vol. II, pp. 4316.
4 Mhnoires de PAcad&nie Royale des Sciences, v., Paris, 1729, pp. 83166.
296 SUPPLEMENT
and (a?  #>) ab is the area, a, b are called the generating numbers (the
numbers from which the triangle is formed) and if a is prime to b, and one
of them is odd and the other even, so that x, y, z have no common factor
except i, the triangle is called a primitive triangle.
If (a?lr)al) is the area of a primitive rightangled triangle and it
is enough to prove the proposition for such each of the three numbers
$  b\ a, b is prime to the other two. If, then, the product is a square
number, each of the three factors must be square, and in that case a?  P
will be the difference between two fourth powers. The theorems
(1) the area of a rightangled triangle in rational numbers cannot be
a square number, and
(2) the difference of two fourth powers cannot be a square,
accordingly state essentially the same fact.
The proof which Frenicle gives of the first of these propositions depends
on the following Lemmas.
Lemma I. If the odd perpendicular of a primitive rightangled triangle
is a square number, there exists a second primitive righta?igled
triangle with smaller sides which has for its odd perpendicular
the root of the said square number.
If"fl a ^ 9 = ^ a , it follows that a* = b* + c*, so that a, b, c are the sides
of a rightangled triangle. The odd perpendicular of this second triangle
is <r, for by hypothesis t? is odd; consequently the even perpendicular is
b, while a is the hypotenuse. The triangle is "primitive* 1 because a
common divisor of any two of the three numbers a, b, c would divide
the third, while by hypothesis a, b have no common factor except i.
Next, the second triangle has smaller sides than the first, since c?,
By this lemma we can from the triangle 9, 40, 41 derive the triangle
3, 4, 5, and from the triangle 225, 25312, 25313 the triangle 15, 112, 113.
Lemma II. If in a primitive rightangled triangle the hypotenuse as
well as the even perpendicular were square, there would exist a
second primitive rightangled triangle with smaller sides which
would have for hypotenuse the root of the hypotenuse of the first,
for odd perpendicular a square number, and for even perpendicztlar
the double of a square number.
Let the sides of the first triangle be a* + b\ a*~Z> z , zab. If tab were
a square, ab would be double of a square ; therefore, since a, b are prime
to one another, one of these two numbers, namely the odd one, would
be a square, and the other, the even one, would be double of a square.
Let a be the odd one of the two, b the even. If now the hypotenuse
tf + lr were a square number <?, we should have a second rightangled
triangle a, b, c which would necessarily be primitive and in which the sides
would be smaller than those of the first triangle; for c?, b<2ab and
? P since
THEOREMS AND PROBLEMS BY FERMAT 297
By means of the above two lemmas combined we can now prove that
the area of a primitive rightangled triangle cannot be a square number.
Let the sides of the triangle be a 2 + P, a*  P, 2ab. If now the area
were square, the product of the perpendicular sides would be double of
a square. But the perpendicular sides are prime to one another. There
fore the odd perpendicular a^ lr would be a square, and the even
perpendicular tab the double of a square. But, if a 2 P 2 were equal to
r 2 , we could (by the first Lemma) find a second primitive triangle with
smaller sides in which the odd perpendicular would be c, the even per
pendicular , and the hypotenuse a. Again, since zab would be double
of a square, ab would be a square, and, since a is prime to 3, both a and
b would be squares. The second triangle would accordingly have a square
number both for its hypotenuse (a) and for its even perpendicular (b).
That is, the second primitive triangle would satisfy the conditions of the
second Lemma, and we could accordingly derive from the second primitive
triangle a third primitive triangle with still smaller sides which would,
exactly like the first triangle^ have a square number for its odd perpendicular,
and for its even perpendicular the double of a square number.
From this third triangle we could obtain a fourth, and by means of the
fourth we could obtain a fifth with the same property as the first, and so
we should have an unending series of primitive rightangled triangles, each
successive triangle having smaller sides than the one before, and all being
such that the odd perpendicular would be a square number, the even
perpendicular the double of a square number, and consequently the area
a square number. This, however, is impossible since there cannot be an
unending series of integral numbers less than any given integral number.
Frenicle proves, by similar considerations, that neither can the area of a
rightangled triangle in rational numbers be the double of d square number.
In enunciating Fermat's problems on rightangled triangles I shall in
future for brevity and uniformity use , >?, to denote the three sides, while
will always represent the hypotenuse and & 17 the two perpendicular sides.
2. To find a rightangled triangle (& TJ) such that
[Since ^^tr + rf, this problem is equivalent to that of finding x t y such
that
which is Question 17 in Chapter xiv. of Euler's Algebra, Part n.]
First method.
Form a rightangled triangle from the numbers x+ i, x\ the sides will
then be
298 SUPPLEMENT
We have then the doubleequation
2.T 2 4 2 X + I = U 2 J
2#" J + 4 X + I = V* j
The ordinary method of Diophantus gives the solution x =  ^ ; the
triangle will therefore be formed from  1 and  ^ or, if we take the
numerators only, 5 and  12, and the triangle is (169,  119, 120) which
is equally the result of forming a triangle from 4 5 and + 12.
But, as one of the perpendiculars is negative, we must find another
value of x which will make all three sides positive.
We accordingly form a triangle from x + 5 and 1 2, instead of' from
5 and 12, and repeat the operation. This gives for the sides
=3^+ io#+ 169, g = x* + lox 119, 77 = 24^+120,
and we have to solve the double equation
169 = # 3 S
Making the absolute term the same in each, we have to solve
#2 + io# + 169 = w 2 ,
1 693? + 57463; 4 169 = 2/ 2 .
The difference is 168^ + 57362;, which we may separate into the factors
14%, i2x + ? 8 f s (the sum of the terms in x being 26% or 2 . 132;).
Equating the square of half the sum of these factors to the larger
expression or the square of half their difference to the smaller, we find
in the usual way
X = 30^8075
The triangle is therefore formed from 2 %gggg 5 , 12, or from 2150905,
246792, and the triangle itself is
4687298610289, 4565486027761, 1061652293520,
the hypotenuse and the sum of the other two sides being severally squares.
Second method.
This is the same as the first method up to the forming of the triangle
from x + 5 and 12 and the arrival at the doubleequation
x?+ IQX + 169=^2^,
x 3 + 342; + i = v\
Multiply the two expressions together, and we must have
44^ + 5 10^15756^+ 169 = a square
this gives, as a matter of fact, the same value of x, namely
 ^ 2048076
THEOREMS AND PROBLEMS BY FERMAT 299
and the triangle is the same as before 1 .
In his note on Diophantus vi. 22 Fermat says that he confidently
asserts that the above rightangled triangle is the smallest rightangled
triangle in rational numbers which satisfies the conditions.
[The truth of this latter assertion was proved by Lagrange 2 . Lagrange
observes that, since + y =y^ 2 + vf = #* 3 say, we have, if we put s for  ?/,
or 2# 4 jj/ 4 = 0,
and, if x, y is any solution of the latter equation,
1 For comparison we may give Enter's solution (Algebra^ Part II., Art. 240; Commen
tationes arithmeticae, II. p. 398).
We have to solve the equations
x+y=u* \
* 2 +y=^f
First make ^ 2 +^ 2 a square by putting xa^b 2 ^ y^iab^ so that
To make the last expression a fourth power put 0=/ 2 ~# 2 , b=ipq, so that
and accordingly
We have now only to make x\y a square.
Now x=a?t> 2 =J>*6j!>*<r z + 4*,
therefore / 4 + ^jPq 6^ 2  ^pif + ^ 4 = a square.
In solving this we have to note that j>, q should be positive, p must be > ^ (for other
wise y would be negative), and a>b in order that x may be positive.
Pat p + 4 /V  6^ 2  4/?3 + 0* = (/ 2  2#? + ? 2 ) 2 ,
and we obtain fiq  6jPf =  4^ 4 6/ 2 ^ 2 , whence jZ>/^ =  .
But, if we put/=3, <7=2, we find *s=  119, a negative value.
To find fresh values, we can substitute for p the expression f q + r and solve for the
ratio ^/r; then, by taking for q the numerator and for r the denominator of the fraction
so found, we find a value for / and thence for x, y. This is Euler's method in the
Algebra. But we avoid the necessity for clearing of fractions if (as in the Comment.
arithw.} we leave 2 as the value of q and substitute 3 + for 3 as the value of p.
We then have / 4 = 8 i + io8z/+ 542/2+ 12^ + 2^,
 2 16  144^  242/2,
 9632^,
+? 4 = 16, '
whence j: +^ = i + 1 482? + 1 02^ + 20^ + z^ = a square = ( r + 74Z/  z/ 2 ) 2 , say ;
and we obtain
1343 = 42^, or v=s2. t and / = 3 + /=, while ^ = 2.
Taking integral values, we put/= 1469, ^=84.
Therefore 0=1385.1553 = 2150905, ^ = 168.1469 = 246792,
and ^=4565486027761, ^=1061652293520,
which is the same as Fermat's solution.
2 N. Mtmoires de FAcad* Roy ale des Sciences et Belleslettres de Berlin, ann^e 1777,
Berlin, ifjg=0euvres de Lagrange> iv. pp. 377398.
300 SUPPLEMENT
He sets himself therefore to find a general solution of the equation
2# 4 y i =s 2 and effects it by a method which is a variation of Fermat's
descente, one of the most fruitful methods, as Lagrange observes, in the
whole theory of numbers. The modified method consists of two parts,
(T) a proof that, assuming that there exist integral values of x, y greater
than T which satisfy the condition 2tf 4 / = s 2 , there are still smaller
integral values which will also satisfy it, (2) the discovery of a general
method of deducing the latter from the former. This being done, and
it being known that x i, y = i are the minimum values, the successive
higher values are found by reversing the process. Lagrange found that
the four lowest values for x, y give the following pairs of values for f, 17,
namely
(1) =i , 17 = 0,
(2) (=120 , ?7 = ii9,
(3) ^=2276953 , >? = 4733 4>
(4) =1061652293520, 17 = 4565486027761,
so that the last pair (4) are in truth, as Fermat asserts, the smallest possible
values in positive integers.]
3. To find a rightangled triangle f, 17 such that
[This is of course equivalent to solving
=t^
* J
Form a triangle from the numbers x * i, i ; the sides will then be
= XT + 2X + 2, = X 2 + 2X, V)  2X + 2.
We have then to solve the doubleequation
X? + 2X + 2 = U Z
Solved in the ordinary way, this gives #= fl; consequently the
triangle is formed from 5%, i, or from 5, 12.
We could proceed, as in the last problem, to deduce a new value for x,
but we observe that the triangle formed from 5, 12, i.e. the triangle 169,
119, 120, satisfies the conditions.
4. To find a rightangled triangle , , 77 such that
where m is any number.
Fermat takes the case where m = 2.
Form a triangle from x, i; the sides are then C =
Tf] = 2X.
THEOREMS AND PROBLEMS BY FERMAT 301
Therefore

must both be squares.
x 2 + 4.x  i
The difference = 2 ^x, and by the usual method we find x = T \.
But f = AT i is negative unless a?> i. We therefore begin afresh and
form a triangle from x + 5, 12.
The sides of this triangle are
tf 2 * io#+ 169, ar'+iotf 119, 24#+i20.
We have therefore to solve the doubleequation
iox + 169 =
Fermat multiplies the two expressions together and puts
cfi + 6Sx? + 870^ + i ioi2# + 20449 = a square
/Tvt^JSSO 6 ^' 626 27 03 .2 \2 coir
\ A 4o "*" "TOT 1 * "" a y a43Tnr"* / a< */ *
tneretore A* = ' 1 4 Y iro u ~^5"i "u >
and the triangle is formed from 103447257961, 17749110120.
The doubleequation could also have been solved by the usual
Diophantine method, as in the next problem to be given.
5. To find a rightangled triangle 4 f, if] such that
 mifj = I? 1 }
m is any number.
Suppose that m = 2.
Form a triangle from x + i, i, so that the sides are
Therefore we have to solve
2X+ 2 =2
XT 2X 4 = f i
Solving in the usual manner, we obtain x=  , so that the triangle is
formed from f^, i, or from  5, 12, and is therefore (169, 119,  120).
We have to replace the value of x by a value which will avoid the
negative sign. Form a triangle, then, from x $, 12.
The sides are ^ 10^ + 169, x ~~ 10^ 119, 24^120.
The doubleequation now becomes
169 = 1^)
121 =#2 j
Multiply the second equation by Iff, and we have to solve
oc* io#+ 169 = t
302 SUPPLEMENT
The difference = ^a 8  % g x = ftx (ftx  fp.).
Equating the square of half the difference of the factors to the smaller
expression (or the square of half the sum to the larger), we have
_ 4 5 9 a 4 5 5
"
*._
x
and the required triangle is formed from 4363225, 552552, the sides being
193430461 13329, 18732418687921, 4821817400400.
Or again in this case we can multiply the expressions x~ iox+ 169
and xr  58.3; h 1 2 r and put their product
#* 68#* + 870^ noi2tf + 20449 = a square
and the result will be the same as before,
, _ 4: o 9 3 4 5 _5_
,. , _
6. To find a rightangled triangle , , 17 such that
=*
= v*} '
number.
Let w = 3. Form a triangle from # n, i ; its sides will be
= tf 3 + 2# + 2, (=X?+2X t f r)
We have therefore to solve the doubleequation
= O 2 j
= ^J '
Solving this in the ordinary manner, we shall find x = y 1 ^.
Hence the triangle is formed from ^4, i, or' (in whole numbers) from
13, 12 ; the sides are therefore 313, 25, 312.
Fermat also finds the solution by multiplying the two expressions and
making the product a square;
#* + lox? + 223? + i2x = a square
= (# 2 + 5tff) 2 , say.
This gives the same value of # as before, x = ^j and the triangle
is 313, 25, 312.
7. Tfc jf^ a rightangled triangle , f, 17 ^^ that
where m is a given number*
Fermat takes the case m = 3.
Remembering that in the corresponding problem with a plus sign we
found the triangle 313, 25, 312 which is formed from 13, 12, we form the
triangle in this case from x  13, 12 ; its sides are
f = # 2 26# + 313, f = # 2 26#+25, 77=24^312.
We have then x 2  26^ + 25 = a
# 2 98# + 961 =?
THEOREMS AND PROBLEMS BY FERMAT 303
Multiplying the first expression by 4^, we have to solve the double
equation
jyyL#2  34i x + 961 = fc' 3
tf 3  98^ + 961 =
The difference = ffix?  m^x =  2 /x (^o
Proceeding as usual, we find #= S 3ilayo 1 ; tne triangle is formed
from x 13, 12, or (in whole numbers) from 23542921, 3820440, and the
sides are
568864891005841, 5396733 6 74i864i, 179888634210480.
The same result is obtained by multiplying the expressions x 2  26# + 25
and xP ySx 1961 and making the product a square; we put
x 4  1 24^ + 3534^ 274363; +24025 = a square
and the result is x = ^fl^g 1 , as before.
8. To find a rightangled triangle f, f, t\ such that
*=
where m is any given number.
Suppose m= 2. Form a triangle from a; + i, i ; the sides are
f = X 2 + 20; + 2,  = tf 3 + 2 a;, >7 = 20: 4 2.
We have then to solve the doubleequation
^c 2 + 6x + 6 
The usual method gives # = J, and the triangle is formed from f , i, or
(in whole numbers) from 5, 4, being the triangle (41, 9, 40).
Since + ?7 = ,xr J +4# + 4 = a square, we have actually solved the problem
Qi finding a rightangled triangle , f, 17 such that
9. To find a rightangled triangle f, f, TJ v4
where m is a given number.
Suppose m = 2.
Since the corresponding problem with a plus sign just preceding has
the solution (41, 9, 40) formed from the numbers 5, 4, we form a triangle
in this case from #5, 4 ; the sides are
9, ^ = 80740.
304 SUPPLEMENT
We have then to solve the doubleequation
AT* icx + 9 =
# 3 26$ + 121 =
De Billy (or Fermat) observes that this doubleequation "seems to admit
of solution in several ways, but it will be found that it is hardly possible to
find a practical solution except by the new method " (expounded earlier in
the Inventum Novuni) of making the absolute terms equal (instead of using
the equal terms in x\ which method gives, in fact, the value x = o). That
is to say, we make the absolute terms in the two expressions equal by
multiplying the first by ^rS and the doubleequation becomes
 121 =Z
The difference = ip* 3  ^x = $x (*x  i).
Equating the square of half the difference of the factors to e> 3 , or the
square of half their sum to ?/ 2 , we find x = $ffi.
Therefore the triangle is formed from 4^ 3 , 4 or (in whole numbers)
from 493, 132, and the sides are 260473, 225625, 130152.
Since 77 = x 2  iBx + 81 = a square, the above actually amounts to the
solution of the problem si finding a right angled triangle , , 77 such that the
three conditions
are simultaneously satisfied.
De Billy (or Fermat) observes however that, while the above one solution
satisfies the conditions of both problems, it is not so with all solutions of
the problem involving the two conditions only; but only primitive triangles
satisfying the conditions of that problem satisfy the additional condition.
Thus the triangle (624, 576, 240) is such that one of the perpendicular
sides is a square and the difference between the hypotenuse and twice the
other perpendicular is also a square, but the hypotenuse minus the latter
perpendicular is not a square.
10. To find a rightangled triangle , , 77 such that
Assume x, ix for the sides f, t\ about the right angle respectively.
This supposition satisfies the second condition.
Again, since fy = xx*, the thifd and fourth conditions are satisfied,
for # 2 = x?, ze/ 3 = i 2 x + so*.
THEOREMS AND PROBLEMS BY FERMAT 305
It remains to satisfy the conditions
and 2 = 2 + ?7 2 = i 2X + 2# 2 = a square)
The difference =2X* x ^xfax 2), and we find, in the usual way,
The triangle is (ft, , fa).
n. To find a rightangled triangle , , t\ such that
= a cube )
Fermat assumes f = i, 17 = #, so that the first condition is satisfied,
i being a cube.
We must now have i \x = ?<f 2 j
and also * = i 2 + if = i + x* = i?]
The difference = x? + ^x = ^x fax + 2), and we find x f .
In order to derive a positive value for x we substitute jKfff for # in
the equations, which gives
Make the absolute term in the first equation equal to that in the latter
by multiplying by ^ , and we have to solve
, _ 1 87917 462543
F09 70"5"2"8200
The difference =/f
We find accordingly
and o
The triangle is then
1 2108 7413881 T 90032 007119
8097^09^8200 J J 809709 2 8 ;j TTtT'
12. To find a rightangled triangle 4 f, rj such that
Form a triangle from the numbers x+ i, x ; the sides are
Thus i 7 = 2^ + $x* + 3^ + i must be a square.
Suppose 2X* + 5^ + 3# + i = (fa; f i ) 2 ,
and we have x =  ^.
H. D. 20
3o6 SUPPLEMENT
Now substitute y ^foix in the expression to be made a square, and
we have
whence 7 = W&, and * =
s& + Iff = a
 V =
The triangle is accordingly found.
13. To find a rightangled triangle f, , v\ such that
Form a triangle as before from x+i, x, and in this case we shall have
23? + 3^ + 3# + i = a square
= (f#+i) 2 , say,
whence # = f .
Substitute y  for x in the expression to be made a square ; thus
sV = a square
whence > = fttf . and * = !f  =
the triangle being therefore
5494337 3463712 426 5025
2"> 122 931^"
14. To find a rightangled triangle , f,
Let f = #, 77 = i ; then ^ 2 = ^ + i = a square]
Also, by the condition of the problem, x* + %x + i = a square]
The usual method of solution gives rc = .
Substitute therefore jy J for # in the two expressions, and we have
the double equation
Or, if we make the absolute term in the first expression the same as in
the second by multiplying by yfff,
5 3 2 9 1; 2 _ 101251 v , 5829 _
T5TFS J "BT5T5"^/ T 2304 ""
The difference =
and we find^ = , so that =>if = '
Therefore the two perpendicular sides of the triangle, in whole numbers,
are 39655, 129648, and the hypotenuse is 135577.
