On 29/04/02 23:36, Paolo Bussotti <email@example.com>, asked for some help :
> I'd like some information on this problem: it is well known that, if p > is an odd number, then > x^p + y^p = (x+y)A(x,y). So, for example, if p=5, > x^5 + y^5 = (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4), > with A(x,y) = (x^4-x^3y+x^2y^2-xy^3+y^4). > I'd like to know whether someone demonstrated that, if p is a prime odd > number, and if x and y are two reciprocally prime integer numbers and > x+y is prime with p, then (x+y) and A(x,y) are reciprocally prime. > While, if x+y is a multiple of p, then x+y and A(x,y) have only p^1 as > common factor.
This is word for word in: LEGENDRE (A. M.) [& GERMAIN (Sophie)] ; THÃÂORIE DES NOMBRES, SECOND SUPPLÃÂMENT, (Paris, Courcier),septembre 1825. This work is an attempt on demonstrating Fermat's Great Theorem. In bottom of p. 5, after quite a clumsy piece of reasoning, it reads: << Il re/sulte de ce qui pre/ce\de que si on fait y^n + z^n = (y+z)Phi(y,z), les deux facteurs y+z et Phi(y,z) auront n pour commun diviseur ou n'en auront aucun >>