CHAP, xiil] SOLUTION OF ax 2 +bxy+cy~ = k. 409 



tinued fraction for Vgi/7 has the quotients 1, 2, 1, 3, 9, 3, 1, 2, 2 (with the 

 period marked). Since 



2+l+3 11' 



x = 15, y = ll, give the minimum 2 of 7x 2 l3y 2 . 

 Given (p. 577) the solution x = a, y = b of 



f=Ax 2 -2Bxy+Cy 2 = c (k = B 2 -AC>0, 

 to find an infinitude of solutions, use the solution of 



corresponding to the Pell problem p = Bq+ ^kq 2 -\-l. Now Af has the 

 factors AxByy-\!k, and < the factors pBqq^k. Hence we use 



(1) Ax-By+y^k = (Aa-Bbb <Jk)(p-Bqq^k) n , 



for any of the four combinations of signs. To find the minimum of / for 

 integral x, y, develop the root (B ^k)/A into a continued fraction and 

 proceed as above. 



G. L. Dirichlet : ' 7 noted that in addition to the infinite set (1) of solu- 

 tions there may exist further similar sets of solutions. Given any positive 

 number a-, we can find one and only one solution x, y of set (1) for which 



where t, q give any positive solution of t 2 kq 2 = l. All solutions of these 

 inequalities can be found by a finite number of trials. Hence we find the 

 initial solutions a, b defining the various sets (1). 



A. M. Legendre 58 discussed the integral solutions of 



(2) Ly 2 +Myz+Nz 2 =H. 



After preliminary transformations, we may assume that z is prime to y 

 and H. Distinguish the cases in which the roots of Lt 2J rMt-{-N = Q are 

 imaginary, real or equal. First, let 4LN-M 2 = B>0. Set x = 2Ly+Mz. 

 Then x 2 +Bz 2 = C = 4LH. Give to z the successive values 0, 1, -, [VC/5] 

 and see whether the resulting value of C Bz 2 is a square x 2 and then 

 whether the resulting x makes Mz^x a multiple of 2L. Second, let 

 4:LNM 2 = B,B positive and not a square. If H<^^B, develop a root 

 of Lx 2 -\-Mx+N = into a continued fraction; if one of the complete 

 quotients (| VB+7)/D has D = H,&t least one of the equations (2) is solvable. 

 But if ff >| V#, we may set y = nz+Hu : where n^H. Thus if Ln 2 +Mn 

 +N is not a multiple /// of H for some value of n between ^H and \H, 

 (2) is impossible; while if such a multiple is found, the equation reduces to 

 fz 2 +gzu+hu 2 =l for g = 2nL+M, h = LH. See Lagrange 76 - 85 of Ch. XII. 

 E. F. A. Minding 59 noted that, if A = b 2 ac is positive and not a square, 

 and if #<f A/A, we can decide whether or not ax 2 -\-2bxy -\-cy 2 = H is 



67 Bericht Akad. Wiss. Berlin, 1841, 280; Werke, I, 628-9. 



B8 Theorie des nombres, 1798, 99-122 (77-98); ed. 2, 1808, 88-110 (68-87); ed. 3, 1830, 



I, 104-129 (81-103); transl. by Maser, I, 105-131 (81-105). 

 69 Jour, fur Math., 7, 1831, 140-2. 