THEOREMS AND PROBLEMS BY FERMAT 307
5* To find a rightangled triangle , , rj such that
This problem is mentioned in Fermafs letters to St Martin of 3ist May
and to Mersenne of ist September I643 1 . The result only is given (in
the letter to Mersenne), and not the solution ; but it can easily be worked
out on the lines of the solution of the preceding problem.
Let = #, 77 = i j we must therefore have
I )
I both squares.
^X + I)
Solving in the usual way by splitting the difference %x into the factors
2x we find x = f$.
Substitute jvf^ for x in the two expressions, and we have to solve
Multiply the last by (f f) 2 so as to make the absolute terms the same ;
and we have to solve
The difference = {( * f ) 2  i } / +
We therefore put
whence
andj, = ^<, so that ^^fj^
The required triangle is therefore (%V^, Vinnnn J ) or, in whole
numbers, (205769, 190281, 78320).
1 6. To find a rightangled triangle & , rj such that
l? + m.%&j = a square,
Fermat takes the case where m  2.
Form a triangle from the numbers x, i ; the sides are then
Thus we must have (^ + i) 2 i 2 x (^ 2 ~ i) a square, that is,
x 4  + 2^ + 2^  2# + i = a square
say,
whence # = J.
1 Oeuvrts de. Fermat \ n. pp. 260, 263.
2O 2
3 o8 SUPPLEMENT
But this value makes a^i negative; so we must seek another by
putting y + % for x in the expression to be made a square.
We have y + $)? + ^jr  ^y + If = a square
This givesj> = fH4, and * =j> + i = fJ
Therefore the triangle is generated (in whole numbers) from 571663
and 436440.
De Billy adds that there is one case in which the problem is impossible.
Tannery observes in a note that this remark seems to refer to the case in
which m = 8.
17. To find a rightangled triangle , , 77 such that
t?^&l = a square.
Form a triangle from x i, 4 ; the sides will then be
Thus (x?2x i5) 2 (4# 4) (x*2xi$) must be a square, that is,
5C 4 8^14^ + ii2x + 165 = a square
= (^4^i5) 2 , say.
This gives x = ^, and accordingly, to find another value, we substitute
^f for # in the expression to be made a square.
We must therefore have
L = a square
This gives ^ = ^, and x =y  ^ = % 5 ^.
The triangle is therefore formed from  6 pnn 4> or (i n whole numbers)
from 6001, 2280.
The sides are therefore 412104015 30813601, 27364560.
1 8. To find a rightangled triangle f, 17
(fif2'ff = a square.
This problem is enunciated in Fermat's note on vi. 22. He merely
that the triangle (1525, 1517, 156) formed from 39, 2 satisfies the
dqnditions, but does not give the solution.
'The solution is however easy to obtain by his usual method, thus.
Form a triangle from x, r, so that
{ = 3*+ I, =flI, T\2X<
Then f^ 2 27 2 = ^2^i a 8^
THEOREMS AND PROBLEMS BY FERMAT 309
This has to be a square ; let it be equal to ($? 2X  5)*, say ; this gives
25,
or
The triangle formed from , i, or from 3, 2, will have one side
negative. To avoid this, we proceed as usual to form a triangle from
Thus =/6# + i3, =/6> + 5, 17 = 47 12,
and ^) 2 2i7 2 = 2
In order that this may be a square, suppose it equal to (jy 2 loy i) 2 or
2oy* + gSy 2 + 2oy + i.
It follows that i o 2j; 2 1 48y = 9 8y 2 + 2 ojy,
The triangle required is formed from jv3, 2, that is, from 39, 2,
and is accordingly 1525, 1517, 156.
Fermat does not tell us in the note on vi. 22 what use he made of this
problem, but the omission is made good in a letter to Carcavi 1 , where he
says that it was propounded to him by Frenicle (who admitted frankly that
he had not been able to solve it), and that it served to solve another
problem which had occupied Frenicle. The latter problem is the
following.
19. To find a rightangled triangle f, , 17 such that
 77 j are all squares.
*J
Fermat does not actually give the solution, but presumably it was
somewhat as follows.
Form a triangle from two numbers x, y \ the sides are then
Now i) = 3? +y*  2xy and is ipso facto a square.
The other conditions give
x* +y* = a square,
and x* ~y*  2xy = (x y) z  2j^ = a square.
These conditions are satisfied by the two perpendicular sides of the
triangle of the last problem, that is, by x= 1517, y 156.
1 Qtuvres $e Ferwat, II. p. 265.
3 io SUPPLEMENT
The triangle required is therefore formed from 1517, 156 and is
(2325625, 2276953, 47334
The present seems to be the appropriate place for a problem contained
in a letter from Fermat to Frenicle the date of which was probably
15 June 1641 \
20. To find all the rightangled triangles in integral numbers such that
the perpendicular sides differ by r.
If a rightangled triangle is formed from x, y, the difference between
the two perpendiculars is either x*j?2xy or 2xy(y?y*\ that is
to say, either (xyfzy* or if  (x  yf. As this difference is to be i,
we have to find all the integral solutions of the equation
2/(*J') 2 = i.
Those who are familiar with the history of Greek mathematics will here
recognise an old friend. The equation is in fact the indeterminate
equation
2?^ = I,
which the Pythagoreans had already solved by evolving the series of
"side" and "diagonal" numbers described by Theon of Smyrna, the
property of which they proved by means of the geometrical theorems
of Eucl. ii. 9, TO.
If x, y are two numbers such that
2 ff 3 y j = + 1,
then the numbers x+y, zx+y will satisfy the equation
2 2 >? 2 = i;
fresh numbers formed from x +y, 2x +y by the same law will satisfy the
equation
2 p_^ = +I ,
and so on.
Take now the equation
2/(*j/) 2 = i,
where x, y are two numbers from which a rightangled triangle has been
formed. We can deduce a rightangled triangle formed from x, y' where
2/ 2  (*'/>* = + * ;
for by the above law of formation we have only to take
whence also x' = 2x +y.
1 Oeuvres de Fermat^ u. pp. 331 sqq.
THEOREMS AND PROBLEMS BY FERMAT 311
Fermat gave two rules for the formation of this second triangle. The
first rule is in the letter above quoted.
First Rule. If^, /, b be any rightangled triangle satisfying the con
dition (h being the hypotenuse, / > b and / b i), then, if a triangle be
taken in which
the least side = 2)1 +f + 2^,
the middle side 2h+p + 2b + i,
the greatest side = 3^ + 2 (p + b\
this triangle also will be a rightangled triangle satisfying the condition.
To verify this from the above considerations we have to consider two
cases, according as 2xy is greater or less than x 2 y z .
Take the case in which 2xy > x 2 y 2 ; then
2y s (tfj/) 2 = ti,
and accordingly
*y> (*/)'= I,
or x'*y' 2 >2x'y'.
The least side, therefore, of the second triangle
2#y 2x(2X f y) = 2 (tf 2 + j/ 2 ) + (2xy) + 2(3? y) ;
the middle side
x' 2 y f2 =2x'y f +i ;
and the hypotenuse
x'* +y 2 = (2x + yf + x? = 3 (x* +y) + 2 (x* y* + 2xy).
The expressions on the right hand are those given by Fermat's rule.
Second Rule.
This rule is given in a letter of 31 May 1643 probably addressed
to St Martin 1 .
Fermat says : Given any triangle having the desired property, then, to
find another such triangle from it, "subtract from double the sum of
all three sides each of the perpendiculars separately [this gives two of the
sides of the new triangle], and add to the same sum the greatest side [this
gives the third side]."
That is to say, the sides of the new triangle are respectively
2 (2^ + 2xy) (x* y 5 ),
2(2^ i 2xy) 2%y,
2 (22? + 2xy) + (x 2 +y).
1 Oeuvres de Fermat, n. p. 359.
3i2 SUPPLEMENT
In fact the three expressions reduce as follows :
yf
and the result agrees with the formation of the triangle from x' 9 y' above.
From the triangle (3, 4, 5) we get (20, 21, 29); from the latter the
triangle (119, 120, 169), and so on. The sixth such triangle is (23660,
21. To find all the rational rightangled triangles in whole numbers
which are such that the two perpendiculars differ by any given number.
To his explanation of the First Rule above, applicable to the case
where the given number is i, Fermat adds in his curt way : "same method
for finding a triangle such that the difference of the two smaller sides is
a given number. I omit the rules, and the limitations, for finding all the
possible triangles of the kind required, for the rule is easy, when the
principles are once admitted."
He adds, however, to his Second Rule 1 its application to the case
where the given number is 7.
There are, he says, two fundamental triangles with the desired property,
namely 5, 12, 13 and 8, 15, 17. [In the case of the former 2xy> x*y\
and in the case of the second x*y* > 2xy.~\
From the first triangle (5, 12, 13) we deduce, by the Rule, a triangle
with the sides 2 . 30 12, 2 . 30  5, 2 . 30 + 13 or (48, 55, 73) ; from the
second a triangle with the sides 2.4015, 2.408, 2.40+17, or
(65> 72, 97)
And so on, ad infinitum*
Next to the explanation of the first of the above Rules Fermat
mentions, in the same letter, the problem
22. To find rightangled triangles in integral numbers , {, 17 (?)
such that
\
L are both squares.
>/]
He observes that alternate triangles of the series in which the two
smaller sides differ by i satisfy the conditions, those namely in which the
smallest side y is 2xy and not ^y 2 ; for x*+y*zxyis a square, and
17, being equal to i, is also a square.
1 Oeuvres de Fermat, 11, p. 359.
THEOREMS AND PROBLEMS BY FERMAT 313
Thus, while 3, 4, 5 does not satisfy the conditions, (20, 21, 29) does,
and, while the next (119, 120, 169) does not satisfy the conditions, the
triangle after that, namely (696, 697, 985), does.
Frenicle naturally objected, in his reply, that the triangles should not
be limited to those in which the smaller square representing the difference
between the perpendicular sides is i, and proposed the problem in the form
To find all the triangles (f, , 77) such that
O X
I are both squares ^
>?]
and one square does not measure the other.
Fermat seems to have, in the first instance, formed the triangle
from two numbers #, y where
x = r ! + i, y 2^2,
and then to have given the more general rule of forming a triangle from
where r, s are prime to one another (Letter from Frenicle of 6 Sept.
1641)'.
It appears from a letter of Fermat's to Mersenne of 2yth January
i643 2 ^ at St Martin propounded to Fermat the problem, apparently
suggested by Frenicle 3 ,
Given a number, to find how many times it is the difference between the
\ferpendicular ?] sides of a triangle which has a square number for the
difference between its least side and each of the two others respectively.
The number given was 1803601800, and Fermat replies that there are
243 triangles, and no more, which satisfy the conditions. He adds " The
universal method, which I will communicate to him if he asks for it, is
beautiful and noteworthy, although I doubt not that Frenicle has already
given him everything on the subject of these questions."
23. To find two triangles, f, f, 77 and ', f, t\ ( > 17, ' > vf) such that
Suppose the two triangles formed from (#, y) and (#', y') respectively,
the sides being
17 =
1 Oeuvres de Fermat, II. p. 233.
2 TZwT., p. 250.
3 Ibid., p. 247.
314 SUPPLEMENT
Then we must have
(*_y) 2 =2/ 2 (*'
and 2f  (x yf = (x' /) 2
which equations show thatj/=y, and that
are three squares in arithmetical progression.
Suppose that these squares are r, 25, 49 respectively; thus y  5 ;
x y  i, so that x = i + 5 ; x' y = 7, so that x f = 5 + 7.
Fermat accordingly gives the rule : Find three squares in arithmetical
progression ; then form the first triangle from (i) the sum of the sides of
the first and second squares and (2) the side of the second, and the
second triangle from (i) the sum of the sides of the second and third
squares and (2) the side of the second 1 .
In the particular case, the triangles are formed from (6, 5) and from
(12, 5) respectively; the triangles are therefore (61, 60, n) and (169, 120,
119) respectively.
For solving the problem of finding three square numbers in arithmetical
progression Fermat seems first to have given a rule which was not general,
and then in a later document to have formed the sides of the three squares
as follows :
Frenicle formed them thus 2 :
the latter form agreeing with Fermat's \ip = r + s, and q = s.
Fre*nicle expresses his formula neatly by saying that we take for the
side of the middle square the hypotenuse of any primitive triangle formed
from /, ^, i.e. p f q*, for the side of the smallest square the difference
between the perpendicular sides of the same triangle, i.e. / 2  f  2pq, and
for the side of the largest square the sum of the perpendicular sides of the
same triangle.
Suppose the primitive triangle is (28, 45, 53) formed from (7, 2).
Then the sides of the three squares in arithmetical progression are 17, 53
and 73, the squares themselves being 289, 2809, 5329. The triangles
derived from these squares and having the above property are formed from
(7> 53) an d from (126, 53) respectively, and are therefore (7709, 7420,
2091) and (18685, 13356, 13067).
1 O&uvres de Fermat \ II. p. 225.
2 Ibid., II. pp. 2345.
THEOREMS AND PROBLEMS BY FERMAT 315
24. To find two rightangled triangles (, , 17) and (', ', i
Form the triangle , & 17 from the numbers x, i ; then
{ = ^+1, ^ = ^ 2 i, 77=2,*;.
Thus f=a*+i; and, since ' i/ = 17 = 3? 2*? i, it follows that
It remains to secure that ' 2 + ^ /2 = (x* + i) 2 + (2* + 2) 2 shall be a square,
that is,
#* + 6x* + Sx + 5 = a square
= (^^3) 2 jS ay;
therefore ^ = J.
Hence the triangle , 1, 17 is formed from , i or from i, 2 j but this
solution will not do, as it gives a negative value for Accordingly
we must find a fresh value for x, which we obtain by forming the triangle
from x + i, 2.
The sides are then
^=^+2^ + 5, ^ = ^+2^3, 77 = 4^ + 4;
thus ' = # 2 +2#+5, i7 / = / (# 2 2#7) = 4tf+ 12.
Therefore (# 2 + 2# + 5) 2 + (4^; 4 i2) 2 must be a square, or
:tf 4 + 4# 3 + 3otf 2 + 1 16^+169 = a square
^(is + fl^^) 2 * sa y
from which we obtain x = / 9 and the triangle is formed from f, 2,
or (in whole numbers) from  979, 1092.
" We can use these numbers as if both were real and form the triangle
from 1092, 979. We thus obtain the two triangles
2150905* 2138136, 234023,
2165017, 2150905, 246792,
which satisfy the conditions of the question."
25. To find two rightangled triangles (, , 77) and (, ^, 77') such that
Form the triangle , , t\ from the numbers x + i, i j then
= x* + 2X+ 2, ^ = ^+2^, 77
Thus ' = # 3 + 2#+2, and 77' = ^+07^= 2x.
3i6 SUPPLEMENT
We must now have ' 2 + i/ 2 = (x*+2X + 2)* + (zxj* a square ; that is,
x* 4 43? 4 123? + Sx f 4  a square
= (3? + 2X + 4) 2 , say ;
whence x =  f .
Accordingly we substitute j> f for #, and we must have
y _ 2 f + lj? *? + * = a. square
This gives ^ = J, and * = f  1 = ,&,
The triangle , f, 17 is therefore formed from f$, i, or from 29, 26, and
is therefore 1517, 165, 1508 ; the triangle ', f, V is 1525, 1517, 156.
Or again we may proceed thus from the point where we found x = f.
The triangle , f, 17 may be formed from  J, i or from  i, 2.
We therefore form a triangle from # i } 2 and start afresh.
The sides are
^ = 3^2^45, ^=^2^3, ^ = 4^4.
Thus ' = X s 2X + 5, and ff = 4 77 f = 4#  12.
Hence (x* 2X 4 5)* + (4#  i2) 3 must be a square ; that is,
^4^+30^ n6x+ 169 = a square
= (i3^ + ^) 3 , say.
This gives # = Y, and the triangle f, , 17 is therefore formed from
f, 2, or from 29, 26, as before.
The remaining problems on rational rightangled triangles in the
Inventum Novum are cases given in Part n. of that collection to illustrate
the method of the TripleEquation due to Fermat and explained by him on
Diophantus vi. 22 as well as, at greater length, in the Inventum Novum.
An account of the method will be found in a later section of this Supple
ment; but the problems applying the method to rightangled triangles
will be enunciated here.
26. To find a rightangled triangle f, f, t\ such that
By Problem 2 above find a rightangled triangle h, p, b (h being the
hypotenuse) in which h,p + b are both squares; the first condition is thus
satisfied.
THEOREMS AND PROBLEMS BY FERMAT 317
To find ft i), put =&, =/#, >/ = <fo.
The three remaining conditions thus give a "tripleequation" in x.
[The numbers would of course be enormous.]
27. To find a rightangled triangle. , , >y such that
#2 w fl/zj; given number.
Fermat supposes m = 2.
Assume for the required triangle (3^ 4#, 5^); we have then the
tripleequation
144^+ 4^=z
144^+ iox = i
the solution of which gives x = ffFDTTrj an< i tlle triangle .is
3
28. To find a rightangled triangle f, f, 17
Suppose w = 2.
Find a triangle (Problem 3 above) in which , 17 are both squares,
say the triangle (119, 120, 169). Put 1190;, i2o#, 1693; for the sides of
the required triangle, and we have the " tripleequation "
166464^+ IK)X =
1 66464^ hi 20^ =
166464^ + 338^ =
29. To find a rightangled triangle ft f, y such that
m is any given number.
* The enunciation has (  }(fy) instead of 2  ^17 ; but (  J^;) is inconsistent with
the solution given, and I have therefore altered it so as to correspond to the solution.
3i 8 SUPPLEMENT
Take a rightangled triangle in which i, p are the sides about the right
angle and are such that i\p is a square (Problem 1 1 above).
Let q be the hypotenuse of the triangle so taken, so that q= x /(/ 2 + I )
and take as the sides of the required triangle x, px> qx ; we thus have
the tripleequation
(i +/ + qf x* + mqx =
30. To find a rightangled triangle , , 17 such that
First find a triangle in which one of the perpendiculars is a square, and
the sum of the perpendiculars is also a square, say 40, 9, 41.
Take 4037, yx, 4ix as the sides of the required triangle; and we there
fore have the tripleequation
SECTION IV.
OTHER PROBLEMS BY FERMAT.
31. To find two numbers , TJ such that
(0 e(eff\
(2) rj  (I?  if) > are all squares.
(3), ( 4 ) fi^^^J
Let +^=i2^, gif)=2x, so that = , vj^2x, and
ev)*=2x 4 x*.
Thus (3), (4) are both satisfied.
The other conditions (i) and (2) give
The difference = 2x = 4x . \ and, putting (2x + J) 2 = 4^ 2  2X + J, we
find tf = .
THEOREMS AND PROBLEMS BY FERMAT 319
The required numbers are therefore J, / T .
Another pair of numbers satisfying the conditions will be obtained by
substituting^ + T ^ for x in the expressions to be made squares, and so on.
32. To find two numbers , y such that
. are all squares.
Let the numbers be \ + x, $x; therefore f^, as well as i 2 ?? 2 , is
equal to 2x. The sum f + rj = i.
Therefore i + 2# must be a square, or we have the doubleequation
Replace a? by jy +7 so as to make i + 2# a square ; therefore
i 2y y~ = a square
whence jy = f , and # = \y* + jj> = f
The required numbers are therefore 4>
33. To find two numbers ( 9 vj suck that
Assume =x, y = 2x', therefore
(^ + 97) (g> + ^) = 2 (20? 4% + 4) = a cube
= (2^^) 3 , say.
This gives # =  ; and to get a " real " value of x we must substitute
y  for x in the expression to be made a cube.
