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



But for lllx - 3Qly = 222, it is simpler to begin with 301 = 3-111 - 32, 

 rather than with 301 = 2-111 + 79. 



Thomas Fantet de Lagny 9 gave examples of a "new method" of solving 

 indeterminate equations. To make m = (19n 3)/28 an integer, double 

 28?i (19n 3) and subtract the result 18n + 6 from 19n 3; thus 

 n 9 is to be divisible by 28. Hence n = 9 + 28/, where / is any integer. 

 Later, he 10 gave (pp. 587-595) the following rule for solving y = (ax + q)/p, 

 where a and p are relatively prime, and (as may be assumed) a < p, q < p : 

 Take a from p as many times as possible and call the remainder r; take r 

 from a as many times as possible and call the remainder t; etc., until the 

 remainder 1 is reached. Then make the same divisions for q and p as were 

 made for a and p, having regard to the signs. According as the last re- 

 mainder is s or + s, we have x = s or p s. 



L. Euler 11 gave a process to find an integer m such that (ma + v)/b is 

 integral, where v > 0. Set a = ab + c. Then A = (me + v)/b must be 

 an integer. Thus m (Ab v)/c. First, if v is divisible by c, we get a 

 solution by taking A = 0. Second, if v is not divisible by c, set b = @c + d. 

 Then m will be integral if (Ad v}/c is integral. Thus we set c = yd + e, 

 etc. Euler remarked that the process is therefore that of finding the 

 greatest common divisor of a, 6, continued until we reach a remainder which 

 divides v. His formula for a solution of ma + v = nb is equivalent to 



(1111 \ .(111 \ 



= oy I + +), m= bv\- - + 3 ')> 



V ab be cd de J \ be cd de / 



in which the series are continued until we reach a remainder dividing v. 

 For the case a, b relatively prime, these results have been given by C. 

 Moriconi. 12 



N. Saunderson 13 (blind from infancy) gave a method to solve ax by = c, 

 where c is the g.c.d. of a, b. Let a = 270, b = 112, whence c = 2. He 

 employed the equations and successive quotients 



la --06= 270, 5a - 126 = 6, 3; 



Oa - - 16 = - 112, 2; 17a - 416 = - 2, 2; 



a --26= 46, 2; 39a - 946 = 2. 



2a - 56 = - 20, 2; 



Divide the term 270 of the first equation by the absolute value 112 of the 

 term of the second, to obtain the quotient 2. Multiply the second equation 

 by 2 and add to the first; we get the third equation. The division of 112 



9 Nouveaux Elemens d'Arithmetique et d'Algebre, ou Introduction aux Mathematiques, 



Paris, 1697, 426-135. 



10 Analyse g^nerale; ou m6thodes nouvelles pour nSsoudre les problemes de tous les Genres & 



de tous les Degrez a 1'infini, Paris, 1733, 612 pp. Same in M6m. Acad. Roy. des 



Sciences, 11, 1666-1699 [1733], annee 1720, p. 178. 

 Comm. Acad. Petrop., 7, 1734-5, 46-66; Comm. Arith. Coll., I, 11-20. 

 12 Periodico di Mat., 2, 1887, 33-40. Cf. C. Spelta, Giornali di Mat., 33, 1895, 125. 

 18 The Elements of Algebra, Cambridge, 1, 1740, 275-288. The solution of the first problem 



was reproduced by de la Bottiere, M6m. de Math, et Phys., pre'sente's . . . divers 



savans, 4, 1763, 33-41. Cf. Lagrange. 22 



n 



