CHAP. XXII] QUARTIC FUNCTION MADE A SQUARE. 643 



formulas can be found giving in general four sets of solutions. In 



consider ax*, etc., as coefficients; we thus have an equation of the first type 

 having now a+-+g = c (an artifice due to Lucas 81 for d=f=g = 0). 

 After dividing such an equation by c and setting X= (p+x)f(p + l), we get 

 an equation in p to which Fermat's method applies. The explicit formulas 

 for the two sets of solutions are very long (each furnishing two sets by 

 changing the sign of z) . 



F. Romero 160 proved that x*-{-x 3 +x 2 -}-x l=y 2 has no positive integral 

 solutions. For, y is odd and the equation becomes 



Thus x = 4n-\-2, and 4n+3 would divide the sum of the squares of two rela- 

 tively prime integers. 



E. Lucas 161 discussed f(x) = y 2 , where f(x) is a quartic with rational 

 coefficients. Set y$(x) = F(x), where <f>=x p +aiX p ~ l -\ ---- with rational a's, 

 while F is of degree p+2. Then F 2 =f<j> 2 is an equation of degree 2p-f4 in 

 which enter 2p+3 unknowns besides x. If we know 2p+3 sets of rational 

 solutions Xi, yi of y 2 f(x~), no two of which differ merely in the sign of y, 

 and determine the coefficients my<}> = F so that it shall be satisfied by these 

 2p+3 sets, these coefficients will be rational. Then F 2 =/< 2 will furnish a 

 new rational x which leads to a new set of rational solutions of y 2 =f(x), 

 We may take two or more of the Xi equal; if x z =x\, we replace 



_ 



by the derivative of =t yf(xi)=F(xi)f4>(xi). Taking all of the Xi equal, we 

 see that one solution of f(x)=y 2 leads to an infinite sequence of solutions. 

 (Cf . Pepin. 158 ) If f(x) has a rational root a, we may take 



If /has a rational quadratic factor q(x), we may take F = q^ p and apply the 

 above method to 2p+l sets of solutions. 



L. J. Mordell 162 assumed that we have one solution of f=z 2 , where / 

 is a binary quartic with the invariants g 2 , 3. Then we can transform / 

 into a quartic with leading coefficient z 2 . The syzygy between its semin- 

 variants (cf. Mordell 176 of Ch. XXI) is g 2 = 4h 3 -g 2 hz*-g 3 z*. Thus g\z\ h/z z 

 give rational solutions of 



It is shown that the knowledge of all rational solutions of the latter leads 

 to all rational solutions of /=z 2 . 



E. Haentzschel 163 treated y 2 =f(x)=a& 4 -\ ----- \-a^. First, let 

 have a rational root r and apply the substitution 



_ x = r+\f(r)l(s-t], <=T 



160 Nouv. Ann. Math., (2), 18, 1879, 328. 



161 Nouv. Corresp. Math., 5, 1879, 183-6. 

 162 Quar. Jour. Math., 45, 1913^, 178-181. 

 163 Jour, fur Math., 144, 1914, 275-283. 