Thus 4/y 2  44y + 125 = a cube
T o c _ vt vin^ O. 1936 *
125 44^ + "VTS
, so that *j;* =
The required numbers are therefore
COR. We observe :
(1) that the numerators 26793, 15799 satisfy the conditions ;
(2) that we have in fact solved the problem To divide 2 into two parts
such that twice the sum of their squares is a cube ;
(3) that we can solve in the same manner the problem To find two
numbers such that any multiple of the sum of their squares is a cube. Thus
suppose that the multiple is 5 ; we then assume x and 5 x for the
numbers and proceed as above;
320 SUPPLEMENT
(4) that we can also deduce the solution of the following " very fine
problem " :
To find two numbers such that their difference is equal to the difference of
their biquadrates or fourth powers.
In other words, we can solve the indeterminate equation
17 = ? +
For we have only to take the two numbers found above, namely 26793
and 15799? and divide by (as a common denominator) the root of the cube
formed by multiplying their sum by the sum of their squares.
This common denominator is 34540, and the two required numbers are
This latter problem is alluded to in Fermat's note to Diophantus iv. 1 1
in these terms : " But whether it is possible to find two biquadrates the
difference between which is equal to the difference between their sides is a
question to be investigated by trying the device furnished by our method,
which will doubtless succeed. For let two biquadrates be sought such
that the difference of their sides is i, while the difference between the
biquadrates themselves is a cube. The sides will, in the first instance, be
^ and f. But, as one is negative, let the operation be repeated, in
accordance with my method, and let the first side be #^; the second
side will be x + ^f , and the new operation will give real numbers satisfying
the condition of the problem V
34. To find two numbers f, 77 such that
4 + 3 1 ? 4 = a square.
Fermat (or De Billy) observes that it must be required that the first
biquadrate ( 4 ) shall not be unity, for in that case the problem would be
too easy, since 1 + 3.1=4 and i + 3 . 16 = 49.
Assume = #, TJXI ; therefore
4#* I2& 8 + iS^c 2 1 2X + 3 = a square
= (2^ 3 ^ + f) 2 , say.
This gives x = ^, x i = f ; and a solution in whole numbers is
= 11, 17 = 3. In fact n 4 + 33 4 = 14641 + 243 = 14884 or I22 2 .
We can also take any equimultiples of (11, 3), as (22, 6) and (33, 9) ;
and the latter pairs of numbers severally satisfy the condition of the
problem.
1 It gives in fact ^123 ^ I 5799 ^ a so i u ti on O f the subsidiary problem, and from this
10994 10994
we can obtain the same solution of the main problem as that given above
M793 5799\
\3454o' 34540;
THEOREMS AND PROBLEMS BY FERMAT 321
SECTION V.
FERMAT'S TRIPLEEQUATIONS.
Format's own description of his method of " tripleequations," which is
contained in his note on vi. 22, is as follows :
" Where doubleequations do not suffice, we must have recourse to
tripleequations, which are my discovery and lead to the solution of a
multitude of elegant problems.
If, for example, the three expressions
x + 4, 2x + 4, 5^ + 4
have to be made squares, we have a tripleequation the solution of which
can be effected by means of a doubleequation. If for x we substitute a
number which when increased by 4 gives a square, e.g. y 1 + %y [Fermat says
oP + 4jc], the expressions to be made squares become
The first is already a square ; we have therefore only to make
2 j/ 2 + 8>' + 4i
$y'> + 2oj + 4 J
severally squares.
That is to say, the problem is reduced to a doubleequation.
This doubleequation gives, it is true, only one solution ; but from
this solution we can deduce another, from the second a third, and so on.
In fact, when w have obtained one value fory [sayj; = #], we substitute
for y in the equations the binomial expression consisting of y plus the value
found [z.<?. y + a]. In this way we can find any number of successive
solutions each derived from the preceding one. 3 '
The subject is developed in the Doctrinae Analytical Inventum Novum
of De Billy already mentioned so often.
It will be observed that the absolute term in all the three expressions
to be made squares is a square. It need not be the same square in the
original expressions j if the absolute terms are different squares, the three
expressions can, so far as necessary, be multiplied by squares which will
make the absolute terms the same, when the method will apply.
We may put the solution generally thus. Suppose that
bx + q 1
have to be made squares (a, b, c or some of them may be negative as well
as positive).
Put ax =y* + 2py,
which makes the first expression a square (or of course we could put
ax = afy z + 2apy\
H. D. 21
322 SUPPLEMENT
Substitute (jr + 2#y)/a for x in the second and third expressions,
Therefore  (f + 2py) + <f 1
must both be squares ; or, if we multiply the first expression by r 3 and the
second by gr (so as to make the absolute terms the same), we have to solve
the doubleequation
The difference = ^t ./ 4 tf   ~^ >
This has to be separated into two factors of the form A/y, p/y + v, where
v must be equal to zqr (in order that, when \ {(A + /*)_y + /} is squared and
equated to the first, or when \{(^\)y v} 2 is equated to the second, of
the two expressions, the absolute terms fr* may cancel each other).
A different separation into factors is possible if bfa and cfa are both
squares ; but otherwise, as Fermat says, the method gives only one
solution in the first instance; the above difference must necessarily be
split into the factors
p(br*ef) , qr
r  2M' and ~y + 2ffr.
aqr * P
Half the sum of these factors
y '
!<%
Squaring this and equating it to (jr* f zpy) + ^V 2 , we have
therefore
tepbi* _ af^ +
~ } ' I a ' "
a p
01
THEOREMS AND PROBLEMS BY FERMAT 323
that is,
y (aY^ + #v' 4 / + #V  2 A^VV 1  zcaffi* 
+
V fr +  aV
r
whence # ( =    ) is found.
V a J
Exx. from the Imvntnm Novum.
(1) 2^+4^j
$x + 4 I to be made squares.
6x + 4 )
Here a = 2, /;6, c^ p gr2\ therefore
__ 4 > 2 . 32 (6 . 16 + 3 . 16 2.16)
^"~i6 2 (4 +36 + 92. 6. 32. 3. 22. 2. 6)
_ 16 . 7
~~^T'
, I/o \ I flI2 2 4,112] 56, N II2O
and ^=(r 2 +4i') = s ~ ~~  h =41124. 23) =  .
2 W ^^ 2l23 2 2 3 J 23 2V * >' 529
(2) a+rj
3^ + 4 ! to be made squares.
2^ + 9 J
Here a = i, <5 = 3, c 2, / = i, q  2, ?= 3 \ therefore
2 +9 8 1 +4. 1612.3616. 366 . 36. 9
43 6
36.46 + 9.8114.16 863'
n^A v _ .,2 i . /Ii4\2 , ,, /144\ _ 203 8
ana # j + 2j'  ( T ^) + 2 (&$ s )  fTTTTnr
The disadvantage of the method is that it leads so soon to such very
large numbers.
Other examples from the Inventum Novum are the following, which,
like those above given, can be readily solved ab initio without using the
above general formula.
(3) To solve i+ x
Put x = j/ 2 + 2j/, and substituting in the second and third expressions we
have only to solve the doubleequation
=74V 3 J *
21 2
324 SUPPLEMENT
The difference = 3 (y* + 2y) = %\> (y + 2).
Equate the square of half the difference of the factors to the smaller
expression; thus
(y l) a =2/ + 4J'+I,
whence y =  6, and x =jr + 21  24.
(4) Equations x + g = v*}
53: l 9 = rv
In this case we put x =j' 2 + 6j/, and we have to solve
. 3(/+6y) + 9^ 2 )
5(y + 6y) + 9=7/ 3 J '
The difference = 2 (/ + 6y) = 2jy + 6) ; we then have
and jv = f, so that # = f~ + 6y =
(5) Equations i+
If we assume x =y* + 2j^, we find y = ^ and x =
There are two other problems of the same sort which are curiously
enunciated.
(6) "To find three cubes such that, if we add their sum to numbers
proportional to the cubes respectively, we may have three squares."
What Ferraat really does is to take three cubes (a s , b*, c*) such that their
sum is a square (this is necessary in order to make the term independent
of x in each of the three expressions a square) and then to assume
aPx, fix, c*x for the numbers proportional to the cubes. He takes as the
cubes i, 8, 27, the sum of which is 36. Thus we have the triple equation
36 +
36+
36 + 27^ = w*
We put x=y*+ i2y in order to make the first expression a square.
Then, solving the doubleequation
36+ 8(y H2J/) = ^ 1
36 + 27 (y + i2j/) = w* ] '
we obtain y = ^ and x =/ + 127 = ^fJHnr 2 .
(7) "To find three different square numbers such that, if we add
to them respectively three numbers in harmonic progression, the three
resulting numbers will be squares."
Fermat first assumes three square numbers i, 4, 16 and then takes
2X 9 33;, 6x as the required numbers in harmonic progression. (He observes
THEOREMS AND PROBLEMS BY FERMAT 325
that, of the three numbers in harmonic progression, 'the greatest must be
greater than the sum of the other two.) We thus have the triple equation
I 4 2X = U 2 "I
4 + 3^z 3 [,
16 + 6x = w* )
or, if we make the absolute terms the same square,
16+ i2x = tf* .
16+ 6x = w*j
Making the last expression a square by putting \f + y for x, we solve
as usual and obtain y =  ^ and x = J (/ + By) = ^^nr 1 
Fermat observes that tripleequations of the form
that is to say, of the form
+ ao; =
can be similarly solved, because they can be reduced to the above linear
form by putting x= i/y and multiplying up by/ 2 .
Examples.
(i) To solve the tripleequation
If x=sify, this is equivalent to
2JV + 4 =
Putting j = l?z 2 i 2$ and solving as usual, we find
* = W^ =
(2) Equations
This is equivalent to
j> 4 1 = w'
We put j; = z + 22 and, solving the doubleequation
27 (^ 3 + 20) + 36 = a 2 )
we find z = ^, j; = 4 ffff^, so that * =
326 SUPPLEMENT
(3) " To find three square numbers such that, if we add their sum to
each of their roots respectively, we obtain a square."
Choose, says Ferroat, three squares such that their sum is a square and
such that the root of the greatest is greater than the sum of the roots of the
other two (the reason for this last condition will shortly appear) ; e.g. let
the squares be 4, 36, 81, the sum of which is 121.
Let 4.T 2 , 36^, BIX? be the three square numbers required; therefore
121 x* + 220 =
1 2 r# 2 + 6x =
The solution, arrived at as above, is x = ^
Fermat actually used his tripleequations for the purpose, mainly, of
extending problems in Diophantus where three numbers are found
satisfying certain conditions so as to find four numbers satisfying like
conditions. The cases which occur are in his notes to the problems
in. 15, iv. 19, 20, v. 3, 27, 28; they are referred to in my notes on
those problems.
De Billy observes (what he says Fermat admitted he had not noticed)
that the method fails when, the absolute terms being the same square, the
coefficient of x in one of the linear expressions to be made squares is equal
to the sum of the coefficients of x in the other two. Thus suppose that
I + 2X, I 4 3#, I 4 5#
have to be made squares. To make the first expression a square put
x = 2j> 3 + 2j. The other expressions then become
i + 6y 4 6y 2 , i + iqy + ioy 2 .
The difference is 4^ + Ay  zy (%y + 2), and the usual method gives
or 6y + 6y = o,
so that jy =  1, and consequently x = 2j/ 2 + zy  o.
It does not however follow, says De Billy, that a set of expressions so
related cannot be made squares by one value of x. Thus i 4 5#, i + i6x
and i +2ix are all squares if x = $, the squares being 16, 49, 64. He
adds ( n) that "we must observe with Fermat " that the tripleequation
i+ x =
I + 2X = Z
I 4 3# =
not only cannot be solved by the above method, but cannot be solved at
all, because "there cannot be four squares in arithmetical progression'' which
however would be the case if the above equations had a solution and we
took i for the first of the four squares.
THEOREMS AND PROBLEMS BY FERMAT 327
The subject of tripleequations has been taken up afresh in a recent
paper by P. v. Schaewen 1 . The following are the main points made,
(i) The equations ax +f = \
can be reduced to the form
by substituting MX* for #, where w is the least common multiple of f, $*, r 1 .
(2) The method of Fermat has the disadvantage that, with one
operation, it only gives one value for x and not by any means always the
smallest solution. From this point of view there is a better method, namely
that of finding the general solution of the first two equations, substituting the
general value of x so found in the third equation and solving the resulting
equation in a new unknown. Consider the equations
i + ax u 2 1
i + ex = ze/ 2
Suppose i + ax  / 2 , some square. Therefore
!+&*= i +(/* 1),
and, multiplying by d\ we have to make
abf 4 a z  db a square.
This is a square if /=i ; and we therefore substitute q + i for/. Thus
abf + zabq + a 1 = a square
say.
Therefore (<z  ^ J ^ = 2 ( a  J ,
2fl# (/ ^)
and ^ abfi* '
whence * = ? +I
.
and
 4 mn \np
(abtfnff
Substituting this value of x in the expression i + ex, we have a biquad
ratic expression in which has to be made a square, namely
m*  4 8 + {4 (a + S) c 2ab] m*n* ' ^abcmtf f 3 ^ 4 .
1 Bibliotheca Mathematica, 1X3, 1909, pp. 289300.
328 SUPPLEMENT
Example. Find x such that
i  x t i + 4Xj i + 7# are all squares.
First find the general value of x which will make the first two ex
pressions squares ; this is
 qtnn (fffl  $mn  40*)
*~ (;;* 2
or, if we substitute for 2n\m,
We have now to make r 4 7# a square j that is,
& 4 14# 4 23/fc 2  14^ + i = a square.
The first solution of this is k = i, and by means of these values we get
the further values k = and k = fj (cf. Euler's solution of the problem of
making #* 4 1 and #41 simultaneously squares quoted in my note on
pp. 84, 85). The corresponding values ofx are respectively
3 120 . I20I20
 , 5 and r .
4 29* 421
Fermat's method gives, as the next solution after f , the value
51294243
X
(3) v. Schaewen observes that the problem of finding x such that three
different expressions of the form mx 4 n are all squares can always be solved
provided that we know one solution ; in this case the absolute terms need
not be squares. I doubt however if he is right in supposing that the
possibility of solution in this case was not known to Fermat or De Billy.
I think it probable that Fermat at least was aware of the fact; for this case
of the tripleequation is precisely parallel to that of the doubleequation
2X 4 5 = I
6x 4 3 = i
given as a possible case by Fermat in his note on Bachet's conditions for
the possibility of solving doubleequations (cf. note on p. 287 above).
Fermat says that the square to which 2% + 5 should be made equal is 16
and that to which 6x 4 3 should be made equal is 36 (corresponding to
#=5j) adding that an infinite number of other solutions can be found.
(4) Lastly, v. Schaewen investigates the conditions under which the
equations
which cannot be solved by Fermat's method, are nevertheless capable of
solution, and shows how to solve them when they htm a solution other
than # = o,
SOLUTIONS BY EULER. PROBLEM i 329
SECTION VI.
SOME SOLUTIONS BY EULER.
PROBLEM i. To solve generally the indeterminate equation 1
Vieta solved this equation on the assumption that two of the four
numbers are taken as known.
[I noted on p. 102 Euler's remark that, if 3 3 + 4 is turned into the
difference between two cubes by the direct use of Vieta's second formula,
the formula gives 3 3 + 4" = (tf) 3  (4ft*) s but not 3' + 4* = 6 s  5*. I ought
however to have observed 2 that the latter can be obtained from Vieta's />.#
formula if we multiply throughout by a 3 + P. The formula then becomes
a* (a* + bj = 3 (a 3 + J) 8 + <r (a?  2^)* + P (26*  P)*.
Putting #=2, 3=i, we have i8 3 =9 3 ti2 3 f 15*, which gives (after division
by 3 s ) 6 3 = 3 s + 4 3 f 5 s . The next solution, obtained by putting a = 3, 3= i,
is' 84^ = 28 + 53 3 + 75 s ; if a  3, b = 2, we have io5 3 = 33 3 + 70 + 92; and
so on. Similarly Vieta's second formula gives
a 8 (a* + 2^) s = a 9 (<fi  3 3 ) 3 + & (a*  P)* + l>* (20* + P)*>
and we obtain other integral solutions; thus
if # = 2, ^=r, we have 20 3 = 7 3 +i4 3 f 17^,
if a = 3, =i, we have &f= 26'* + 55 3 + 78*;
and so on.]
(i) A more general solution can be obtained by treating only one of
the three numbers x, y, z as known.
To solve
put x
therefore
and we obtain, after dividing out
1 JV. Comment. Acad. Petfop. 175657, Vol. vi. (1761), pp. 155 $qq.
arithm. l. pp. 193206. Cf. pp. 1012 above.
3 See Nesselmann's " Anmerkungen zu Diophant" in the Zeitschrifl fiir Math. u.
Pkysik, xxxvii. (1892), Hist. litt. Abt. p. 123.
330 SUPPLEMENT
 f)  >* (P
+ r =
where a, r and the ratio p : q may be given any values we please.
(2) A more general solution still is obtained if we regard none of the
first three cubes as known.
Suppose that, in the equation
x = mt +///, v = nt + qu, s= ut + m.
Therefore
+ 3V r + ytf \ + <? \
+ r)
n .
Put now p = ;^/ +  ^L_ ~ y //,
w
and we have, after division by w 3 ,
3^ {mf + n(f r*}} + (/ + ^ +
whence, neglecting a common factor which may be chosen arbitrarily, we
have
/ = m* (f + f + r*)  \m*p + tf (q + r)}\
u = 3^ 3 {w^ + a (^ 4 r)Y  3^ 6 {^/ + n (f  r 2 )},
or, if we divide by the factor q + r,
t = m* <
so that ^ j/, j and y can be written down.
The solution is, however, still not general.
(3) General solution.
To find generally all the sets of three cubes the sum of which is a cube.
Suppose A* + & + C 3 = H\ or A 3 + * = &* C\
and assume A=p+q, #=/# C^rs, D = r + s.
Then A 9 + & = 2/ 1 + 6^, Z^  C 3 = 2^ + 6r*s,
so that
SOLUTIONS BY EULER. PROBLEM i 331
This equation cannot subsist unless p* + 3^, j 2 + $r* have a common
divisor. Now it is known that numbers of this form have no divisors
except such as are of the same form.
To find them, we introduce six new letters to take the place of
A ?> r, s, thus : let
whence f + 3^ = ( 3 + 3 P) (tf + ^), s> + ^ = (<f + 3 r) (or 1 + ay*),
and our equation, divided by x 2 + 37*, becomes
(a* + 34?) (tf 2 + 3^ 2 )
so that f  " & (** +
j " "~a(a* + 3#) + </ (^ + 3?) '
and we may put x =  $nb (or + 3^) + $nc (d* + y a ),
Hence the values of/, ^, r, s are found to be
3^,
3^)  n (a +
and A = n ($ac + 3^  ad + $bd] (d* + 3^)  n (a 1
^ = (3^  3^ + ad + tfj) (d* + 3^) + n (a 2
j>= (^ 3 + 3^) 2 + n ($ac  $!>c + ad
These values satisfy the equation
and, since no restriction has been introduced, the solution is capable of
giving all the sets of three cubes which have a cube for their sum.
More special forms for A, 2?, C, D can of course be obtained by putting
zero for one. of the letters a, &, c, d, and still more special forms by com
bining with the assumption a = o or b = o the assumption d=c, or com
bining with the assumption c= o or d o the assumption b = a.
Two cases are worth noting.
First, suppose b o, d c^ and we have
A = Sna<? na\ J3 = i 6nat? + na 4 , C =
If further we write za for a and /i6 for , we have
A = na (<*</), B=na(2<*+<?), C=*e (<*<?), D = nc
which is equivalent to Vieta's solution of his second problem.
Secondly, suppose d= o, b  a, and we have
A~\ &ncu? 1 6//# 4 , B =
or, if we write !# for ,
*, ^ = na*, C =
332 SUPPLEMENT
which, if n = a  c  i, gives the simplest solution of all
A = S, B=i> C = 6, Z>= 9 , and i* + 6 3 + 8* = 9*.
In proceeding to other solutions we have to remember that, while
A, B, C, D must be integral, they should all be prime to one another; for
those solutions in which A, B, C 9 D have a common factor are not new
solutions in addition to that from which the common factor is eliminated.
