CHAP. II] SOLUTION OF ax-\-by = c. 45 



Multiply the second equation (1) by the coefficient 5 in (3), and eliminate 

 the term 5 -4 by use of (3) ; we get 



(4) 43-7 = 5-60 + 1. 



Finally, multiply the first equation (1) by the coefficient 43 in (4) and 

 eliminate the term 43-7 by use of (4) ; we get 



(5) 43-67 = 48-60 + 1, 



so that the least X is 43 and the corresponding Y is 48. 



Bachet's first step, leading to (1), is Euclid's algorithm for finding the 

 greatest common divisor of A and B. His next steps are the elimination 

 from equations (1) of the terms in 3, 4, 7, respectively, in a special way 

 so that negative quantities are not introduced. 



John Kersey 7 treated Problems 18 and 21 of Bachet, 6 but " without 

 following Bachet's very tedious and obscure method of solution." To 

 solve 9a + 6 = 76, start with 6 and by successive additions of 9 form the 

 series 15, 24, 33, 42, ; next, form similarly the multiples 7, 14, 21, 28, 

 35, 42, of 7; the common number 42 yields a = 4, b = 6. Another 

 method is used for 49a + 6 = 136; find the multiple (65) of 13 which just 

 exceeds 49 + 6; divide 49 by 13; since in 55 = 65 - 10, 49 = 39 + 10, 

 we have remainders differing only in sign, we add and get 104; then 

 b = 104/13, a = 2. If one remainder had been merely a divisor of the 

 other remainder, we first multiply one of the equations. Neither of these 

 cases arises for 121a + 5 = 936. Then 126 = 186 - 60, 121 = 93 + 28, 

 and we seek c and d such that 93c + 60 = 28d. After the latter is solved 

 by the former process, we deduce a and 6 as in the preceding case. In a 

 new type of problem, the constant term occurs in the member with the 

 smaller coefficient, as in 71a + 3 = 1736. Take 2-71, which increased by 3, 

 gives a sum < 173. Since 145 = 173 - 28, solve 173A + 1 = 71B as 

 above to obtain A = 16, B = 39. Multiply the latter equation by 28 

 and subtract the former. Thus 173(16-28 + 1) = 71 -39 -28 + 145, whence 

 6 = 16-28 + 1 = 449, a = 1094. 



Michel Rolle 8 (1652-1719) gave a rule to find integral solutions; he 

 applied it as follows. For I2z = 22lh + 512, divide the larger coefficient 

 221 by the smaller 12; the largest integer in the quotient is 18. Set 

 z = ISh + p- we get 12p = 5h + 512. By the same method [dividing 

 12 by 5], h = 2p + s, 2p = 5s + 512. By the same method, p = 2s + m. 

 Then 2m = s + 512, and we have now reached a coefficient which is unity. 

 Eliminating s and p from 



s = 2m 512, p = 2s + m, h = 2p + s, z = ISh + p, 

 we get the desired solution 



z = 221m - - 47104, h = 12m - 2560. 



7 The Elements of Algebra, London, I, 1673, 301. 



8 Traite d'Algebre; ou Principes generaux pour resoudre les questions de mathe'matique, 

 Paris, 1690, Bk. 1, Ch. 7 ("eviter les fractions"), pp. 69-78. 



