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204                    HISTORY OF THE THEORY OF NUMBERS.            [CHA**- VI1
A. Cunningham137 gave five primes p for which there is a number of exponents to which the various numbers belong modulo p.
On exponents and indices, see Lebesgue174"6 and Bouniakowsky179 ;   also Reuschle69 of Ch. VI, Bouniakowsky111 of Ch. XIV, and Calvitti48 of Ch.
Bhascara Achdrya149 (1150 A. D.) found y such that 2/2~30 is by 7 by solving t/2=7c+ 30.   Changhig 30 by multiples of 7, we reaett a perfect square 16 with the root 4.   Hence set
7c+30=(7n+4)2,         c = 7n2+8n-2,          y = 7n+4.
Taking n = 1, we get y = 11. Such a problem is impossible if, after abrading the absolute term (30 above) by the divisor (7 above) and the addition of multiples of the divisor, we do not reach a square.
Similarly for the case of a cube, with corresponding conditions for impossibility (206, p. 265). For t/3 = 5e+6, abrade 6 by the divisor 5 to jgd> the cube 1; adding 43-5, we get 216 = 63. Hence set y = 5n+6.
An anonymous Japanese manuscript150 of the first part of the eighteen*^ century gave a solution of xnky=a by trial. The residues Oi,. . ., afc_i of ln,. . ., (k l)rt modulo k are formed; if ar=a, then x=*r. It was noted that ajb_r0r or kar according as k is even or odd, and that the residue of rn is r times that of r*"1.
Matsunaga,150* in the first half of the eighteenth century, solved a2 -\-bx- y* by expressing b as a product mn and finding p, q and A so tttat mpnq=l, 2pa^A (mod n). Then x= (Am 2a)A/n [and y=a  rrib>\. But if Am=2a, write A+n in place of A and proceed as before. Or write 2a+b in the form 6Q+#, whence z=2a+&-(Q+l)JK. To solve 6O-f-llx=y2, consider the successive squares until we reach 52=3 (mod 11). Write 2-5+11 in the form 1-11+10. Then for a=5, 6=11, Q=l, #= 1O, the preceding expression for x becomes 1, whence 52+ll-l = 62. Then write 2-6+11 in the form 2-11+1. Then 23-(2+l)-l = 20 gives 62-+-20-11=162, and a?=(256-69)/ll = 17.
L. Euler151 proved that, if n divides p  1, where p is a prime, and if a = cn+/bp, then (by powering and using Fermat's theorem), a(p~1)/n 1 is divisible by p. Conversely, if am-l is divisible by the prime p = mn-i- 1, we can find an integer y such that a  yn is divisible by p. For,
and the differences of order mn-n of Q(l), Q(2), . . ., Q(mri) are the saxrxe
137Math. Quest, and Solutions (Ed. Times), 3, 1917, 61-2; corrections, p. 65.
149Vfja-ganita,  204-5; Algebra, with arith. and mensuration, from the Sanscrit of Brahmegxip-ta
and Bhascara, transl. by H. T. Colebrooke, London, 1817, pp. 263-4. uAbhand. GescWchte Math. Wiss., 30, 1912, 237. uoJ&wl, 234-5. ulNovi Comm. Acad. Petrop., 7, 1758-9 (1755), p. 49, seq., 64, 72, 77; Comm. Arith.,    1,
270-1, 273.   In Novi Comm., 1, 1747-8, p. 20; Comm. Arith., 1, p. 60, he proved the firat
statement and stated the converse.