Thus, while giving any values, positive or negative, to the numbers
, b, c, d in the formulae
y = nd (d* + 3^) + na (d 1 + 3^),
we have to choose for n such a fraction as will make x 9 y prime to one
another. We then form
and, after again eliminating any common factor, we put
A=p + y> B=pq, C=rs, D =
and we shall have A* + B* + C* = D 3 .
(The cases in which one of the three cubes A s , *, C 3 is negative will
give the solutions of the equation a 8 +j?  z* + tf.)
While any values of #, , c 9 d may be taken, it is necessary, if we want
a solution in which A, B^ C, D will be small numbers, to choose a, b, c, d
so that d 1 + 3^, d* + 3^ may have a common factor. Euler accordingly
sets out a table of all numbers of the form m 2 + 3 2 less than 1000 (giving
m values from i to 31 and n values from i to 18), and then chooses out
cases in which or + $b\ d* + 3^ have a tolerably large common factor.
Now, assuming that a 2 + $&  mk t
we have (supposing further that ac+ bd=f, $bcadg)
In these formulae/, g may be either positive or negative, the signs of
t>, t, d being ambiguous ; and we may put
either /=(^ +w ) I or /= ("
But, if /changes sign while ^remains unaltered, we get numbers of the
same form, only in different order ; therefore we may confine ourselves to
the positive sign In/
SOLUTIONS BY EULER. PROBLEM i
Example i. Let
a?+ 3& 2 = 19, so that a = 4, b i,
d? + 3^ = 76, so that ^ = i ) or //= 7 ] or d= 8
* = 5J ^=3J ^=2
Then ;// = i, = 4, = 19,
The following values for/ ^ result, viz.
I. /=2i, II. /=i 9 , III. 71
333
IV. /= S ,
 /=i6, VI. /= 0f
^=26, ^=
while, since m = i, n = 4, /' = 19,
^  304 + 3/^
The values (VI) /= o, g= 38 are excluded because, if/= o, A =  B
and C=Z>.
The values (I) give
= 233 + 44, that is,
271^44
241 + 11
T = 277
189 or
= 315
#=227
(7=230 I (7=252
= 378
^367 + 11 > = 35 6 # = 378 >=
The values (II) and (III) give, after division by 19,
^ = 11 + 4, that is, A = i$orA = $ A= 7
= 11 + 4, that is,
^=13+4
D= 19 + 1
The values (IV) give
A 41 + 148, that is,
B^ 79+148
A = i 5
B= 9
Z> = i8
(7=4
7=14
= 189 or A 63
37
Lastly, the values (V) give
^=173+104, that is,
^ = 211 + 104
=256+ 26
^ = 69
C= 252
Z>= 282
^ = 277
.#= 107
= 23
7= 84
?= 94
Ai^
B= 227
jC= 336
^= 356
= 326
^= 69 or <<4 = 23
(7=282 C= 94
72 = 378 Z>=is6
334
SUPPLEMENT
Thus from the one assumption for a a + 3^, d* + $c we have the
following solutions :
107* 4 230* + 277"'= 326* 23*+ 94 3 =
7 3 H 14 s + 17' = 2o 3
3 + 4 + 5 ;I = 6 8
Example 2. Assuming
^ + 3^= 28, so that rt=i]or0 = 4)ortf = 5
84, so that ^=3} or d=6\ or ^ =
d= 3 1 or d=6\ or ^=9}
<r=5J <r=4J 6=iJ'
we have ^=28, mi, n 3, and the following solutions will be obtained :
I2=
I0
0'* H
PROBLEM 2. To find three numbers x, y, z such that
xy, xz, ys,
are all squares.
First solution?*.
Assume that x y p^ oc z  ff, y s = r^ ;
therefore y = ocp\ z = xf, and ^ 3 =/f ^.
The first three formulae now become
x +y 2X p\ x + %  zx  <f, y + z = 2X fP  g*.
Suppose that zxp^f = / 3 , so that zx = f 2 + $* + // , therefore we
have to make ^ + f and f +p* squares, while in addition f =/ 2 4 r*.
Let ^ = a a + /; 2 , p^a*P, r~*ab\
then f 2 + (^ 3 + P)* = ? + &' + &* + *cPP 1
and p + (d*  //) 2 = p + a * + ^ _ 2 a^ 2 J
must be made squares.
Comparing now / 2 + 4 + 4 with ^ + ^ and zaW with 2<r4 let us
suppose f=fl^=/y##, ^=/y, ^=^ 2 /^, a 2 / 3 ^ 3 , ^ 2 =^ (or a=//^,
3 =^) ; then the assumption / 2 + /z 4 + 3 4 = r 2 H^ 2 will assume the form
or
1 Algebra, Part II. Art. 235.
SOLUTIONS BY EULER. PROBLEM 2
335
. Hence the problem is reduced to finding the differences of two pairs of
fourth powers, namely / 4 / 4 and^ 4 ^ 4 , the product of which is a square.
For this purpose Euler sets out a table of values of m^n 4 corre
sponding to different values of m, , with a view of selecting pairs of
values of m 4 n* the products of which are squares.
aw 3
11*
4
I
9
I
9
4
16
i
16
9
2 5
i
25
9
49
i
49
16
64
i
81
49
121
4
121
9
121
49
144
25
169
i
l6g
Si
22 5
64
*
W 2 r W 3
*
3
5
3 5
8
10
16. 5
5
13
5 . 13
15
17
3
5 r 7
7
25
25 7
24
26
16.
i6
34
16.
2.17
48
50
25
16.
2. 3
33
65
3
5
1113
63
65
9
5
7i3
32
130
64.
5 1 3
117
125
25
9
5 J 3
112
130
16. 2 ,
5
7i3
72
170
144.
5 I 7 :
ir 9
169
169 .
717
168
170
16. 3.
5
7.17
88
250
25'
16.
5 1 "
161
289
289.
723
One solution is obtained from/ 3 = 9, & = 4, g* = 81, >^ 2 = 49, whence
= a 2 ^ 2 = 117,
therefore
Therefore
= ^ 2 I b~ = 765/^ = 2^=756;
p= 434657]
y = Xf~ 420968 I.
5 = x ^ = 1505 68 J
The last number may be taken positively; the difference then becomes
the sum and the sum becomes the difference ; therefore
#=434<557, x+y= 855625 = (925)*, a?
7 = 420968, x + z = 5 8 5225 = (7<55) 3 >
0=150568, J ; 4 0= 571536 = (756)2,
We might also have taken / 2 = 9, #4,
equally have given a solution.
13689 =
~z = 284089
0 = 270400 = (52o) 3 .
2 = 121, ^ 2 = 4, which would
336 SUPPLEMENT
Second solution 1 .
This later solution (1780) of Euler's is worth giving on account of the
variety of the artifices used.
We can make x+y and x y squares by putting x=f + f,y^2.pq.
Similarly x + z, x  z will be squares if x = r* + s 3 , z = 2rs.
Therefore four conditions will be satisfied if only / + f = r* + r .
Now [cf. Diophantus in. 19 and pp. 1056 above] if we put
x = (a* + l?)(c* + d*\
x can be made the sum of two squares in two ways ; in fact
p  ac + bd) radb &*,
q~adbC) s = ac bd>
and
y = %fg = 2 (c?cd + obd*  abc*  &cd\ z=2rs = 2 (tfcd + abP  abd*  IPcd),
so that y + 2 = ^cd (a  P), yz $ab (d*  r).
These latter expressions have to be made squares.
First make their product y* z z a square ; this means that
ab (tf 2 ^ 2 ) . cd(& ^) must be made a square.
To effect this, let us assume that cd(d* c*) rPab^cP b^) \ we may
further, since the question depends on the relations between the pairs of
letters a> b and c, d, suppose that d= a.
We have then c (a*  <*) = tfb (a 2  F),
whence a 2 = ^ , which fraction has accordingly to be made a square.
wo ~~ c
373
Suppose that a = l>c, so that ? .  fi 2fc + c\ and we have
b
therefore
Put b  n? i 2 and c= 2n 2 + 1 ; therefore a = i 1? = d.
As we have now made the product of the expressions ab (d*  <?) and
cd^P) a square, it only remains to make either of them singly a square,
say ab (d*  ^).
But ab (d* <?) = ab(d c) (d + c) = $n* (ri*  i) (n 2 + 2) 2 .
We have therefore only to make 3 (w 2  i) a square, which is easy, since
if i has factors ; for we have only to put
which gives 3 (  i) = ( + i), or = ^j^ 
1 Mim&ires dd T Acadtmie Impfriale des Sciences de St Ptter$bourg> 181314, VI, ( i8r 8),
pp. 54 sqq* = Commentationes arithmeticae, n. pp. 3925.
SOLUTIONS BY EULEIL PROBLEM 2 337
[Euler had previously tried the supposition a  b + <, which would
require ^(n z +i) to be made a square, which is impossible.]
All the conditions are now satisfied, and we have to find a, b, c\ d etc.
in terms off, g:
As the whole solution depends on the ratios of a, b, c> d^ we can
multiply throughout by the common denominator, 'divide by 3, and put
whence p =  8/y (/ 4 + 9^), r =/ 8 + 3 q/V +
[Euler took # to be 2 i instead of i  2 and consequently obtained
positive signs for the values of/ and s , he also has q (/ 4  Q^ 4 ) 2 , which
appears to be a slip.]
Assuming therefore any values for f 9 g in the first instance, we first find
values for a, b> t, d, then values for /, q , r, s, and lastly values for x, y, z.
It is to be observed that it is a matter of indifference whether we get negative
values or not; for positive values can be substituted without danger.
Euler gives four examples.
If/= i, g= i, we find that x, y have equal values; this solution there
fore does not serve our purpose.
The same is the case if/= 3, g = i.
Suppose then that/= 2, = i ; therefore a= d~ 16, b = 17, c= 33; and
(taking positive signs) we have
/ = 8oo, ^ = 305, r=8i;, ^=256,
and # = 733025,^=488000, 2 = 418304,
If/= i, g= 2, we have a = d= 16, b= 137, c= 153, and
^ = 4640, ^=20705, r= 21217, ^=256,
leading to large numbers for x, y, 2.
Euler adds that, if x, y, z satisfy the conditions of the problem, another
solution is furnished by X 9 Y, Z where
H, D. 22
338 SUPPLEMENT .
PROBLEM 3. To find three squares such that the difference of any pair
is a square^ or to find x, y, z such that
x? y, 3?  # 2 , y*  z 2 are all squares.
Any solution of the preceding problem will satisfy this, but the numbers
would be large and we can get smaller solutions 3 ..
y? v 2
Dividing by s 2 , we have to find three squares, 5 ,^ and i, such that
are all squares.
The last two conditions are satisfied if we put
said y
and
, , i , * i i
and we have only to make 3  = ( ^_ ( z  /o_ ( 2 a square,
Now
Therefore (ffi)(/f) or (/V~i)  i has to be made a
square.
(r) The latter expression is a square if
' a *
And /^ .  = ^, a square ; therefore
2 ^2 , Z2
. ~ prj^ (/ 2 h^ 2 ) . hk (tf + ^) must be a square,
If /=# + , g=ab, hc + d^ k = cd, the expression becomes
4 (ff 4 4 ) (^  </ 4 ), which must be a square.
From the Table to the last problem we may take the values
which make the expression a square.
Then /= 5,^= i, >fc=i6, /5 = 2,/^ =  1 /, ^// = ^6o.= , so that / = ^,
^ = ^, and therefore ^ = 4.
Therefore *.*** * . J .1 is the solution .
2r / 2 ~i 9 0^1 153
 1 Algebra^ Part II., Arts. 236, 237. '
SOLUTIONS BY EULER. PROBLEM 3 339
To obtain whole numbers, we put 2=153 and then ^ = 697 and
7=185.
Thus ^ = 485809] and tf 2 .^ 451 5 84 = (6 72)*,
/= 34225V /* 3 = io8i6 = (io 4 ) 2 ,
z*= 23409] ^^ = 462400 = (68o)l
(2) Without using the Table, we may make (j?f i) (  i) a square
in another way.
Put qjf = m or q = mf, and (m^  i) (;w 2  i) has to be made a square.
This is a square when / = i substitute therefore i + s for p and we
have
Dividing out by (w 2 i) 2 and, for brevity, putting a for
we have
which has to be made a square.
Equating this to (i +.A t^J 2 ) 2 , let us determine^ g so that the first
three terms disappear ;
therefore 2/= 40, or /= 20,
and 6a = 2g +/ 2 or g = (6a f 2 ) $a 20 2 .
Lastly, the equation gives 4*3 + as= zfg+g\ so that
4^ 2 fg 40 i 2# 2 + Ba? __ 4 ;
Now Win the expression for # may have any value.
Ex. i. Let m = 2, so that a =  ;
therefore ,. 4 .iJS, , = 2, ,g;
f^iSf
Ex. 2. Let = , so that a = ;
!?.; 260 249 747
therefore j = 4  f^ = , ^ = ~ ^ , ^ = ^>
whence x\z>y\z are determined.
Euler considers also the particular case in which # = m*/(m*  1) is a
square, #* say.
The expression i + 4&*s + 6^^ + 4^ + ^j 4 is then equated to
_ . 122 , .
and we obtain ^ =  r  and p
r  7 
22  2
34 o SUPPLEMENT
Ex. a is a square if m  f , and in that case b = . Therefore p =  
q^=.mp  ^, and accordingly
^^689 jy = 433
js;~"nr' 2? 145*
PROBLEM 4. 70 ^^ //4r^ square numbers such that the sum of each
pair is also a square, i.e. to find numbers x,y, z such that
are all squares 1 .
Dividing by s , we have to make
all squares.
The second and third are made squares by putting
x f i
an
z
, v / i
and = 
and it only remains to make
 ')'+/" (f  ^> a square.
This can hardly be solved generally, and accordingly we resort to
particular artifices.
i. Let us make the expression divisible by (p + i) 2 , which is easily
done by supposing/ + t q 1, or q =p + 2, so that q+ i becomes/ + 3.
Thus (/ + 2) 2 (p  1) 2 +/ 2 (p + s) 2 , or 2/ 4 + 8/ 3 + 6/  4^ + 4, must be a
square.
Suppose 2/ + 8/ 3 + 6/  & + 4 = (^/ 2 +$ + 2 ) 2 ,
and let us choose/, ^ such that the terms in/,/ 2 vanish ; therefore/=  i,
and 4^+1 = 6, or ^=f.
We now have 2/ + 8 =/ 2 / + 2^
=!!/!,
so that/ =  24, and ^=  22, whence
fs^LU^.STS y = f~*^ 4B3
5f 2/ 48 ' s 2^ "" 44
Making z = 16 . 3 . u, the least common multiple of 48 and 44, we have
the solution
#=11.23.25 = 6325, ^=12.21.23 = 5796, 3 = 3. 11.16 = 528,
and
1 Algebra, Part II., Art. 238.
SOLUTIONS BY EULER. PROBLEMS 3, 4 341
2, 3. Euler obtains fresh solutions by assuming, first, that
?i = a(/+i),
and, secondly, that ^ I= !0**)
4. Lastly, he makes our expression divisible by both (J> + i) 2 and
(/i) 2 at the same time.
For this purpose he takes
'>  1 " '>
whence y+ r 
Substituting in the formula ^(f *f+JP(f i) 2 the value of f in
terms of/, ^ and then dividing by (/ 2  r) 2 , we have the expression
and we have to make (ft + i) 2 (/ + /) a +/ 2 (/ 2  i) 2 a square,
or ty + 2* (/ 2 + i)f + {2^ 2 + (/ 2 + i) 2 + (/' 2  i)*}/ + 2 /(/ 2 + i)/ + / 2
must be a square.
We now equate this to {# 2 + (/ 2 + i)f  ^p,
whence we have
{2t* + (t*+ l) 2 + (/ 2  l) 2 }/4 2/(/ 2 + l) = {(/ 2 + i) 2  2/ 2 }/ 2f(f* + I),
which gives J4^ 2 + (* a  i) 2 }/ + 4^ (/ 2 + i) = o,
and t
therefore jtf + i = , f + / =
,
and * =
where t can be chosen arbitrarily.
Ex. Let/=2j then ^= , q^ and
_ a
Z 2/ 80 ' J5 2^ 44 "
Putting 5 = 4.4.5.11, the least common multiple of 80 and 44,
we have
# = 3. 11.13 = 429,
j = 4. 5. 9. 13 = ^2340,
z= 4.4.5*11= 88oj
342 SUPPLEMENT
and a?+y = 3 a . if (121 + 3600) = 3 2 . i3 2 . 6i 2 ,
=n. Q,
2
2 = 20 2 (13689 +1936) = 20 2 .I25 2 .
, PROBLEM 5. (Extension of Dioph. iv. 20 to five numbers.)
To find five numbers such that the product of every pair increased by unity
becomes a square' 1 .
Euler had already shown (see pp. 1 8 1, 182 above) that, if mn+ i =/ 2 ,
then the following four numbers which we will call a, b, c, d have the
property, viz.
am, b = n, cm^n^zl^ d = 4/(t+m)(t+n).
If now z is the fifth required number, the four expressions
i + az, \+bz, i + cz, i+dz
must all be squares.
If, says Euler, we had to satisfy these conditions singly, the difficulties
would be insuperable. But here too it happens, as in the former case,
that, if we make the product of the four expressions a square, the
expressions are all severally squares.
Let the product be i +pz + qz* + r&* + s^
where accordingly
r  abc + abd + aed+ bcd^ s = abed.
Suppose now that
therefore, since the absolute term and the terms in z, z* vanish, we have
r + SZ =p(\q\f) + (\q\ffz,
whence 2 lzik
wiiCiiLC Z  7^  ; rr
Now it will be found (see the proof lower down) that
ki/=H*+i);
the denominator of the fraction will therefore be \ (s i) 3 ; that is, the said
denominator fortunately turns out to be a square ; if it were not so, the
single expressions j + az, i + bz> i + &, i+dz could not have been made
squares.
As it is however, substituting for \q\f its value in the numerator
and denominator of the fraction for *, we have
s ~
1 Commentcttiones arithintticae t "II. pp. 5052.
SOLUTIONS BY EULER. PROBLEMS 4, 5 343
and 'all the conditions will be fulfilled, that is, all the expressions
a^ + i, ac+i, ad+i, fo+i, bd + i,
cd+ij az+i, bz+i, cz + i, dz + 1
will be squares.
Lemma. To prove the fact (assumed above) that
For brevity, put m + n + J=f, l(l+m)(I+n) = k, so that k=
and, since a = m> b  n, c=f+ /, d^ we have a + b + c= 2f, and therefore
/=2/+4& :
Again, since q (a + b + c) d + (a + b) c + ab^
q = %fk H (m + nY^c 2l (m + n) + mn \
and, since mn M = / 2 , the latter expression becomes
Moreover, s = abed = qmnk (/+ 1) ,
therefore '1+^ + ^ = 8^ +/ 2 + qmnk (/+ /),
and we have to see whether the righthand expression is equal to \p.
Now ' ' ' J/=/ a +4/5 + 4^.
Assume then, as a hypothesis, that
or, if we divide throughout by 4^,
/+ mn (f+ l)=k =// 2 + Imn, from above ;
that is, " f+fmn =// 3 ,
which is of course true, since mn + i = l~.
Consequently it is proved that
i+? + j = 4^, or <^4^) = H^+0
Ex. i. Assume w = i, # = 3, so that /=2; therefore
ai t ^ = 3, ^ = 8, ^=120,
whence ^ = 132, ^=i47S> ^=4224, ^ = 2880,
and we deduce that
__ 4. 4224 + 264. 2881 777480
* = i 28792 ~ ~ 8288641 '
The conditions are satisfied, for
344 SUPPLEMENT
Ex, 2. To get smaller numbers (since we must put up with fractions)
let us put m~\^ # = !, so that /=; therefore
whence ^=57> =45 I T ^=93 I T
4.9311 + 114.361 44880
and *=  
PROBLEM 6. Euler has a general solution of the problem of Dioph. in.
15, viz.
To find three numbers x, y, z such that
xy + x +y, xz + x + z, yz+y + z
are all squares' 1 .
