642 HISTORY OP THE THEORY OF NUMBERS. [CHAP, xxn 



Set Q = y[x, 2w=zfx, c'v+b' = u(x. Then 



u* = Aa?+By*+Cz*, A = b' 2 -a'c f , B=-c r , C = c'e, 



which can be given the form Ai(u 2 x 2 )=Bi(y 2 z 2 ) by choice of k. The 

 solutions for u, x, y, z are evident. Substitute these in the quadratic in 

 u, x, y obtained by eliminating v between Q = y/x, c'v+b' = u/x. For 

 example, if ti 4 -2 = io 2 , then w 2 =(v*+k 2 )-2kv 2 -k 2 -2. Set v 2 +k = u/x, 

 v = y/x, w=z/x. Then 



u 2 = (k 2 +2)x 2 +2ky 2 +z 2 , u 2 -z 

 for A; =4. It has the solutions 



u, z = %(aypd); 12x, 8y = 

 Substitute these in ux = kx 2 +y 2 (obtained by eliminating v). Thus 



with coefficients quadratic in 7, d. Taking L = we get four sets of solu- 

 tions a, |3, 7, d; likewise four from N = 0. 



S. Bills 156 made f=x*+kc 3 +8x 2 +7x+6= D by noting that /=4 for 

 x= 1 and setting /=Q 2 , Q=x 2 +2x-\-k, where k is chosen so that Q= 2 

 for x= 1. 



T. Pepin 157 made use of the notations of Euler, 140 viz., (1) and 



00) =f+gx+hx 2 , F(z) =/(z) - 2 (z) = a -/ 2 + . + (e - W}z*. 

 Pepin took xi, x 2 , x 3 arbitrary but distinct, and determined /, g, h, x by 



(5) e(xd=eif), 5 = 1, x = -x 1 -x,-x s (i = l, 2, 3). 



Then x\, x z , x s , x are the roots of F(z) =0. Hence if x\, x 2 , x 3 are three 

 solutions of f(x) = D, then/, g, h are rational and x is a new solution. Next, 

 let Xz=Xi; then F'(XI) =0, and (5) for i = 3 is to be replaced by the deriva- 

 tive of (5) for i=l. Finally, for i=2= 3 , we use (5) for i = l and its 

 first and second derivatives, and so obtain a second solution from a first. 

 Then the preceding case gives a third solution and (5) a fourth solution. 



Pepin 158 noted that if a quartic f(x) can be transformed into a square 

 by replacing x by a rational function, then F=y 2 f(x)=Q is a unicursal 

 curve and hence has three double points, whence the partial derivatives of 

 F with respect to x and y vanish, showing that / has a double root. The 

 problem is then to make the remaining quadratic factor a square. The 

 problem to make a product of two binary quadratic forms a square is 

 treated by means of a congruence. Conditions are given in order that a 

 reciprocal quartic shall never be a rational square for a rational value of the 

 variable. 



A. Desboves 159 noted that if x, y, z is a set of solutions of 



166 Math. Quest. Educ. Times, 22, 1875, 91-2. 



167 Atti Accad. Pont. Nuovi Lincei, 30, 187&-7, 211-37.. 



168 Atti Accad. Pont. Nouvi Lincei, 32, 1878-9, 166-202. 



" B Comptes Rendus Paris, 88, 1879, 638-40, 762 (correction). Cf . Desboves 148 of Ch. XXI. 



