CHAP, xxvi] FERMAT'S LAST THEOREM. 749 



A. Desboves 103 noted that aX m +bY m = cZ n has integral solutions if 

 and only if c is of the form ax m -\-by m ; we can find a function c of a, b and 

 as many parameters as one pleases such that integral solutions exist. 

 Next, let n = m. Then we can find a, b, c so that there are two sets of 

 solutions and these determine a :b : c. There exists such an equation 

 with three sets of solutions if and only if 



P m +Q m +R m = U m +V m +T m , PQR^UVT 

 have integral solutions ={=0. We can solve X 4m a 2 F 4m = Z 2 if 



(x + yi) 4OT (x yi) 4m 



a = - 2i - ' 

 viz., byX = rM-2/ 2 , 7=1. 



A. E. Pellet 104 considered, for p a prune, the congruence 



At m +Bu n +C=Q (mod p}, ABC^O (mod p). 



Let d be the g.c.d. of m, p 1; di that of n, p 1. Set #="*. Then x must 

 satisfy the two congruences 



Ax 



\(p-i)^i "1 



J -1 J=0 (mod p). 



Conversely, to each of the ju common roots of the latter two congruences 

 correspond dd\ sets of solutions of the proposed congruence, which therefore 

 has fjddi sets of solutions. For m n = 2, the two congruences have at 

 least one common root, since the second is not z (p ~ 1)/2 +l = 0, being of 

 higher degree. Hence Atf-}-Bu 2 -}-C=Q (mod p) is solvable (Lagrange, 9 

 etc., of Ch. VIII). 



R. Liouville 105 claimed that X n +Y n = Z n is impossible if n>l and X, Y, 

 Z are polynomials in a variable t. Set a = XfZ. Then 



is a polynomial hi Vl n = Y/Z. Since dU/dt is the argument of the 

 second integral, 



l (}__* (IT*! 



dt\z)- * \x) dt 



\x 



must be the product of Y by a polynomial A. Hence 



"~ 2 d Y\ 



Thus X n ~ l divides Z*d(Y/Z) . Call the quotient B and set P = Y/Z. Then 



" 



But in the latter, the left member is infinite for a root of P n = 1, while the 



s Nouv. Ann. Math., (2), 18, 1879, 481-9. 

 104 Comptes Rendus Paris, 88, 1879, 417-8. 

 106 Comptes Rendua Paris, 89, 1879, 1108-10. 