(i) Put x+i = A, y + i=, z+i = C, so that ABi, ACi and
BCi have to be made squares.
Let AB=f+i, AC=f+i, jBC = r*+i,
therefore ABC= ,/{(/ + i) (f + i) (^ + i)}.
To make this expression rational, let us regard /, q as given and put
(p z + i ) (^ f i) = w 2 + n\ so that m=pqi^ n =/ + ^ ; therefore
Put the latter root equal to mr + n + / (r m) ; therefore
nrm = 2mrt + 2nt+ nrPmf*,
. m(t*i)2nt
and " ?"= rs  r  v

_ , . , ._.
Therefore r 2 + i = ,75 ^  35 and ^BC = ^5 ^  ~ ; ;
{n (f  i) + 2;/2/} a 7? (^  i) 4 s;///
thus, since BCr^+i^ we have
and, since m* + n* = (f + i) (^ + i),
where z ==/^ i> =/ + ^
This solution is very general, inasmuch as we may choose /, ^ as we
please, thus equating ABi, 'ACi to any given squares; and, as t
can be chosen arbitrarily, we have an infinite number of square values for
(2) Euler adds two methods of obtaining solutions in integers, the
second of which is interesting.
1 "Considerationes circa analysin Diophanteam," Commentationes arithmetic^ n. p. 577.
SOLUTIONS BY EULER. PROBLEMS 5, 6
345
Take two fractions r and ^so related that adbc i: and form a
o a
""
third fraction ,"" >> which is similarly related to either of the former
a +
fractions.
Then the following three numbers will satisfy the conditions :
For, since adbc=i,
(Cf. Dioph. in. 19.)
Simple solutions are seen thus :
a
b
c
d
a+c
r i
i 7 7+i
1 ~/~ r
1 f
and so on,
if I 2/+ I
B
(3) If two of the numbers A, B are given such that AB~ r =/ 2 , we
can find an infinite number of values for a third, C, which with A> B
will satisfy the conditions.
For, since AC i and^Ci have to be squares, take their product
ABC* ~(A+) C+i and equate it to (mC+ i) 2 ; we have then
.
and
,
=~TV   whence
ABm* '
^ 2
ABm*
rs r
ABm*
Therefore we have only to make AB  m* a square ; that is,
/* + i  m* = a square = # 2 say, so that m* + 3 =/ 2 + i,
Take now two fractions a and a such that a* + a 2 = i, and let z = a^ + a,
;z = a^ a ; then
6
where a;, a are determined by giving any values whatever to f, g in the
expressions
346 SUPPLEMENT
PROBLEM 7. To find four numbers such that the product of any fair
plus the sum of that pair gives a square; or> in other words ; to find four
numbers A, B, C 3 D such that the product of any pair diminished by i is a
square, that z's, such that
ABi, AC I, AD i, BCt, BDi, CDi
are all squares*. (Cf. Diophantus iv. 20.)
Let us regard two of the numbers A, B as given, being such that
AB  i p\ or AB=p* + i.
Let a, a be such fractions that a 2 + a 3 = i, and put
A + B + 2 (ap + a)
c  (a.pdf : .
Similarly let * + /P= i, and put for the fourth number
A+B + 2(bp + ft)
(ftp^bf
Thus five conditions are satisfied, namely, that
AJBi, AC 1, BCi, AD i, BDi are all squares,
The sixth condition, that CD  i shall be a square, gives
(A + #y+2 (A+JB){(a + S)f+ a + /3} +4 (ap + a) (fy + /3)
 (a/  a) 2 (ftp  l>f = a square,
where AB has at the same time to be equal to/ 2 + i.
Regarding a, a, fr, ft and/ as given, we have
and the expression to be made a square becomes the following expression
in powers of A,
4* + 2A*(a + b)p + 2A*(/ + i) +2A (p*+ i) (a + b)p + (/ + i) 2
i) (a + ft)
this to the square* of
' , ; ..

and we have
whence A is found.
Euler goes on to some particular cases, of which the following may
be given.
1 Commentotianes arithmeticae^ii. pp. 579582.
SOLUTIONS BY EULER. PROBLEMS 7, 8 347
Suppose b = a and /? =  a ; we then have
A + J3 + 2 (ap + a)
and the expression above in A which had to be made a square becomes
This can be put in the form
{^  (/ + I)} 2 + ^ 2 (^  0) 2 {4  (a/  a) 2 },
by virtue of the relation a 2 + a 2 = i.
Our expression is clearly a square if 4 (a/  a) 2 = o, or
that is, / = (2 + )/a, and
a a "
and in that case
p _
~
where A can be chosen quite arbitrarily.
Putting = (/ 2 ^)/(/ 2 + /), =2j/(/ 2 + 2 ), we obtain the following
as a solution, where m, n can be any integers whatever.
 j X? /. 5
infg 2mfg
C =
Ex. Suppose /== i, *=2,
therefore A=ft, ^ = f^ 3
and
PROBLEM 8. 2> find Jour numbers such that the product of any pair
added to a given number n gives a square 1 .
(i) A particular solution is found in this way. Let A, B, C, D be
the required numbers, and, since AB + n has to be a square, put
so that AB=(nacbcCf n(ad~bcf. [Cf. ^the Indian formula above,
p. 282.]
1 Commentatimes arithmeticae, JLJL. pp, 5023.
348 SUPPLEMENT
The condition that AB + n is a square is therefore fulfilled, provided
that adbc = i] therefore we have to take fractions T, ^ such that
#* &r = i ; and, when this is done, the fractions j, and 73 will have
the same property in relation to either of the former fractions.
We accordingly put
Thus five conditions are satisfied, and it only remains to make CD + n
a square ; that is,
' n 2 (a 2  <*2n(abcdf + (b*d*f }
 zn (ad  bcf r = a square.
+ n '
or, since (ad bcf = i,
2 (a 2  ff  n {2 (ab  cdf + 1} + (^ 2  ^ 2 ) 2 = a square.
(2) We obtain a general solution by the same method as that applied
above (p. 345) in the problem of making AB ' i, 2?C~ i, etc. squares.
Put ABp^n\ then, to make AC+n, BC^n both squares, take the
product of these expressions and equate it to (n + Cxf\ therefore
so that (ic 2  AB}jn must be a square.
Let then x*AB= tff + n =ny*, or * y" =/.
Similarly let us put ^  js 2 =/ 2 n, so as to get
~ A+B2X
and it remains to make CD + n& square,
that is, (A H B^ ~ 2 (x + 0) (^ + .#) + ^V + 4^
must be a square.
But, since B = and ^4 + B = ~ H , the expression becomes
(after multiplication by A 2 )
SOLUTIONS BY EULER. PROBLEMS 8, 9 349
which must be a square = {A' 2  A (x + v)  (/ 2  n)f say; therefore
A*
so that
or
(3) A particular solution is obtained by assuming that z> = #, so that
=j, and
zx A + J3+2X
+ (/  nf a square ;
 4)  a square.
while AB =j>*n = x*~ nf.
For then we have to make
that is, (A* p* + ^) 2 +
This is satisfied if we put/ = 2, so
Suppose p = x t> and we have
* 2 + 3 j ^ 3 s
x =   and / =  T
2t * 2t
*r>
and hence AB
We may therefore put
ji
or / =
*
2tu
,
and # =
2tu
2gtU
n =
It will be seen that in this solution C+ D =
PROBLEM 9. To find four numbers such that the product of any pair
added to the sum of all gives a square^.
First find four numbers A, B^ C, D such that the product of any pair
increased by a number n gives a square (Problem 8).
Take as the numbers sought mA, mB^ mC> mD, and, since n(AB+ri)
is a square or nPAB + nfiti is a square, we have only to make m*n equal to
the sum of the four numbers or m (A + B + C + JD\ whence
A+B+C+J)
1 Commentationes aritkmetuae,, II. pp. 5835,
350 SUPPLEMENT
But, since in the other problem C+D=%(A+\ this gives
3 (.4 + 2?) _.3
where n as well asf>&f and u can be chosen as we please.
Since n may be chosen arbitrarily, take / 2 = x*  3;;, as in the last
problem, so that n = (x? /), and AB=fn =
Accordingly we may put
therefore
ar
andhence ^ 2 ^ + ^)/ + (T 3 ^ 3^ V ?
n_
and
therefore
Now two of the numbers, A, B, can be chosen arbitrarily, and
,  ;
therefore / and
sothat ^i(^
while m
2H
If, in order to get rid of fractions, we put A = 40^, B  $bfg, we have
Ex. Let/==. 2,^=1; therefore
^_ 12(0+0) '_ i2(a + b)
SOLUTIONS BY EULER. PROBLEMS 9, 9 A, 10 351
The following are simple cases :
(1) a= 5, <$=r, whence ^'40, '#= 8, C= i, ^=23, m^^.
(2) a~n, /5=2, whence^4=i88, .#=16, C= i, > = $i, m = %.
If/= 5, g= i, we can obtain integral solutions, thus.
Assuming then a = 19, b  7, we have
^ = 380, ^=140, =248, 2) =12, *,
so that the required numbers are
the sum of which is 975.
We can also 'solve the corresponding problem : . . .
9 A. To find four numbers such that the product of any fair minus the
sum of all gives a square.
For we have only to give m a negative value.
PROBLEM 10. To find three numbers x, y, z such that
x 4jy + \
yz + zx f xy I are all squares 1 .
xyz }
(This may be expressed as the problem of finding ^, q , r such that the
equation" ? f * + '<? r=o has all its roots rational while/, ^, r are
all squares.)
Take nx, ny, nz for the numbers required, so that
n(x+y + z) I
w 3 (xy + xz +yz) [ must all be squares.
ffxys ) ,
If the first and third conditions are satisfied, we must have, by
multiplication,
xyz (x + y + z) = a square.
Put therefore xyz (x+y + z)v* (x +y + z)\
whence xyz  v* (x +y + z), and z  ^ y .
xy ijr
Since xyz =  xy ^ x /^, we must have, in order that nxyz may be
a square,
n = m*xy (x +y) (xy  ^). '
i Commentarii Acad. PetropoL 176061, Vol. VIII. (1763), p. 64 sqq. 5= Commen
tationes arithmeticae, i. pp. 239244.
35 2 SUPPLEMENT
If the values of z, n thus found be taken, the first and third conditions
are satisfied, and the three numbers will be
nx = mWy (x +y) (xy  fl 2 ),
ny = m*xf (x +y)(xy  z> 2 ),
nz = nPtfx
The second condition requires that
iP (
xy + z(x+y)^xy +  = a. square.
Suppose for this purpose that xytf=u z (this introduces a restric
tion because there are doubtless plenty of solutions where xyv*is not
a square); therefore
# _j_ ^ ^_ ^
and xy^tf + u*, x + y =  , so that we must make
, 2 (^
# f w 2 +   2^   a square.
Put x = ^, so that j; =  , and
o 2
z/ 2 + ^ 2 + * v ;  L =a square,
or / 3 wV +W + v* (f 2 + i) 2 + 2V (/ 2 + i) + ^ = a square,
/.*. zr 4 (^ + i) 2 + wV (3^ + 2) + w 4 (^ + i) = a square
= {^(/ 2 +i)+^ 2 } 2 , say.
Therefore ^(3^+ 2) + ^ 2 (/ 2 + i) = 2^(^+ i) + A 2 ,

and ^ = rrn  r  5 = a square.
Z* 2 2j(/ 2 +l)3/ 2 2 H
Further, let s = tr, and we shall have
W 2 2/ 8  (2^f 3) / 2 + 2^ 2 (r + i) '
Multiply the numerator and denominator by zrt  r* + 1, and we have
The problem is accordingly reduced to making the denominator of this
fraction a square. If we suppose this done, and Q to be the square root,
while / and r are determined as the result of equating the denominator to
<?, we shall have
v 2rtr*+i
=  g , and * = *,> =
whence we can derive the numbers required.
SOLUTIONS BY EULER. PROBLEM 10 353
Now the denominator to be made a square is easily made such if the
coefficient of /*, or the absolute term, is a square j and the absolute teim
is a square if 2 (r i) is a square.
Case L Suppose /= i; the coefficient of t* is then a square and the
absolute term vanishes.
We have 4/ 4  ictf 3 + 4?  8/ = Q\ while v/u = 2t/Q.
Suppose Q zt* Y, and we have
we therefore put # = 36, 2^=173, f = ^/, and #=&r=i28; further
_ 3 I22 5 _ 2 5 1249
~ 128 r28 '
173 128
and, since xy zf* = u\ the required numbers will be
____ i28 a .25. 1249.47609. i73 2 _ 3
flX _, ft TrP
128.128 '
128. 25 2 . 1 249 3 . 47609.
~ i2T 2 .rT8
36 s . 128. 25.
=
In order to get rid of fractions, put ^ = 1 ^, l and we have
nx= i28 2 . i73 2 . 1249. 47609 = 490356736. 59463641,
j/ = 5*. i73 2  I249 2  47609 =934533 2 5 594^41,
5 = 36^ . 1249 . 476o9 2 = 61701264. 59463641.
The product of the three numbers is obviously a square j their sum is
found to be 25 . 5 9463 64 1 2 ; and the sum of the products of pairs
= i73 2 .5 94 6364i 3 . 18248924559376
59463641
I2 *" 5
Case If. Put r = ; then = " ,
and
whence 6^4^^ 3 = ^/ 4 ^, and f
Accordingly <2 = ^^ 6 , ^ = H and we P ut
Therefore ^ = ^ = 60, y
H, D. 2 3
354 SUPPLEMENT
and the required numbers are
That is, ## = 705600 . 2315449 = 1633780814400,
#y= 109172*2315449= 252782198228,
ng = 1500677 . 2315449 = 3474741058973.
The product of the three numbers is clearly a square; their sum will be
found to be 23I5449 3 ; and the sum of the products of pairs
= I4 2 . 2315449* . 6631333489
= (14. 2*315449. 81433)2.
These numbers are much smaller than those first obtained.
If fractional numbers are admitted, we may divide those found in the
last solution by 2315449*, and the solution will be
705600 190 SCI
TSTToTTTTJ TT5T> TST
PROBLEM n. To find four numbers x, y, z, u such that
& +y + z + u
are all squares 1 .
^
xyz
xyzu
zux + uxy
A general solution being apparently impossible, some particular as
sumption simplifying the problem had to be made. Euler therefore
assumed as the four numbers
Mob, Mbc, Mcd, Mda,
which assumption, although five letters are used, involves the restriction
that the product of the first and third numbers is equal to the product of
the second and fourth.
We must therefore have
+ dcPb + 2 abed)
all squares.
1 Novi Commentarii Acad. Petrop.^ 1772, XVII. (1773), pp. 24 $*$\.=>Commentationes
arithmeticae, I. pp. 4505.
SOLUTIONS BY EULER. PROBLEMS 10, n 355
The above assumption therefore has the advantage of making the
product of the four numbers automatically a square and also of making the
third formula take the form
M*abcd(ab + be + cd+ da).
Since the first formula M(ab + bc + cd+da) must be a square, it follows
that abed must be made a square.
In order to make the first and third formulae squares, take
M abbc + cd + da,
or, if the latter expression has a square factor, say/ 2 , put
M= (ab + t>c + cd + da)/f 2 .
We now have only two conditions remaining to be satisfied, namely
~& square .............................. (i),
bcd=& square ........ ,...(2).
The expression in (2) reduces to
(a 2 + ?)bd+ ac (P + d*) + zabcd,
or bd (a 2 + r) + (d + f ac = a square.
We have therefore only to find numbers a, b, c> d satisfying these two
conditions. It is further to be noted that a, c are connected by a relation
similar to that between b, d, and the whole question depends on the ratios
a : c and b : d. We may therefore assume a, c prime to one another and
likewise , d prime to one another, for, if either pair had a common divisor,
it could be omitted and the relation would still be satisfied.
Consider now the second condition as being the more difficult. Although
two ratios a : c, b : d are involved, neither can be arbitrarily assumed. For
suppose e.g. that b : d 2 : i; then 2# 2 + 2^ + 9^ would have to be made
a square; this however is seen to be impossible, for, if we put a=f + q,
cjp q, we obtain the expression i^S^ 2 , which cannot be made a
square. The same impossibility results if we put b : d=$ : i. Therefore
the ratios a : c and b : d can only be certain particular ratios.
Obviously the first class of ratios adapted for our purpose are square
ratios. Assume then that b : d f : q\ and put
ff (a* + <?)*ac O 2 + ^ = (fqa + J *)*, say ;
therefore 2 (/ 2 + s ) 2 a + tfpfc = zmnpqa

50 that 7 ~
or, if m = kpq,
_ , }
d <?* c n*(j? + ff2knfq** v '
232
356 SUPPLEMENT
Now, if values could be found for k, n, p, q such as would make ac or
n (&  n*) {n (J> 2 + ffi 2/fc/V 2 } a square,
we should have a solution of the problem, since, bd being already a square,
abed would then be a square. Euler however abandons the investigation
of this general problem as too troublesome and as certain, in any case, to
lead to very large numbers; and, instead, he proceeds to seek solutions
by trial of particular assumptions.
Particular values of ajc in terms of /, q are the following, which are
obtained by putting ^ = 2, =i; = 3, =i; etc.
g _
" ~~~ / o . 0\o . .0 o 1 A4
in IV _
5 '
V  VT ~
~ ~
VII " VIIT
222 ' Vil *
Taking now the simplest values of ^/^=/ 2 /^ 2 , let us write down the
simplest corresponding values for a/c:
if _ 1 hprnm^c 3 445 157 7 8.
*?"~ T ' "^ DeCOineS ^" J ""TJ T> y"ir> ""1~ T1TJ 7TZT> '$?>
if J $, f becomes , $, ^, f> T B F> ifr, ^, fi
if J = $, f becomes ft, ^, f, ^,
if t, 5 becomes ^, f, f, ^,
The last assumption gives, "praeter exspectationem," two cases in which
a/c becomes a square ; and these give two solutions of our problem.
1. Putting a 64, b = 9, ^=49, ^=4, we have
J^"=^ + ^ + / + ^af = 576 + 441 + 196 + 256
= 1469,
and the four numbers are
1469.196, 1469.256, 1469.441, 1469.576.
2. Putting a = 64, = 9, <r= 289, ^ = 4, we obtain
$f= 576 + 2601 + 1156 + 256 = 4589,
and the four numbers are
4589.256, 4589576, 45891156, 4589.2601.
SOLUTIONS BY EULER, PROBLEM n 357
Again, the form of the expressions abed, bd (a 2 + <?) + ac (b + df to be
made squares shows that any values for ajc obtained by the above process
may be taken as values for bfd. Also a and c may be interchanged. Euler
accordingly sets down as values oft/dto be tried the following:
4 o 8 12 13 20 28 32 88 At ~
S> T> F TI TTJ TF> TJ T> ~8~ etc 
He obtains no solution from the assumption /</=f, but he is more
successful with the assumption b\d^.
Putting b\d^) we have to make
20 (a 2 + <?) + Biac  a square.
This is satisfied by a\c\\ let therefore <z/V = i + #, and we have to
make 20 {(i + #) 2 t i} + 8i (i + x) a square; that is,
121 + I2i# + 20^ = a square = (n +^y) 2 , say;
, , 121 220; ,a y 2 
therefore x = 5  , and  =^
and, by putting w = 5, n = i we obtain a/^= ^.
This solution serves our purpose, since it makes abed a square.
Putting 0=16, 3=s, f=5, ^=4, we have
80 + 25 + 20 + 64 189
^=  ya  = 7""'
and, if f= 3, Jl/'^: 21 ; the four numbers are therefore
21.20, 21.25, 21.64, 21.80.
This is a solution in much smaller numbers ; and
the sum of the numbers =9. 2i 2 ,
the sum of products of pairs = no 2 . 2i 2 ,
the sum of products of sets of three = i2o 2 . 2i 4 ,
and the product of all four = i6oo 2 . 2i 4 .
When one solution is known, others can be found. Take, for example,
the last solution in which, for 3/^=, we found that
a ^JF 3 22J/4 IOI
c ~~ y*  20
In order that abed may be a square, we must have
5 ( < y ! 2o)(y22jy+ ioi) = a square.
