(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Children's Library | Biodiversity Heritage Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "History Of The Theory Of Numbers - I"

CHAPTER II.
FORMULAS FOR THE NUMBER AND SUM OF DIVISORS, PROBLEMS OF FERMAT AND WALLIS.
FORMULA FOB THE NUMBER OF THE DIVISORS OF A NUMBER.
Cardan1 stated that a product P of k distinct primes has l+2+22+.. +2*"1 aliquot parts (divisors <P).
Michael Stifel2 proved this rule and found3 the number of divisors of 243352P, where P=7-lM3.17.19.23-29, by first noting that there are 1+2+.. .+64 divisors <P of P according to Cardan's rule and hence 128 divisors of P. The factor 52 gives rise to 128+128 more divisors, so that we now have, 384 divisors. The factor 33 gives 3.384 more, so that we have 1536. Then the factor 24 gives 4.1536 more.
Mersenne4 asked what number has 60 divisors; since 60 = 2«2-3'5, subtract unity from eaqh prime factor and use the remainders 1, 1, 2, 4 as exponents; thus 32-24-7-5=5040 (so much lauded by Plato) has 60 divisors. It is no more difficult if a large number of aliquot parts is desired.
I. Newton5 found all the divisors of 60 by dividing it by 2, the quotient 30 by 2, and the new quotient 15 by 3. Thus the prime divisors are 1, 2, 2, 3, 5. Their products by twos give 4, 6, 10, 15. The products by threes give 12, 20, 30. The product of all is 60. The commentator J. Castillionei, of the 1761 edition, noted that the process proves that the number of all divisors of ambn.. .is (m+l)(n+l).. .if a, b,.. .are distinct primes.
Frans van Schooten6 devoted pp. 373-6 to proving that a product of k distinct primes has 2*—l aliquot parts and made a long problem (p. 379) of that to find the number of divisors of a given number. To find (pp. 380-4) the numbers having 15 aliquot parts, he factored 15+1 in all ways and subtracted unity from each factor, obtaining abed, a*bc, a?bz, a?b, a15. By comparing the arithmetically least numbers of these various types, he found (pp. 387-9) the least number having 15 aliquot parts.
John Kersey7 cited the long rule of van Schooten to find the number of aliquot parts of a number and then gave the simple rule that a\"... an*w has (ei+1)... (en+l) divisors in all if alt..., an are distinct primes.
John Wallis8 gave the last rule. To find a number with a prescribed number of divisors, factor the latter number in all possible ways; if the
*Practica Arith. & Menaurandi, Milan, 1537; Opera, IV, 1663.
'Arithmetica Integra, Norimbergae, 1544, lib. 1, fol. 101.
•StifePa posthumous manuscript, fol. 12, preceding the printed text of Arith. Integra; cf. E. Hoppe, Mitt. Math. Geeell. Hamburg, 3, 1900, 413.
4Cogitata Physico Math., II, Hydravlica Pnevmatica, Preface, No. 14, Paris, 1644. (Quoted by Winsheim, Novi Comm. Ac. Petrop., II, ad annum 1749, Mem., 68-99). Also letter from Mersenne to Torricello, June 24, 1644, Bull. Bibl. Storia Sc. Mat., 8, 1875, 414-5.
'Arithmetica Universalis, ed. 1732, p. 37; ed. 1761, I, p. 61.    De Inventione Divisorum.
•Exercitationum Math., Lugd. Batav., 1657.
7The Elements of Algebra, London, vol. 1, 1673, p. 199.
8A Treatise of Algebra, London, 1685, additional treatise, Ch. III.