CHAP. II] SOLUTION OF ax+by = c. 47 



by 46 gives the quotient 2; the product of the third equation by 2 when 

 added to the second gives the fourth equation; etc. But on dividing 6 by 2 

 we use 2 and not the exact quotient 3, since the latter would lead to an 

 equation 5Qa 1356 = with constant term zero. 



Our sixth and seventh equations each solve the problem. Other solu- 

 tions follow by adding to either equation 56a 1356 one or more times. 



The process must succeed since the formation of the constant terms is 

 identical with Euclid's process to find the g.c.d. of a, 6. 



The determinant of the coefficients in any two successive equations of 

 the above set is 1. From the pairs of coefficients form the fractions* 



1 2 5 12 41 94 

 1' 0' 1' 2' 5' 17' 39' 



They are alternately less than and greater than a/6 and converge to it; 

 if / and F are two successive fractions of the set, a/6 lies between them and 

 differs less from F than from /. Also a/6 is nearer to F than to any fraction 

 whose denominator is less than that of F. This method of approximating 

 to fractions is attributed to Cotes and is said to be simpler than the methods 

 of Wallis and Huygens. 17 



L. Euler 14 proved that if n and d are relatively prime, a + kd (k = 0, 1, , 

 Ti 1) give n distinct remainders when divided by n, so that the remainders 

 are 0,1, , n 1 in some order. Since one remainder is zero, a + xd = yn 

 is solvable in integers. 



W. Emerson 15 used the first method of de Lagny 9 to solve ax by + c. 

 Let d and / be the remainders obtained by dividing 6 and c by a. Subtract 

 some multiple of (dy + / )/a from the nearest multiple of y. The resulting 

 " abridged" fraction or some multiple of it is to be subtracted from the 

 nearest multiple of y, etc., until the coefficient of y is unity. Thus 

 x = (14y 11) /1 9 is subtracted from y; the product of the difference by 4 

 is subtracted from y, we get (y + 6)/19, an integer p, whence y = 19p 6. 

 The same rule and same example was given by John Bonnycastle. 16 



J. L. Lagrange, 17 to find integers pi and qi satisfying pqi qpi = 1, 

 where p, q are relatively prime, reduced p/q to a continued fraction ( 29, 

 p. 423). As noted by Chr. Huygens, De scriptio automati planetarii, 

 1703, we get a series of fractions converging towards p/q, alternately less 

 than and greater than p/q. Hence take pi equal to the numerator and qi 

 equal to the denominator of the convergent immediately preceding p/q. 

 Then pq l qp l = + 1 or 1 according as pifqi < "or > pfq. To apply 

 ( 8) to py qx = r, where p, q may be assumed relatively prime, multiply 

 the former equation by r and subtract. Thus 



x = mp d= rpi, y = mq rg t . 



* The last is replaced by a/b if the final quotient had been taken as 3. 

 14 Novi Comm. Acad. Petrop., 8, 1760-1, 74; Comm. Arith. Coll., I, 275. 

 16 A Treatise of Algebra, London, 1764, p. 215; same paging in 1808 ed. 



16 Introduction to Algebra, ed. 6, 1803, London, 133. 



17 M<m. Acad. Berlin, 23, ann<5e 1767, 1769, 7; Oeuvres, 2, 1868, 386-8. 