This is satisfied by y = 5. Substitute 5 + 5 for y, and we have
5 (z 2 + 105; + 5)(z a  i2z + 16) = a square, 
or 400 + 5002 49 50 a  1 0/4 5^ = a square.
358 SUPPLEMENT
Equate this expression to (20 + % 5 ^% 2 ^^) 2 , and we have
/52i 2 \ 25.
( *r  5 ) * = ~^~
\3 2 / 3
, A
whence
3212705
=  =
77  >
266321 24211 2201 2201
therefore y = z + 5 = VaVi 5 > anc * t* 16 resulting values of a, c are large
numbers which Euler does not trouble to develop. As a matter of fact,
?= 55696 = 4
.? "109465205 ~5
It follows that
/W= 5 . 55696 + 5 . 109465205 + 4 . 109465205 + 4 . 55696
= 278480+ 547326025+ 437860820+ 222784
= 985688109;
and, putting/= 9, we have
M= 12168989.
The four numbers are therefore
12168989 . 278480, 12168989 . 547326025,
12168989. 437860820, 12168989.222784.
PROBLEM 12. To find three numbers x, y> z such that the expressions
are both squares 1 .
In order to satisfy the first condition, we have only to put
ffs/s + ^r 8 , y=2pr, z = zgr,
for then x* +7 2 + z* = (/ 2 + f + r t f = /^ 2 , say.
The second condition requires that
therefore, since / + f = 4^ (^ 2 + f>),
or
In order to get rid of the sixth power of / and so make /* the highest
power of/, suppose that r=.p nq (which introduces no restriction);
therefore
GY4 (f 
or
Scient. Imp. PetropoL, 1779, Vol. ill. (1782), pp. 30 sqq. = Commenta
tiones arithmeticae, n. pp. 457 sqq.
SOLUTIONS BY EULER. PROBLEMS n, 12 359
Let the latter expression be denoted by JP, so that Q~zq(p~ nq) R\
and
.tf 2 = 4 (i H
This may be made a square in two ways, either (i) by taking advantage
of the fact that the last term is a square, or (2) by making the first term,
i.e. making i + ;/ 2 , a square.
(i) Put IP=\(\n*)q* + 2npq + aLf}\ and make the term in f
disappear by choosing a so that i + 6 3 + 4 = 4 2 + 2a(i ^ 2 ), or
4 . .
; we then have
whence (15 35 2+ r 3 4  ^ 6 )/ = 8 (i ^ 2 ) (3 
this divides throughout by 3  ;z 2 , and
p_
q 5
Let/ = 8// (i  n*), q= 5  io 2 + ^ 4 ; then r=pnq = n
while
and #, j>, can be expressed in terms of n.
Ex. i. Suppose n2, then
^ = 48, ^ = 19, r=io, ^ = 7035, <2=3 8o
ic = 2565, y~ 960, 2? = 380, or (dividing throughout by 5) ^ = 513,^= 192,
0= 76 (in which case Q= 106932, P= 553).
Ex. 2. Suppose # = 3; then / =  192, ^ =4, r =  180, or (dividing
by 4)
^ = 48, ^=i, ^=45> J?=i4i2o> <2= 1270800;
^ = 280, ^ = 90.48, 2 = 90.
Dividing the values of x, y, z by 10, we have
#=28, J> = 43 2 * = 9> and (2=12708, ^=433
(2) To make the first term in K* a square, suppose i + n* = m*, which
is the case whenever n = ( 2 b*)J2ab.
We have then
Euler solves this in three ways.
360 SUPPLEMENT
First, he puts R=*mfnmpq + (in*)q*\ and from this, by taking
a 2, = i, so that # = f , m = i> he obtains the particular solution
or a = 196, y= 693, s= 528.
Secondly, he puts R = zmf + 2#/f + (i  2 ) ^, and deduces, by the same
particular assumptions,
# = 936, j>=74, * =
or # = 468, jy = 37, *=
Thirdly, he supposes
J? = zmfi  mnpq +  ^ ^
r ^ ^m *
where however the last term should apparently be 
Euler's son, J, A Euler, gave, in a Supplement to his father's paper,
another solution as follows.
We know that
(/i) 2 + 4/ 2 =:(/hi) 2 and (^i^ + ^M^+i) 2 .
Multiply the first equation by 4^ and the second by (/ 2 + i) 2 ; this gives
or (^ i) 2 (/^ i) 2 + 4^ (/ 2  i) 2 + i6ff = (f + i) 2 (^ + i) 2 .
Therefore the three numbers
satisfy the first of the conditions.
The sum of the squares of the products of pairs of these numbers must
now be a square; after dividing out by 4^, this gives
(f O 2 (/* i) 2 + 4^ (f  i) 2 (/ + i) 2 + i6/V 2 (/  i) 2 = a square,
therefore (/*+ i) 4 (^ i) 2 + i6/y (/ 3  i) 2 must be a square.
For brevity, let ^4 2 = (/ 2 + i) 4 , ^ 2 = i6/ (/ a  1) 2 , and
A*(fif + B*f, or
must be a square.
Put AY + (* ~ 2^ 2 ) ?* + ^ 2 = <
whence s
SOLUTIONS BY EULER. PROBLEMS 12, 13 361
Now both the numerator and denominator of this fraction are squares
if v* = A* B\ for the numerator becomes B* and the denominator
which is the square of A +
But, putting for A, B their values in terms of/ as above, we have
^ 2 ^ 3 =/
therefore
'
and the numbers required will be
(fif
or, if we multiply by (jP  i) 2 ,
The sum of the squares of these numbers turns out to be (/ 2 + 1) 6 ,
which is not only a square but a sixth power, while the sum of the squares
of the products of pairs is found to be
or i6/ 2 (f  i) 2 {(/ i) 4 + 16/} 2 .
Ex. Put p = 2, and we have
= 1 6 . 4 . 9 . 337 2 = 8o88 2 ,
the solution being in smaller numbers than Euler's own.
PROBLEM 13. To find 1 three positive integral numbers x, y, z such that
x +y + z =
To make ^ 2 +y j + 2f 2 a square, put A; = a 2 f IP  1?, y = 2ac } z=2fa, and we
have * 2 +y + s 2 = (a 2 + ^ + ^) 2 .
We have now to make a 2 + ^ + ^ a square, and we put similarly
we have then x*+y* + z* = (p* + $* + r 2 ) 4 .
Now let us express #, jy, z in terms of^, ^, r; this gives
1 Commentationes arithmeticae, II. pp. 399400.
362 SUPPLEMENT
therefore
(i) Arrange this according to powers of/, and
In making this a square, we have to see that /, #, r are all positive, and
also / 2 + f > r*. Also a? + P must be > <?.
Equate the expression to \f + (q 4 r) 2 } 2 , and we have
whence S/^r 2 = 1 20V + S^r 8 , or / =  ^ 4 r.
Therefore a^^f + tfr, b^^qr+2^ czqr^
where both the letters ^, ^ may be given any positive values.
Ex. r. Suppose q 2, r i \ therefore
/ = 4, a=jg> 3 = 8, <r=4;
accordingly the numbers are
#==409, j/=i52, g = 64,
and  #+jy + z = 625 = 25 2 } rc 2 +y + 2 = 194481 =44i 2 =
Ex. 2. Let q = 2, r = 2 ; therefore
and ^'=961, j/ = 4oo s ^ = 320;
therefore a:+^ + s= 1681 =4i 2 , jc 2 fj/ 2 + 2 2 = 1185921 = 33*.
(2) Arrange the expression for x + y + z according to powers of q ;
this gives
In order that the terms in ^ and f and the absolute term may vanish,
equate the expression to
(?+2qr(f + r*)}\
. , . 2pr(p+r)
whence we obtain q = ^ .r__ . 3
Ex. Suppose/ = i, r = i ; therefore
^ = 4, a =i6, 6 = 2, =8,
or (if we divide by 2) a = 8, = i, <r= 4 ;
therefore # = 49, ^=64, = 8;
and x + y + z=n 2 , x?+y* + ^ = 6561 = 8i 2 = 9 4 .
These numbers are no doubt the smallest which satisfy the conditions.
The case of three numbers is thus easier than that of two (see p. 299,
note). Euler solves the same problem for four and five numbers, and shows
how the method may be extended to six numbers, and so on indefinitely.
SOLUTIONS BY EULER. PROBLEMS 13, 14 363
PROBLEM 14. To find three numbers M, y, 2, positive and prime to one
another, such that both x+y + z and x?+y 2 + z 2 are fourth powers 1 .
As in the above problem, put
x = ct? + 2 c*, y = zac> z = zbc,
and further a =/ 2 + ^  r*, b ipr^ c  zqr^
and make the expression in /, q, r for x + y + z a square by equating it
to {/ 2 + (ff + rfY as before. This gives p^g + r', but we have now, in
addition, to make / 2 + (q f rf a square.;
Put + +
therefore ^ 2 (^ + r) = 2^/ +/ 2 (^ + r).
Substitute ^g + r for/, and this becomes
whence
The problem may therefore be solved in this way.
Take q =/ + 2^^ 2 and r = ^ 2 ~ $fgf*,
so that / = ^(/ 2  < T 2 ), tnen finc ^ ^ ^ ^> anc ^ tnen a g^ n ^> ^ ^> ^ n terms
Ex. Let /= i, g= 3 ; therefore
= _2, ^ = 1, ^ = ~4,
or ^ = flj r=i l / = 4.
Thus a = 19, <5 = 8, <: = 4,
and # = 409, jj/==i52, 2 = 64;
so that x+y + = 625 = 5*, ^ 2 +jp 2 f ^= 194481 = 2i 4 .
To find limits for the values of f, g, change the signs of ^, r, putting
In order that ^ may be positive,
gff> r + \^2
and, in order that r may be positive,
Suppose e.g . that /= 2,^=5 then
ff=*> ^=9 / =
or in integers q = 2, r=i&, p = 21 ;
hence a = 121, = 756, ^=72,
^c = 580993, y = 17424, ^ ~ 108864,
= 707281=29*, # 2 H/ + 2 = 34977 8 3 2 3
1 Commentationes arithmeticae^ II. p. 402. .
364 SUPPLEMENT
PROBLEM 15. (Problem in Fermat's note on vi. 13.)
To find a rightangled triangle (in rational numbers) such that either
of the sides about the right angle less the area gives a square'*.
2X V XV
Let the perpendiculars be ,  , so that the area is 4 ; and
Z Z Z"
2X XV
4 Or 2XZXy
Z Z* '
y xy
 
as
well as  ~ or 4^ +y*
"
have to be made squares.
Since the first two expressions are to be squares, their product must be
so also ; therefore
zxyz* 2x*yz  xy*z + xPy* = a square
and, after dividing by yz, we have
a* xy zp0 xy
2p
2x~ xy = ~ xy +
.
whence z =
Thus 2z  v 
inus 2z y
~
and
an<1
2 a* 
Therefore the two expressions are squares if 20*3? p*xy is a square,
Put 2fx?~p*xy = r i x*}
therefore (20* ~ r*) x = p*y, or x\y = f\(tf  1?).
It is sufficient for our solution to know the ratio x/y, since a common
denominator z has already been introduced.
Therefore we may put
vi Commentarii Acact. Petrop.> 1749, Vo1  IL ( r 75 r )> PP 49 soR
arithmeticae, I. pp. 6272.
SOLUTIONS BY EULER. PROBLEM 15 365
whence *
and z
It only remains to make 4# 2 +y* a square ; that is,
4#* + 4 4 4V + ^ must be a square.
A general solution of this equation giving all possible values of/, g, r
is impossible. We must therefore be satisfied with particular solutions.
Particular solutions (i) and (2).
Put 4^ 4 4 4 ^
therefore 4^ 
and / = + ^(^
that is, either
~^>, or (a)
(i) Now / =  A/(^ ^) is satisfied by q  <? + d\ r = zed, whence
or we may put
and we thus find values for
Ex. i* Suppose c 2, d i ; then
/ = 53 =:I 5> ? = 45 = 20 > ^=44=16,
= 544, ^
and the triangle is
2# = i44 J^ = 4352
= =
89 ' 2 25.89' 25.89"
Ex. 2. If c 3, ^= r, we get the triangle
2X _ 32 9 j^ _ 81 . 41 N /(4^+/) = 988i
* " 185 ' 5 "25. 185' z 25.185"
And so on.
3 66 SUPPLEMENT
(2) In this case p =  ^(f  f\ and we have to put
2cd(<?~d?\
r = + d*, q = 2cd, whence p = ^ ^ ' ,
or
while
Here, since 2^ must be > r 3 ,
8<r^ 2 > (^ + ^ 2 ) 2 and
therefore d 2 >(c~d ^2)", and
d>c d J2, so that  >
^r d>d>j2c, so that  </  .
^ \/2 I
If therefore rf=i, either r< N /2 + i or ^> N /2i. The second
alternative is satisfied by ^> i.
Ex. Let r= 2, d i, and we have
/ = 4.3 = i2, ^ = 4.5 = 20, ^=5.5 = 25;
therefore
JX75,
The triangle is therefore
2*^288.25^450. y = 25^75^4375. r= =
z 4048 25 3 ' z 4048 4048 J 4048 4048
Particular solution (3).
Put 4^ + 4^  4 ?V + H = (2/ 2 2^) 2 ;
therefore H  4^^ = 8/y, and / 2 =
therefore either / = ^(2^  8^), or p = ~
The first value is however useless, since zf  r* > o, or 2^
while
We have therefore / = ~
/) = 2/ 2 2^, and z =
Since 8^ 2^ must be a square, put
SOLUTIONS BY EULER. PROBLEM 15 367
therefore 4^ %r = = (zq + r), or ^d*q 2d*r = 2c
whence q = ^ j 2// 2 , ^ = 4^ 3 2^,
2# + r= %d\ ^(S^ 2  27 s ) = 80/, and therefore p = 
Multiplying by 2d* + ^, we have in integers
while ^;, j, 5; have the values above stated.
Ex. i. Put c= i, ^= i ; therefore
j  126 . 2% 2O7
and 2=16 +  ~^. = L.
36 2
The triangle is therefore
2jc _ 64 j_252 ^(4^ +y) _ 260
z 2oj } z 207' z 207'
This is the triangle in the smallest numbers satisfying the conditions, as
Euler proves later.
Ex. 2. Since 2^>r 8 , it follows that c\d>2 ^2 ; but it does not
matter whether 2$ > <* or not, since /, ^, r may be negative as well as
positive.
Put then d 2, c = 3 ; therefore 2d? 3  r* = i, and 2d' 2 + ^ = 17.
We then have / = 24, ^ = 289, ^ = 34;
^ = 576, j; = 2. 7 . 4 i.i7 3 , N/(4^+/) = a SS33I3. ^
The triangle is therefore
2_* = 2304 J^ = 28.4I.17 a ^(4^ +/) = 4. 5. 53. 313
2 28118255' z 28118255 ' z 28ri8255
It is to be observed that in all the above examples it matters not
whether c, d are negative ; it will only result in the values of p or q or r
becoming negative, but the values of x and y will not be thereby changed.
Only z will vary, since z may be either
y(Pqf y(P + g?
x+ or x+ jr ^^ y/ .
After remarking that the problem of making
^ + 4^  4fV + f* or ^ + (2^ 2  r 3 ) 2 a square
may be solved generally by equating it to
Euler passes to his general solution.
368 SUPPLEMENT
General solution.
If ,  are the perpendicular sides of the triangle, let
z z
y^cPP} the triangle is then
2ab
and the area is
Now we found above, at the beginning of the investigation, that
*f& + f3y*pg*y xy(Pq?
2fxp*y ' *fxpy*
or, since q can be taken positively as well as negatively,
where x = a& 9 y = a 3
And we took
whence z = x
We have therefore only to satisfy the equation
or 3y=p**)>
and, since xyab(c^ b\ we have to find such numbers for a, b that
aJ(a*_?) may be of the form 2f 2 g 2 or (2/ 2 ^ 2 )/^
Suppose now that such numbers a, b have been found that
Then, since x = ab,
and a natural inference is suggested, namely that
abq abr ,
ff*, j=g*.
Let now/ = ab, and accordingly
t
q=fa r=gh, z
the triangle is then
zab
+ (a 2  P) (ab
 ff) (ad flif >
(a 2  2 ) (^ fKf '
SOLUTIONS BY EULER. PROBLEM 15 369
Also from any particular values of a, b any number of triangles can be
derived satisfying the conditions.
For, if p ab^ and
we have (2/ 2  g 2 ) ft? =2$* r*,
or
Put 2 (fh + q} = (gh t r), and fh q = (gh  r) ;
therefore q  
or, in integers, p = (2??  m~) ab>
q zmngh
r=(2?i* + m^) gh
i,i *
while
TI, A *  i 2 a* a +  i
Thus the triangle ( ,  ,  is known.
\ z z z J
Lastly, to find suitable values for <z, 3, Euler writes down all the
numbers from i to 200 which are of the form 2/ 2 u\ including all the
squares arising from the supposition that u t, and all the doubles of
squares corresponding to u o. Inspection shows that the table contains
(within the limits) all the prime numbers of the form Sm i, and no other
primes, the doubles of the primes, the products of the primes into all
squares and into one another, and the doubles of those products.
Now, since the product a.&(a*rb)(a Z>) is to be of the form 2/ 2  w 2 ,
and the factors <z, b, a + 3, a  b are either prime to one another or at the
most have 2 as a common divisor, while 2 is itself contained in the form
2* 2 u 2 , the several factors must all be of that form, in which case the
product will be of that form.
We have therefore first to take some value of b in the table and then see
whether there are in the table three other numbers a 1>, a, a + b differing
by b. Euler gives a second table showing values of a corresponding to
values i, 7, 8, 9, 16, 17 etc. of b.
The values of a in the table corresponding to <5=i are 8, 17, 63,
72, 127.
Ex. i. Take b = i, a = 8 ; therefore
<tf=8, # 2 ^ = 63, ^(a 2 ^) = 8.9. 7=4.9. 14,
and 4. 9. 14 = A 2 (2/ 2 ^ 2 ), so that >fc = 6, 2/ 2 ^=14, and accordingly
/=3> ^=2.
H. D. . 24
370 SUPPLEMENT
We have therefore in this case
or, dividing by 2,
whence #r2=i2mnion*i3m*, p + q = \2mn 26^
whil , * = 8 + ^l)!.
Thus there are any number of values of z from which triangles may be
obtained satisfying the conditions.
The simplest value is found by putting m= i, n o, whence
and either * = 8 + . 25 = f,
or 2 = 8 + J. 169 = 1^5..
The first value gives the triangle in smallest numbers above found
(P 367),
2a _ 64 a? fr 2 _ 252 g a + ^ 2 260
" ""
_
207 2 207' 2; ""207"
Substituting 1215 for 207, we have the sides of the triangle corre
sponding to the second value of z.
The particular triangles are also directly obtained from the values of
a > &> ft & h without bringing in m, n ; for
that is,
i2
Ex. 2. Take b = 41, a  112 ; therefore
ab=i . 16.41, a 2 F= 71.9. 17,
and <z(<z 2  ^) = 16 . 9 . 7 . 17 . 41 . 71 = (2/ 2 
whence ^ = 12, and 7.17.41.71 = 2/ 2 ^ 2 .
The simplest solution is/=4i7, ^=37.
Thus z
is easily found, and consequently the triangle
2ab_ a*P a 2 + 1>*
2 ' ~^~' T"
[Euler finds values for^ g by using the formulae
(2a a /3 2 )(2/a 2 ) = (2a 7 ^) 2 2(^y
and op  2y* = 2 (x yf (x
SOLUTIONS BY EULER. PROBLEMS 15, 16 371
He does not actually give the steps leading up to the particular
solution /= 4 1 7, ^=37, but it can be obtained thus.
Since 7 = 2 . 2 2  i and 17 = 2 . 3 a  i, we have
7 . 17 = (2 . 2 . 3 + i . i) 2  2 (3 . i + 2 . i) 2
= I 3 2 2. 5 2 = 2 ( I3 5)2(l32.5) 2 =:2.8 2  3 2 .
