50 HISTORY OF THE THEORY OF NUMBERS. [CHAP. II 



The number of integral solutions x, ^ x < c, is 



2u(ax + b)T 



cos 



C x=0 u=0 C 



A. L. Crelle, 28 after proving the existence of solutions of a 2 #i = ^1^2 + k, 

 where GI and 02 are relatively prime [Euler 14 ^j, solved it by setting 



also by the modified equations in which the left members are all a 1} or Xi. 

 There are given three more such methods. The sixth method uses a prime 

 factor 0:1 of a\ = i/3i, and a primitive root TTI of G>I. There exists an integer 

 ei such that dzTrl 1 = Zicxi 1. Multiply the proposed equation by Trl 1 . 

 Thus ZiXi = fiiTrl'Xz + #3, where x 3 = (knl 1 =F x^fai is an integer. The 

 latter gives x\ = T (aiX 3 /CTrl 1 ) . Here x s must satisfy (hx 3 = T f3iX 2 + zjc, 

 which is treated as was the initial equation. 



Crelle 29 considered ay = bx + 1, where a, b are relatively prime and > 1. 

 If x , ?/o give the least positive solution, the general solution is = jua + x , 

 2/ P = fjb + 2/o (M = 0, 1, 2, ) If 2/0 < 6/2, the numerators of 



yo y-i yi y-z yz 



XQ' X-i' X\ X-z' Xi' 



increase alternately by b 2y and 2y , and the denominators alternately by 

 a 2x Q and 2xo, and no one of these fractions differs more from a/6 than the 

 next fraction. There are similar theorems on series of fractions involving 

 only positive or only negative subscripts. If yjx^ b/a = k > 0, 

 v/u b/a = X > 0, where v < y^ +l \ , \u\ < \ x^ +l \ , then X > k. 

 If X < 0, he found the number of fractions v/u for which k > X, ju being 

 given. 



J. P. M. Binet 30 treated ax Ay = 1, A > a, by a process for finding 

 the g.c.d. of a and A in which A is always the dividend. On dividing A by 

 a, ai, a 2 , , let p, p iy p 2 , - be the quotients and a\, a z , a 3) 

 the remainders. Then 



(6) app^ -pi-i = Of + A {1 + Pi-i + Pi-ipi-z + + Pi-i- -Pipi}- 



Let a n be the divisor when the remainder is zero. Since a n divides A, 

 it is the g.c.d. of A and a if it divides a. But if a n is not a divisor of a, 

 proceed as above with a and a n and call the remainders 61, 62, , 

 b n >, the last corresponding to the remainder zero. Then a n , b n >, c n , 

 form a rapidly decreasing series and one of them will be 1. If a n = 1, 

 (6) for i = n gives a relation of the form aP = d= 1 + APi. 

 E. Midy 31 used Euler's 14 result to solve by cz = a by trial. 



28 Abh. Akad. Wiss. Berlin (Math.), 1836, 1-53. 



29 Ibid., 1840, 1-57. 



80 Comptes Rendus Paris, 13, 1841, 349-353; Jour, de Math., 6^ 1841, 449-494. 



31 Nouv. Ann. Math., 4, 1845, 146; C. A. W. Berkhan, Lehrbuch der Unbest. Analytik, Halle, 1, 

 1855, 144; A. D. Wheeler, Math. Monthly (ed., Runkle), 2, 1860, 23, 55, 402-6; L. H. 

 Bie, Nyt Tidsskrift for Mat., Kjobenhavn, (4), 2, 1878, 164; J. P. Gram, ibid., 3, B, 

 1892, 57, 73; E. W. Grebe, Archiv Math. Phys., 14, 1850, 333-5. 



