384 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xn 



Set D = b z -4ac, p=(K+ba)'f(2d). Then # 2 = aA 2 +(cr-a). Hence we 

 have to assign to A and a such values that the latter sum is a square. 



To apply (pp. 338-346) this method to the congruence ?/ 2 ^c (mod b), 

 where b is a prime, we have (17) for a = 0. Then (19) holds for A = b+2pa if 



_ bw(v -\-fiw) v 

 v 2 cw 2 w 



The first denominator may be made equal to b if 5 is a quadratic 

 residue of c. Then a = =F w ( VQ + w o) 



Kunerth 180 continued the same subject. , Let i, &\ be a solution of 

 r = ma-i-n(3. Then a = cti np, (3 = (3i-\-mp. Substitute these in (18), with 

 7 = w, 5= m. After several reductions, we get 



ex f = (np 2 2aip e)x (mp 2 +2j3 1 p+). 



Then (17) has an integral solution if and only if p can be chosen to make the 

 value of x for which the preceding vanishes an integer. 



A. B. Evans and others 181 proved that, if p n /q n is the last convergent in 

 the first period of the continued fraction for \Cl, and r is the largest integer 

 ^ VA, then p n =Tq n q n -i. Hence we can derive x from y in a solution 



J. de Virieu 182 used the final digits to show that xy is divisible by 5 in 

 (20) 24x 2 +l = y\ 



E. Lionnet 183 stated and M. Rocchetti and F. Pisani 183 proved easily that 

 three successive sets (x i} j/,-) of solutions of (20) or 2x 2 +l = 3y 2 satisfy 

 x n+l = Wx n -Xn-i, y n +i = lQy n -2/n-i, with (x i} 2/0 = (0, 1) or (1, 1), (x 2 , yj 

 = (1,5) or (11, 9), respectively. For solutions x of the second equation, 3x 2 

 +2 is of the form 360n+5 and is simultaneously a sum of three consecutive 

 squares and a sum of two consecutive squares. For x 2 +l = 2y 2 , x n = Qx n -i 

 -x n -z, y n = 6y n -iy n -z, (xi, 2/1) = (l, l), (x z , yz) = (7, 5). 



S. Realis 184 used x 2 -ky 2 =(a 2 -k^)(A 2 -kB 2 ) 2 , where 



x = aA 2 -2kpAB+kaB 2 , y= -pA-+2aAB-kpB 2 , 



to derive a new solution of x 2 ky 2 = h from a given solution , |8 and a 

 solution of A 2 -kB 2 =l. 



H. Poincare* 185 noted that, if m is odd, and a, b give the least integral 

 solutions of a 2 w6 2 = l and c, d give the least odd integral solutions of 

 c 2 wd 2 = 4, then 



V 

 J = 



Several 186 proved easily that x n +p=2xpX n Xn- P) y n +p 2x p y n yn-p, if 

 a;, 2/ B be the nth set of positive integral solutions of x~ Ny 2 = I[.T O = 1, 2/o = 0]. 



is" Sitzungsber. Akad. Wiss. Wien (Math.), 82, II, 1880, 342-75. 



181 Math. Quest. Educ. Times, 30, 1879, 49. 



182 Nouv. Ann. Math., (2), 17, 1878, 476. 



^Ibid., (2), 18, 1879, 479, 528; (2), 20, 1881, 425-7, 373-4. Cf. Pisani" of Ch. VII. 

 184 Nouv. Corresp. Math., 6, 1880, 306-312, 342-350. 

 186 Comptes Rendus Paris, 91, 1880, 846. 

 186 Math. Quest. Educ. Times, 34, 1880, 114. 