Again, since 41 = 2 . 5 2  3 2 and 7 1 = 2 . 6 s  i 2 , it follows that
Therefore, by multiplication,
= 797 2 ~ 2 . 3 8o 2 = 2 (797  380)* (797  2 . 3 8o) 2 = 2 . 4i7 2 ~ 37 2 ]
PROBLEMS 16. "De problematibus indeterminatis, quae videntur plus
quam determinata 1 ."
We have seen that by means of certain " Porisms " stated without
proof Diophantus is able to obtain relations between three numbers
x, y, z which have the effect that, when they are satisfied, a quite
appreciable number of symmetrical expressions in #, y, z are auto
matically (as it were) made squares.
It is clear therefore, says Euler, that, if a general method of finding
"porisms" of this kind can be discovered, the whole subject of Diophantine
analysis will be appreciably advanced. Accordingly he proceeds to discuss
such a method.
The method depends on a Lemma the truth of which is evident
Lemma. If values have been found for the letters z, y, x etc. which
satisfy the equation WQ, where Wis any function of those letters^
and P, Q, JR. etc. are other functions of the letters such that PW,
Q W> R W etc. are squares ', then, if the values of z, y, x etc. are
taken which satisfy W= o, the resulting values of P, Q, R etc. will
also be made squares.
Cor. P 9 Q, R etc. will similarly be made squares if P+ a W> Q + ft W,
or, more generally, if
are squares.
Conversely, If such values for g 9 y, x etc. have been assigned as will
satisfy W= o, all formulae such as P* + a W, <2 3 + ft W, R* + y W etc. will
at the same time be made squares.
1 Nffvi Commentarii Acad. Petro$oL> 175657, vi. (1761), pp. 85 sqq.
tionts arithmeticae, I. pp. 245259*
24 2
372 ' SUPPLEMENT
And, as the number of such formulae is subject to no limit, it is clear
that an unlimited number of conditions can be prescribed which are all
satisfied provided that the one condition W o has been satisfied.
The same Lemma can be extended to the case of cubes or any higher
powers i for, if W= o has been satisfied by certain values, all expressions of
the form P^ + aW will thereby be made cubes, all expressions of the form
a.W will be made fourth powers, and so on.
While it is plain that, if values for z, y, x etc. are found which satisfy
the condition W= o, all the expressions P* + a W, Q z + ft W, * + y W etc.
will be made squares by the same values of x, y, 0, the difficulty will be,
when a number of expressions P* + a W, Q* + fiW etc. are given which are
capable of being made squares in this way, to identity and separate the
expression W the equating of which to zero will make the rest of the
several expressions automatically squares* It would indeed be easy so to
hide away the composition of the expressions as to make this separation
itself a most arduous problem. On the other hand it is easy and in
teresting to begin with W= o, and then investigate the simpler formulae
which can by this means be made squares. Before proceeding to the
particular cases, Euler observes further that it is convenient to take for W
an expression in which 0, y, x etc. enter symmetrically and can be inter
changed ; for then, if P* is such a square that F* + a W is a square, and
*, y, oc etc, are interchanged so as to turn jP 2 into Q*, R* etc., Q* + a W^
R* + o,W etc. will also be squares. Also, since solutions in rational
numbers are required, 2, y, x etc. should not enter in any higher power
than the second into the expression W. Euler begins with expressions
containing two unknowns $, y only.
Problem (i). Given W=y + 2 a = o, to find the more simple formulae
which by means of this equation can be made squares.
When the equation y + zao is satisfied, it is clear that the general
formula P* + M( a +y + z) will become a square whatever quantities are
put for P and M. Accordingly Euler, by giving P, M various values,
obtains without difficulty 44 different expressions which become squares
when jy h = o.
He supposes M= 2, 2, 2*, y, zy,
(y + z + a)(yz + a)(zy + a), and 3 and n 2  i times the last expression
respectively, and with each of these assumptions he combines one or more
forms for P. I need only quote a few expressions which are thus made
squares, e.g.
(y i) 2 + 2 (a+y + z) =y z +
(z  1 ) 2 + 2 ( a +y + z) = z z + 2y +
I 2O\
I 20)
SOLUTIONS BY EULER. PROBLEMS 16 373
(y~z + i) 2  2 ( a+y + z) = (yzf^z f i + 20}
(0 j;+ i) 2  2 ( a +y + 2) = (y z) 2  4y +i + 205] '
= + znz + n z zna^
= ay + az,
(yz nY + n(y + z + a)(a+y + z) f z* + nf + nz* + n*
(y+g + a ) (y z + a) (sy+ a) (~a+y + )
and so on. Wherever a new expression can be got by interchanging z
and y, this may be done.
Taking the more particular case of Wy + *  1 = o, Euler obtains the
following expressions which are thereby made squares,
gyz,
which indeed are easily seen to reduce to squares if we put y + z=i or
y = i  z.
The fact that f& +y + 2; 2 is a square if j/ = x  * or, more generally, if
j/ = i* is included in the Porism in Dioph. v. 5. Similarly
is made a square if we put y = az.
The last expression but one in the first of the above lists, namely
fz z + ny* + nt* + n*~ na\ becomes a square whatever value n has. If a = i,
it becomes
That this is a square when z =y i is part of Diophantus' assumption
in v. 4 (see p. 104 above).
Euler*s Problems (2) and (3) similarly show how to find a number of
formulae which are all made squares by values of jy, z satisfying the
equations W=yza(y + z) + = o and J^=/ + 2 2  znyza = o.
He then passes to the cases where there are three unknowns.
Problem (4). Given W x +y + z  a = o, to find the more noteworthy
formulae which can be made squares by satisfying this equation.
In this case the general expression JP* f M(x +y + z  a) becomes a
square.
374 SUPPLEMENT
Put M= in and P= one of the expressions x  n, y  , z  n, or one
of the expressions y + z n> z + xn, x+y~n.
These assumptions make the following expressions squares :
z 2 * 2# (x+y) + n 2 2na, and the two other similar expressions,
,
If M 2?iyz, P=yz ny nz, and so on,
jyV + 2nxyz + ri*y* + n z z* + zn (n a) yz
and the two other similar expressions are squares.
~ (a + x +y + 0), Py + zx and corresponding expressions,
qyz  tyx are all squares.
In particular, if n = 2^, a = J, the following six expressions are made
squares by putting
If a = 2, we make the expressions
i xy xz \
i yz yx Y
izx xy )
all squares by putting x +y i z = 2.
Problem (5) finds expressions which are made squares if
is satisfied.
Problem (6). Given W $ +y* + z* 2yz  2zx 2xy a = o, to find
the more simple formulae which can be made squares by means of solving that
equation.
Here the general formula will be
P 2 + M(x* +y* + 2 2yz 2zx 2xy a).
IfM= i, P=x+y + z,
4yz + 4zx + 4xy + a = a square*
If Jlf=~ i, ^=JF + zx, etc.,
f 1
4jgf^ + a f are squares.
J
> =jy^, etc,,
a + 2 (y + z) x x* ~\
a + 2 (z + x)y y 2 j are squares,
a + 2 (x +y) zz* )
SOLUTIONS BY EULER. PROBLEMS 16 375
In the particular case where a = 4^, so that
tf 2 +y* + z* = 2yz + 2zx + 2xy + 4^,
yz + n \
zx + n\
r are all squares :
xy + n ^ '
yz f zoc 4 xy + n J
or our formula gives the means of solving the elegant Diophantine
problem :
Given any number n, to find three numbers such that the product of any
pair added to n gives a square, and also the sum of the products of the pairs
added to n gives a square.
By solving the equation
we obtain z = x +y 2 *
We assume, therefore, such numbers for x,y as will make xy + n a
square ; suppose xy + n = ^ } and we then have two values for z, namely
z = x +y 2u, each of which along with #, y will satisfy the conditions.
In fact, if z = x +y 2u, while u = J(xy + n)>
J(xz n) = ^(x + z y) = xu,
(Cf. Eule^s solution of Dioph. in. 10, p. 160 above.)
Problem (7). Given
to find the more noteworthy expressions which can be made squares by
satisfying this equation.
The general expression here is
P* + M{x z +y + 2  2yz  2zx  2xy  2 a (x + y + *)  b}.
If M= i, JP=x+y + z + a, we have
(a) 4yz + 4zx + 4x3? 4a(x+y + z) + a? + & = a. square.
(b) 4/yz + 4%x + ^xy + 2 + b = a square.
etc.,
4yz + 4& (y + z) + a? 4 b \
4^rc + 4a (z + x) + a* I b r are all squares.
'
376 SUPPLEMENT
If M=~i, P=y + zx a, etc.,
4yz + ^ax + a 2 + 1 \
(d) 4200 + ^ay + o? + b L are all squares.
4xy + 4&z + a? h b )
Cor. i. In order to solve the problem represented by (<;), equate the
expression $xy + 40 (# + j>) + a? 4 to a square 2 , whence
4 (x + a) (y + a) = u*  b + 3 a 2
or (# + a) (y + #) = (u*b + 30?) \
x andj> are then determined by splitting J( 2 3t3 2 ) into two factors
and equating x + a,y + a to these factors respectively. Next, solving,
for 5, the equation
3? +J; 2 4 2 2  2^2  22WC  2A7J;  2$ (x + j; + s)  ^ = O,
we find, since ^xy + 40 (x + jy) + a 2 + ^ = w 2 , that
= # + y 4 <z + ?<f.
Cor. 2. If = at 2 , then, by solving the equation
op tjy 2 + = 2yz + 2zoc + zxy + 2a(x +y + z) a?,
we make all the following formulae severally squares,
yz + a (y 4 z),
zx 4 ^ (z + #),
^ + a (x f j/), iqy + az,
yz + zx + xy + a (x +y + 2),
by assuming
where (itr + ) (y + a) is put equal to u z + a 2 .
An interesting case of this last problem is that in which a i ; and
from this case we can deduce a solution of a new problem in which the
corresponding expressions with x 2 , y\ 2 in place of x, y^ z are all squares.
The problem is
To find three square numbers suck that (i) the product of any two added
to the sum of those two, (z) the product of any two added to the third, (3) the
sum of the products of pairs , (4) the sum of the products of pairs added to
the sum of the numbers themselves, all give squares.
We have to find values of # 3 , y, which will make all the following
expressions squares,
SOLUTIONS BY EULER. PROBLEMS 16, 17 377
As we have seen, all these will be squares if
We have also seen (Problem (i) above) that x?y* + x* +y* becomes a
square if only y = % + i. Put then y x + i, when we have
S 2 = 2X 3 + 2X + 2 2*J(X*+2X 3 .+ $X? + 2X + i) ;
that is, s 2 = 4 (ar 2 + x + i).
It only remains to make tf 2 + x 4 1 a square. Equate this to ( x + ^) 2 ,
and we have
whence  2 V(# 3 + * + i ) =
Therefore the roots of the required squares are
' ~ 2t+I
Or, putting t(rq)\2q, the values become
Let ^ = i, ? = 2, and we have x = , ^ = f , a? = J ; or, if we put / = 2 in
the values expressed in terms of /, the values are x = , jy = f, z = l.
PROBLEM 17. To find two fourth powers A\ B* such that thtir sum
is equal to the sum of two other fourth powers*.
In other words, to solve the equation ^? 4 + J5*= C* + jD\ or (what is
the same thing) ^ 4 Z? 4 = C 4 ^ 4 .
It is proved, says Euler, that the sum of two fourth powers cannot be
a fourth power, and it is confidently affirmed that the sum of three fourth
powers cannot be a fourth power. But the equation A* + B* C 4 = D 4 is
not impossible.
First solution*
Suppose Ap + q^D^pq, Cr^s^Brs\
thus the equation A*D 4 =C 4  4
becomes pq (f + f) = rs(r* + .j 3 ).
Put/ = ax, q fy, r = kx and s y^ and we obtain
vi Commentarii Acad. Petrofol., 1772, Vol. XVIL (1773), pp. 64 sqq.
tationes arithmetics, i. pp. 4736; Mtmoires de FAcacZ. Imp. de St Pttersbourg^ XI. (1830),
pp, 49 sq. = Comment, aritkm.^ rr. pp. 4506.
245
378 SUPPLEMENT
v a $ _ a? ft
therefore ^= ^ r , which fraction has therefore to be made a square.
scr air  k
One obvious case is obtained by putting k = ab^ for then
j) 2
'
whence j> = tf 3 *=i, so that/ = a, ^ = a, r = a, s = a, and the result is
only the obvious case where J> s^ q = r.
Following up this case, however, let us put k = ab(i+s).
We then have
f __
"
therefore, multiplying numerator and denominator by Piss and ex
tracting the square root, we obtain
X &* I Z
To make the expression under the radical a square, equate it to
and assume /, g such that the terms in 0, z* vanish.
In order that the term in z may vanish, /= (3^  i), and, in order
that the term in s* may disappear,
whence
The equation to be solved is then reduced to
* (3  4) ajfc =(** + *)
,
Now b can be chosen arbitrarily; and, when we have chosen it and
thence determined 0, we can put
X = P  I  0, J/ = ( 3  I +/0 +2 2 ),
and accordingly
where we may also divide out by a.
If A*, jy have a common factor, we may suppose this eliminated before
/, gr, r, 5 are determined.
SOLUTIONS BY EULER. PROBLEM 17 379
Ex. i. Let 3=2 (for b cannot be i, since then g would be oo ).
Therefore /=^, ^=f, * = JMfr.
As a does not enter into the calculation, we may write i for it;
therefore
6600 2187 ii 6600 25 /66oo\ 2
/y _ ij __ ^^____ _ * iti __ ,j i . ._ * I _______ \
2929 2929' * 6 2*2929 24
= 55407.1100 ^ 3 . 28894941
^ /rt//\2 2Q2Q 2
But the ratio x : y is what we want, and
?.  3 28894941 __ 28894941 _ 3210549 _ 1070183
x "~ 2187.2929 2929. 729 ~ 2929.81 ~~ 27. 2929'
so that we may put
# = 79083, y 1070183.
Further ^ = 2 (l + K ) = iii59 = 9058 .
' 2929 2929
Therefore
P = 79083, r = 27 . 19058 = 5145^6,
^ = 2.1070183 = 2140366, s 1070183.
Consequently
= 2219449, C=r + j= 1584749.
&=#_* =  2061283,
Ex. 2. Let b = 3 ; therefore /= 13, g= , s =
200 8 . 144
further *= ^Br l^'
e_ L 200 /' T , J .5 2oo\_ a 200 2447 _ 8. 89736
 J> = 8 + il^V I3 + 4 > ^/~ + rf9'~rf9~ i6 9 8 '
m , * 8.144.169 6.169
Therefore _._4__9._.,
and we may put x = 1014, y = 3739.
Accordingly p = 1014, r = IOI 4 =
.and therefore ^=12231, (7=10381,
^=2903, Z> =  10203,
and again A 4 + j5 4 = C 4 + D\
3 8o SUPPLEMENT
Another solution in smaller numbers.
In the second of the papers quoted Euler says that, while investigating
quite different matters, he accidentally came across four much smaller
numbers satisfying the conditions, namely,
.4 = 542, = 103, C=359, 0 = 514,
which are such that A 4 + 5 4 = C 4 + D\
He then develops two methods of analysis leading to this particular
solution; but, while they illustrate the extraordinary ingenuity which he
brought to bear on such problems, they are perhaps of less general interest
than the above*
INDEX.
[The references are to pages.]
I. GREEK.
q.v.
adtivaros, "impossible," 53
&\oyo$ ( = "undescribed" apparently),
Egyptian name for certain powers, 41
s, indeterminate: vXijOos (jLovdduv
, an undetermined number of
units=the unknown, dpifl/i6y, i.e. #, 32,
H5> * 3 5 ^ T V dop/ory, indeterminately,
or in terms of an unknown, 177
apLOptrriKfi distinguished from XoywriKij, 4
dpt0yit6s, mimber, used by Dioph. as techni
cal term for unknown quantity (=#),
32, 115* 130; symbol for, 3237, 130
( = *lx) and sign for, 47, 130
"absurd," 53
, doubleequation,
square," used for square of un
known (=# 2 ): distinguished from rerpd
7WZ/OS, 3738 ; sign for, 38, 129 ; TerpctTrX?)
Sifrajus, "quadruplesquare," Egyptian
name for eighth power, 41
a/w5iWAy, fourth power of unknown
(=^), sign for, 38, 129
a.fJLodvvafJi.offTdj') submultiple of SwapoSv
vafits ( = i/,* 4 ) and sign, 47, 130
a/j6oos, "squarecube" (=^ 5 ), sign
for, 38, 129
* submultiple of Svva^Kvfios
and sign, 47, 130
1 submultiple of 5vVa^w (= r/ar 2 )
and sign, 47, 130
, "species," used for the different terms
in an algebraic equation, 7, 130, 131
Xei^w, "deficiency": to &Xety<ri TWO,
etdy, "any terms in deficiency," i.e.
"any negative terms," 7, 131
"existent," used for positive
terms, 7, 130
brdvOwa. ("flower" or "bloom") of
Thymaridas, 114116
fcros, equal, abbreviation for, 4748
"cubecube," or sixth power
of unknown (=.%*), and sign for, 38,
129
KvpoKvpoffTdv (si/jc 6 ) and sign, 47, 136
/cujSos, cube, and symbol for cube of un
known, 38, 129; /cujSos efeXwcrds, Egyp
tian term for ninth power (^), 41
Kvf$o(rr6v (=zi\a*) and sign, 47, 130
Xeforet?, to be wanting : parts of verb used
to express subtraction, 44 ; \elirovra, etSi],
negative terms, 130
Xetyw, "wanting/* term for subtraction
or negation, 130; Xetyet (d&t.)*= minus*
sign common to this and parts of verb
Xeiireiv, 4144
\oyt<TTiKT?i, the science of calculation, in ;
distinguished from Apt^rwciJ, 4
TJ h X<&7^, 132^., 144^.
s, "part,'* an aliquot part or sub
multiple ; fj^p^t " parts," used to describe
any other proper fraction, 191
jtwyXJTTjs tiptoes (from ^Xoy, an apple), 4,
113
juovds, "unit," abbreviation for, 39, 130
MoptcwmKd, supposed work by Diophantus,
34
/jLopLov, or b> juoply, expressing division or
a fraction, 46, 47
irpcOn;, Seur^pa, 4748
382
6/w>ir\i]07} (effy), (powers of unknown) ffupeta,
"with the same coefficient," 7 units, 37
JT, value of, calculated by Archimedes
and Apollonius, 122
jrapw^nys, TrapurbrijTos Aywyr}, approxima
tion to limits, 95 sq., 207, 208, 209 #.,
211
tfXcMTjiwmwJj', "formativum" (Tannery),
meaning of, 140 .
side, = square root, 65 n.
u, "number" or "multitude," used
for "coefficient," 64 n.
proposition, 9
INDEX
"heap" or collection of
<rwp<5s, "heap," 37
<rrepe6s, solid, used of a number with
three factors, 183 n. '
SWOJLUS, "quadruplesquare, 5 *
Egyptian term for eighth power, 41
;y, "existence," denoting a positive
term (contrasts with Xetyas), 41
tiirdpxovra (elfo?), "existent" or positive
(terms), 130 w.
&pidp6s (from 0t<X?7, a bowl), 4,
of Apollonius, 122
, determinate, 115
"3
II. ENGLISH.
Abu'lFaraj, i
Abu'1Wafa alBuzjani, 6, 19
Achmim Papyrus, 45
Addition, expressed in Dioph. by juxta
position, 42; Bombelli's sign for, 22;
first appearance of + , 49 n.
Ahmes, 112
Alfraganus, 20
Algebra: three stages of development,
4951
Algebraical notation : Diophantus, 3239,
4144 ; Bombelli, 22, 38 ; earlier Italian
algebraists, 38 ; Xylander, 38, 48 ; Bachet
and Fermat, 38; Vieta, 38, 39, 50 n. ;
beginnings of modern signs, 4950 n.
