CHAP. Xlll] Ax' i +2Bxy+Cy*+2Dx+2Ey+F = 0. 413 



where B = p z -4ay, f=p8-2ae, <7 = S 2 -4af. Set A=f~-Bg. Then 



By+f=u, u* 



u-f t-8 (3(u-f) 



B ' 2a 2aB~ 



Hence the rational solutions of (1) follow from the rational solutions of 

 u- = A+Bt 2 . The latter depend on the integral solutions of Ar 2 = p 2 Bq 2 , 

 discussed by Lagrange. 110 



To obtain the integral solutions of (1), it is necessary that not only 

 u and t be integers, but also that it / be a multiple mB of B, and that 

 td pm be a multiple of 2a. If B is negative, u i Bt- A has only a 

 finite number of integral solutions, which can be found by trial. This 

 is not true when B is positive, as will be assumed henceforth. We may set 

 u = ap, t = ffq, where p, q are relatively prime. By Lagrange 75 of Ch. XII, 

 the solutions of pBq^ = A/ff 2 are given by 



whence 



Here a, b, X, Y are given integers for which X 2 BF 2 = 1. We may 

 restrict attention to the case X 2 BF 2 =+1, to which the contrary case is 

 easily reduced. The problem is now to choose positive integral values of 

 the exponent n for which the resulting values of x, y are integers, viz., for 

 which <rp /is a multiple of B, and <rg 5 j8(=t<rp f)/B is a multiple 

 of 2a. These two questions are special cases of the general question of 

 the divisibility of 

 (2) 



by R = r m r"i l , where r, TI, are distinct primes. It is easily shown that 

 (XY^lB) p 1 is divisible by r, where p = 2r if B is divisible by r, p = rl 

 if JB (r - 1)/2 l is divisible by r, and p = r if r = 2. Then (XFV5) e -l is 

 divisible by r m for e = r m ~ 1 p- Hence if n = ke-\-N, (2) is divisible by r m if 

 and only if r m divides the function obtained from (2) by replacing n by N, 

 so that we need only test the values <e of n. Similarly we need only test 

 the divisibility of (2) by r'f 1 for n<r^ -1 pi. Suppose that the test succeeds 

 for n = N and for n=Ni } etc., in the respective cases. Then determine n 

 so that it shall have the remainder N when divided by r m ~ l p, the remainder 

 NI when divided by r" ll-1 pij etc. We saw that also a second expression 

 F \+Gip -\-Hiq had to be divisible by a certain number RI. The conditions 

 on n are similar to those just stated. Hence the method leads to all the 

 (infinitude of) integral solutions of (2) when it is solvable. 

 Lagrange 80 multiplied (1) by 4a and set 



u = 2ax+(3y+d, a = j3 2 4ay, b = @5 2ae, c=5 2 -4af. 



We get u 2 = ay 2 -\-2by +c. Multiply by a and write t = ay+b, R = Wac. 

 Hence t 2 au' 2 = R. Assume that it has a known solution t = P, u = Q. 



80 Miscellanea Taurinensia, 4, 176&-9; Oeuvres, I, 725-31. 



