CHAP. II] SOLUTION OF ax+by = c. 51 



J. A. Grunert 32 solved bx ay = 1 by a process for finding the greatest 

 common divisor of 6, a in which the divisor is always a, while the dividend 

 is the sum of 6 and the preceding remainder, a process due to Poinsot, 

 Jour, de Math., 10, 1845, 48. 



V. Bouniakowsky 33 would solve ax + by = k by adjoining a'x + b'y = h f , 

 whose coefficients are arbitrary. Set D = ab' ba', p = b'/D, q = h'/D, 

 r = a! ID. Then x = kp bq, y = aq kr, subject to ap br = 1. 



A. L. Crelle 34 gave over 4000 pairs of positive integral solutions Xi < a\ t 

 x 2 < a 2 , of diX 2 = a z xi + 1, for i ^ 120, < a 2 < a ly with 2 prime to 



and indicated methods used to simplify the calculation of the table. 



V. Bouniakowsky 35 integrated by parts 



f (ax + V) m - l (a'x + V} n ~ l 

 tJ 



dx 



to obtain an identity giving a solution of b m X b' n Y = 1, where x = a', 

 y = a is a particular solution of bx b'y = 1. For m = n = 2, the 

 identity is 



(3aV6 - a 3 b')b' 2 - (3aa'V - a' 3 b)b* = (a'b - ab')*. 



H. J. S. Smith 36 reported on a recent method to solve ax = 1 + Py 

 [no reference]. Join the origin to the point (a, P). No lattice point 

 (i. e., with integral coordinates) lies on this segment; but on each side 

 of it there is a point lying nearer to it than any other. Let (1, 771) and 

 (2, 772) be these two points and let $1/771 < $2/772- Then the 's and 77's are 

 the least positive solutions of 0771 Pi = 1, 0772 P& = 1. 



G. L. Dirichlet 37 solved ax by - 1 by continued fractions, using the 

 algorithm due to Euler. 38 



C. G. Reuschle 39 found the general solution of ax + by = cby combining 

 it with ax + &y = m, where m is an arbitrary integer, while a and are 

 integers determined so that a/3 ba = 1 [cf . Bouniakowsky 33 ]. 



J. J. Sylvester 40 noted that the number of positive integers < pq which 

 are neither multiples of p or q nor can be made up by adding together mul- 

 tiples of p and q is %(p l}(q 1) if p and q are relatively prime. 



H. Brocard 41 solved ax + by = 1 by a process of reduction. It suffices 

 to find the residue of a modulo a b to obtain an equation x + y = / 

 consistent with the given one. Thus, if b = 563036, a = b + 7, then 

 o = 6 = 5 (mod 7), 3-5 = 1, and the given equation may be combined 

 with x + y = 3 to get integral solutions x, y. A table gives the successive 



32 Archiv Math. Phys., 7, 1846, 162. 



33 Bull. Acad. Sc. St. PStersbourg, 6, 1848, 199. 



34 Bericht Akad. Wiss. Berlin, 1850, 141-5; Jour, fur Math., 42, 1851, 299-313. 



35 Bull. Cl. Phys.-Math. Acad. Sc. St. Petersbourg, 11, 1853, 65. 



36 Report British Assoc. for 1859, 228-267, 8; Coll. Math. Papers, I, 43. 



37 Zahlentheorie, 23-24, 1863; ed. 2, 1871; ed. 3, 1879; ed. 4, 1894. 



38 Comm. Acad. Petrop., 7, 1734-5, 46 (Euler 96 ). Novi Comm. Acad. Petrop., 11, 1765, 28; 



see Euler, 72 Ch. XII. Cf. Gauss. 24 



39 Zeitschrift Math. Phys., 19, 1874, 272. Same by J. Slavik, Casopis, Prag, 14, 1885, 137; 



V. Schawen, Zeitschrift Math. Naturw. Unterricht, 9, 1878, 107 [194, 367]. 



40 Math. Quest. Educ. Times, 41, 1884, 21. 



41 Mem. Acad. Sc. Lettres Montpellier, Sec. Sc., 11, 1885-6, 139-234. See p. 153. 