Aljabr, 64
alKarkhi, 5, 41 n.
alKhuwarazml, Muhammad b. Musa,
34 5<>
Almukabala, 64
Amthor, 122
Anatolius, 2, 18
Andreas Dudicius'Sbardellatus, 17, 25
Angelus Vergetius, 16
Anthology, arithmetical epigrams in, 113
114; on Diophantus, 3; indeterminate
equations in, 114
Apollonius of Perga, 5, 6, 12, 18, 122
Approximations: Diophantus, 9598; Py
thagoreans, 117118, 278; Archimedes,
278279
Arabian scale of powers of unknown
compared with that of Diophantus,
40, 41
Arabic versions and commentaries, 19
Archimedes, n, 12, 35, 278, 279, 290;
Cqdex Paris, of, 48; Cattle Problem,
1211.24, 279; Arenarius, 35, 122
Arenarius of Archimedes, 35, 122
Arithmetiea of Diophantus : different titles
by which known, 45 ; lost Books, 512 ;
division into Books, 5, 1718; notation
in, 3253; conspectus of problems in,
260266
Arithmetical progression, summation of,
248249
Ars rei et census^ 20
Aryabhata, 281
Auria, Joseph, 15, 1 8
Bachet, 12, 16, 17, 21, 22 ., 25, 2629, 35,
38, 45, 48, 8082, 87., joi, 107 .,
109, no, 140 #., 173 ., 196197 .,
2i3., 220., 230 ., 232 ., 234235*.,
246, 271, 273, 287, 293
"Backreckoning," 56, 89, 93
Baillet, 45 n.
Bessarion, Cardinal, 17, 20
Bhaskara, 281
Bianchini, 20
Billy, Jacques de, 28, 165 ., i66#., 184 .,
221 ., 267, 304, 308, 320, 321, 326
INDEX
333
Bodleian MSS. of Dioph., 15, 34, 35 ;
MS. of Euclid, 35
Bombelli, Rafael, n, 27; Algebra of,
21, 22 ; symbols used by, 22, 38
Brahmagupta, 281
Brancker, 'Thomas, 286 n.
Brouncker, William, Viscount, 286. 288
Camerarius, Joachim, 21
Cantor, Moritz, 3 ., 6, 6$n., 112, n8#.
120 ., 125 ., 281
Cardano, 21, 23, 40
Cattle Problem of Archimedes, n, 12,
121124, 279
Cauchy, i88#., 274
Censo, or Zensus, = square, 40, 41
CharmideS) scholiast to, in, 113, 121
122
Chasles, n
Cleonides, i6.
Coefficient, expressed by 7rXiJ0os, multitude,
39, 64 .
Colebrooke, 6, 281 .
Cosa, =the unknown, 22, 40
"Coss," 23
Cossali, i, 21 ., 40, 41, 140 n.,
lion.
Cracow MS. of Dioph., 5*., 14, 18
Cube: Vieta's formulae for transforming
the sum of two cubes into a difference
of two cubes and vice vers^ 101103;
Fermat's extensions, ibid. ; a cube cannot
be the sum of two cubes, i44#. ; Euler's
solution of problem of finding all sets of
three cubes having a cube for their sum,
329334; sign for cube of unknown or
**, 38, 129
" Cubecube " (= sixth power of unknown,
or xfi), sign for, 38, 129
Cubic equation, simple case of, 6567,
242
Cuttaca ("pulveriser"), Indian method of,
283
Definitions of Diophantus, 32, 38, 39,
129131
"Denominator," 137
Descartes, 271, 273; notation, 50 n.
Determinate equations : of first and second
degrees, 58 ; pure, 5859 ; mixed quad
ratics, 5965; simultaneous equations
leading to quadratics, 66; a particular
cubic, 6667
"Diagonal" numbers, 117, 118, 310
Dionysius, 2 #., 9, 129
Diophantus : spelling of name, i ; date,
12; epigram on, 3; works, 313; in
Arabia, 56, 19; "Pseudepigraphus,"
12,31; MSS. of, 1418; commentators
and editors, 1831 ; notation of, 3253 ;
methods of solution, 5498; porisms
of, 3, 810, 99101 ; other assumptions,
103 sqq. ; theorems in theory of numbers,
105110; on numbers which are the
sum of two squares, 105106; numbers
which are not the sum of two squares,
107108; numbers not sum of three
squares, 108109; numbers as sums of
four squares, no; Dioph. not inventor
of algebra, inir6; nor of indeter
minate analysis, 115124; his work
a collection in best sense, 124; his ex
tensions of theory of polygonal numbers,
127
Division, how represented by Dioph.,
4447
Doppelmayer, 20 .
Doubleequations (for making two ex
pressions in x simultaneously squares),
n, 7387, 9192; two expressions of
first degree, 7380, 8082 n. ; two ex
pressions of second degree or one of first
and one of second, 8187 ; general rule
for solving, 73, 146; double equations
for making one expression a square and
another a cube, 9192
Dudicius Sbardellatus, Andreas, 17, 25
Egyptians: hau, sign for, 37; names for
successive powers, 41 ; beginnings of alge
bra, ^^calculations, 111112; method
of writing fractions, 112
Eisenlohr, ii2.
Enestrom, 63^., 286*1.
Exanthema of Thymaridas, 114116
Epigrams, arithmetical, in Anthology, 113
114; on Diophantus, 3; one in Dio
phantus (v. 30), 124
Equality: abbreviation for, 4748 ; sign in
Xylander, 48 ; the sign = due to Recorde,
5o n.
Equations, see Determinate, Indeterminate,
Double, Triple, etc. ' / . .
Eratosthenes, 121
Euclid, 8, n, 12, 19, 63, 117, 124, 132 .,
i44., 191
Eudoxus, 124
Euler, 56, 7172 ., 8385 ., 86., 90 .,
384
INDEX
ioo#.,io2tt., 107, no, 145 .,*i
160 ., 162 #., 178 #,, 181182 #.,
1 88 ., 224 #., 236 #., 241 ., 242 #.,
268, 272, 274, 275, 286, 288292,
294, 297, 299 ., Supplement, 329379
passim
Euler, J. A., 360
Eutocius, 5, 6
Exponents, modern way of writing due
to Descartes, 50 .
Fakhri, 5, 41 n.
lt False supposition," use of, in Egypt,
112113
Fermat, 28, 29, 30, 38, 78, 90, 101,
102, 103, 106, 107, 108, 109, no,
144145 ., 163 ., 173 #., 179180 n.j
182 ,, 183 #., 184 ., 188 ., 190
191 #., 197 ., 202 ., 204 ., 205 a.,
213214 ., 2l8 #., 220 ., 223 #.,
229 ., 230 ., 231 ., 232 ., 233 .,
35 ^3 6 i 239 HO ., *4 r
242 #., 246, 254 n., Supplement, 267
328 passim, 364; "great theorem of
Fermat," 1441 45 #,; Fermat on num
bers which are, or are not, the sums of
two, three, or four squares respectively,
106110, 267275 ; on numbers of form
jf9 2 y or ix* fi 276277, of form
xP+^y*, 275, and of form # 2 + y 2 , 276,
277; on equation jpAy*= i, 285287 ;
jt*y*=zZ cannot be solved in integers,
224, 293297; problems on rightangled
triangles, 204205 #., 218219 ., ssow.,
229 ., 230 #., 231233^., 235 ., 336 .,
339240*2., 297318; Fermat's " triple
equations," 321328
Fractions : representation of, inDiophantus,
4447; sign for j, 45; for , 45;
sign for submultiple, 4547
Frenicle, 102 ., 276, 277, 285, 287,
39> 3io, 313, 314
Gardthausen, 35, 36
Geminus, 4
Georg v. Peurbach, 20
Georgius Pachymeres, 18, 19, 31, 37
Girard, Albert, 30, 106 n*
Gnomons, 125
Gollob, 14, 1 8
Grammateus (Schreiber), Henricus, 497*.,
50 .
Greater and less, signs for, 50 n.
Giinther, 6, 278., 279??,
Hankel, 6, 5455, io8., 281, 283, 284,
286 .
Harriot, 50 n.
Hau ( = "heap"), the Egyptian unknown
quantity, 37, 112
Heiberg, 35, 48^., 118, 205 .
Henry, C., 13, 28 n.
H^rigone, 50 n.
Heron, 12, 13, 35, 36, 43, 44, 45, 63,
129 n.
Hippocrates of Chios, 63, 124
Holzmann, Wilhelm, see Xylander
Hultsch, 2., 3, 4, 9, TO, n, 12, I9.,
35*36, ST^ywj 63 ri8 " I22 > 2 53
Hydruntinus, loannes, 16
Hypatia, 5, 6, 14, 18
Hypsicles, 2 ; on polygonal numbers,
125126, 252, 253
lamblichus, 2, 3, 37 ., 49, 50, n 5116, 126
Ibn abl Usaibi'a, 19
Ibn alHaitham, 19
Identical formulae in Diophantus, 104, 105
Indeterminate equations : single, of second
degree, 6773 ; of higher degrees, 87
91 ; how to find fresh solutions when one
is known, 6870; doubleequations for
making two expressions simultaneously
squares, u, (i) two expressions of first
degree, 7380, 8082 ., (2) two of second
degree, or one of second and one of first,
8187 ; double equations for making one
expression a square and another a cube,
9192 ; rule for solving doubleequations
in which two expressions are to be made
squares, 73, 146 ; indeterminate equations
in Anthology, 114; other Greek ex
amples, 118121; 2^ 2 .y 2 =i solved
by Pythagoreans, 117118, 278, 310
"Indian method," 1213, 2IW 
Indian solution of ^ 2 if^2=i, 281285,
290, 292
Iwventum Nwum of J. de Billy, 28,
165 ., 184 #., 198 n., 204 n., 205 #.,
221 ., 230 ., 231 M 239 ., Supple
ment, 267328 passim
loannes Hydruntinus, 16
Ishaq b. Yunis, 19
Italian scale of powers, 40, 41
Jacobi, io8., 288
b> Arabic term for cube of unknown,
41 .
INDEX
385
alKarkhi, 5, 41 .
Kausler, 31
Kaye, G. R., 281
Kenyon, 45
Konen, 278 ., 279 ., 281 n. 9 285 .,
286 ., 288, 292 w.
Kronecker, 288
Krumbiegel, 122
Kummer, 145 n.
Lagrange, 72 #., no, 188 #., 272, 273,
374 275, 276, 277, 285, 287, 288,
290, 292, 299, 300
Lato, "side," 40 .
Legendre, 107 n., 109 #., i88#., 273
Lehmann, 35
LejeuneDirichlet, 145 n. t 288
Leon, 124
Leonardo of Pisa, ri, 41 #., 120
Less and greater, signs for, 50 .
Limits : method of, 57, 94, 95 ; approxi
mation to, 9598
Logistica speciosa and Logistica numerosa
distinguished by Vieta, 49
Loria, 62 ., 157 ., 168 ., 175 #.,
i?6n. t 195 w., 197 ., 207 ., 240 #.,
241 #.
Lousada, Abigail, 31
Luca Paciuolo, 21, 40
Madrid MS., 14, 15, 16
Mai, Arabic term for square, 41 n.
Manuscripts of Diophantus, 1418
Maximus Planudes, 13, 14, 19, 21, 31,
43 44. 45. 46, 48
Measurement of a circle (Archimedes),
122
Mendoza, 17
Metrica of Heron, 43, 44, 45, 63, 129^.;
MS. of, 118
Metrodorus, 5, 113
Minus, Diophantus' sign for, 4144,
130 ; same sign in Heron's Metrica,
43, 44 ; BombelU's abbreviation, 22 ;
modern sign for, 49 n.\ Vieta's
sign for difference between (=for ~),
50*.
Montchall, Carl v., 18
Montucla, 28
Moriastica of Diophantus, 34
Muhammad b. Musa alKhuwarazm!, 34,
50
Multiplication, signs for, 50 .
Murr, Ch. Th. v., 20 n.
Negative quantities not recognised by
Diophantus as real, 5253
Nesselmann, 610, 21 ., 25, 26 ., 29,
33> 34 49~5i 5558, 67, 87, 89, 93,
108 ., 140 ., 173 ., 204 ., 207 .,
25* ., 329
Nicomachus, 2, 126, 127
Notation, algebraic: three stages, 4951 ;
Diophantus' notation, 3249, 5152
Numbers which are the sum of two squares,
105107, 268271 r numbers which are
not, 107108, 271272; numbers which
are the sum of three squares, 272273 ;
numbers which are not, 108109, 3 73 ?
numbers not square are the sum of two,
three or four squares, no, 273, 274;
corresponding theorem for triangles,
pentagons, etc., 188, 273
Numerusj numero, term for unknown quan
tity, 38, 40
Nufiez, 23
Oughtred, 50 n.
Ozanam, 288
Pachymeres, Georgius, 18, 19, 31, 37
Paciuolo, Luca, 21, 40
Pappus, n, 13
Papyrus Rhind, 112 ; Berlin papyrus
6619, 112
Paris MSS. of Diophantus, 15, 16, 18
Pazzi, A. M., 21
Pell, John, 31, 286 n., 288
" Pellian " equation, origin of this er
roneous term, 286
Peurbach, G. von, 20
Philippus of Opus on polygonal numbers,
125
Planudes, Maximus, 13, 14, 19, 21, 31,
43> 44 45. 46, 4^
Plato, 4, 38^., in, 113, 116, 125
Plus t signs for, 22, 49^.; expressed in
Diophantus by juxtaposition, 39
Plutarch, 127
Polygonal Number^ treatise on, 3, 1112,
247259; sketch of history of subject,
124127; began with Pythagoreans, 124
125 ; figured by arrangement of dots,
125 ; Hypsicles on, 125126, 252, 253 ;
Diophantus 1 extensions, 127
Porisms of Diophantus, 3, 810, 99101,
201, 202, 214
Poselger, 30, 98
Powers of unknown quantity and signs
386
INDEX
for, 3739, 1^9; Italian and Arabian
scale (multiplicative) contrasted with Dio
pbantine (additive), 4041 ; Egyptian
scale, 41
Proclus, 4., 113, 116, ii7#., 1 1 8, 242 n.
Psellus, 2, 14, 18, 41, in
Ptolemy, 18, 44
Pythagoreans : 3 ; on rational rightangled
triangles, 116, 242 . ; on polygonal
numbers, 124125; on indeterminate
. equation 2# 2  < 7 2 = i, 117118, 278,
310
Quadratic equations in Diophantus, 78,
5966; in Hippocrates of Chios, 63;
. in Euclid, 63; in Heron, 63, 64
Quadratic inequalities in Diophantus,
6063 ; limits to roots, 6063, 65, 95
Qusta b. Luqa, 19
RaeUce (=#), 40
Radix (=#), 38
Rahn, 50 #., 286/2.
Ramus, i6n.
Rationality, Diophantus' view of, 5253
Recorde, Robert, 50 #.
Regiomontanus, 5, 17, 20, 23, 49
" Regula falsi" in Egypt, 112113
fieZafo, Italian term for certain powers
of unknown, 41
Res, alternative for radix^ in sense of
unknown quantity, 38
Rhind Papyrus, 112
Rightangled triangles in rational numbers
in Diophantus, 9394, 105106 ; method
of "forming," 9394; other methods of
forming attributed to Pythagoras, 116
117, and to Plato, 116117; Euclid's for
mula for, 117, 1 20 ; Pythagorean formula
once used by Diophantus, 242 ; Greek
indeterminate problems on, other than
those of Dioph., 119121; Fermat's
theorems and problems on, 204205 .,
218219 #., 220 ., 229 tt., 230^., 231
.. 232 **., 235 ., 236 ., 239240**., 293
,318, 364371
JRodet, 34, 35
Rosen, 50
jRudio,, 63 #.
Rudolff, Christoff, 23, 50 n.
Salmasius, Claudius, 17
Sandreckoner of Archimedes, 122
Saunderson, N, 27 n.
Schaewen, P. v., 327,. 328
Schmeisser, 31
Schone, 43, 45, 118
Schreiber, H., see Grammateus
Schuler, Wolfgang* 24
Schulz, 9, u, 18, 30, 31, io8., 140 n.,
zign.
Sebastian Theodoric, 24
Serenus, 12
1 ' Side " = square root, 65 n.
"Side" and "diagonal" numbers, Py
thagorean solution of 2x z y*=i by
means of, 117118, 278, 310
Simon Simonius Lucensis, 25
Simplicius, 63 n.
Sirmondus, J., 27
Smith, H. J. S., 292
" Species" (eftij) of algebraical quantities,
7, *3 9 131
Speusippus on polygonal numbers, 125
"Squarecube" (=^), sign for, 38,
129
Square root, sign V f r J & n '\ =wXcv/o<
(side), 65 .
"Squaresquare" (=J^), sign for, 38,
, 129
Squares : numbers as sum of two, three,
or four, no, 273, 274; of two, 105107,
268271; not of two, 107108, 271272;
of three, 272273 ; not of three, 108,
109, 273
Stevin, Simon, 29, 30 n.
Stifel, M., 23, 49., 50 n.
Submultiples, sign for, 4547; decom
position of fractions into, 46, 112;
submultiples of unknown and powers,
47
Subtraction, symbol for, 4144
Suidas, i, 1 8, 22
Surdesolides, sursolida or supersolida, 41
Surds, 2324
Suter, H., 19 .
Tannery,?., 2#., 3, 5, 6, 8, 1072, 1419,
25, 28., 31, 3237, 4344, 45, io8.,
in, 118 ., 125 #., 135 138 .,
144 ., 148 ., 150 n., 156 ., j6o .,
198 ., 219 ., 234 n., 256, 278, 279,
280, 281, 290, 308
Tanto, unknown quantity, in Bombelli, 22
Tartaglia, 21, 40
Theaetetus, 124
Theon of Alexandria, 2, 18
Theon of Smyrna, 2, 36^ 117, 126, 310
INDEX
387
Theudius, 124
Thompson, D'Arcy W., 37
Thymaridas, Epanthema of, 114116
"Tripleequations" of Fermat, 163 .,
179 ., 1827*., 202ft., 223/2.) 224 .,
"Units" (/*ovd5es)= absolute term, 3940;
abbreviation for, 39, 130
Unknown quantity (=#), called in Dio
phantus <i/>i0^s, "number": definition
of, 3 2 H5> I 30J symbol for, 3237,
^o; signs for powers of, 38, 129;
signs for submultiples of unknown and
powers, 47, 130 ; Italian Arabian and
Diophantine scales of powers, 40, 41 ;
Egyptian scale, 41; sc first used by
Descartes, 50 . ; other signs for, i, used
by Bombelli, 22, 38, N (for Numerus]
by Xylander, Bachet, Fermat and others,
38, R (Radix or Res), 38, Radice, Lato,
Cosa, 4.0 n.
Vacca, G., io6n.
Valla, Georgius, 48
Vatican MSS. of Diophantus, 5 ;*., 15, 16,
17
Vergetius, Angelus, 16
Vieta, 27, 3839, 49, 50^., 101, 102,
214 ., 285, 329, 331
Vossius, 31
Wallis, 40 ., 286, 287, 288, 289
Weber, Heinrich, 3 n.
Weber and Wellstein, 107 ., 145*.
Wertheim, 30, no ., 137 ., 138 .,
r 45 J5 1 i 161 ., 209 n., 211 .,
212 ., 216 ., 217 ., 254 ., 256, 257,
286 ., 294, 295
Westermann, 125 .
Widman, 49 .
Wieferich, 145 n.
Woepcke, 5 n.
"Wurm's problem," 123
x for unknown quantity, originated with
Descartes, 50 .
Xylander, 17, 2226, 27, 28, 29, 35, 38,
107108 ., 140 .; Xylander's MS. of
Diophantus, 17, 25, 36
Zensus (Censo\ term for square of un
knownquantity, 38
Zetetica of Vieta, 27, 101, 285
Zeuthen, 118121, 2057?., 278, 281 #.,
290, 294295
CAMBRIDGE I PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS.