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Topic: [HM] number theory problem
Replies: 2   Last Post: May 1, 2002 4:30 PM

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 Udai Venedem Posts: 199 Registered: 12/3/04
Re: [HM] number theory problem
Posted: May 1, 2002 4:30 PM

On 29/04/02 23:36, Paolo Bussotti <crbpm@tin.it>, 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)] ; THÃÂORIE DES NOMBRES, SECOND
SUPPLÃÂMENT, (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 >>

this "supple/ment" is appended to Legendre's ESSAI SUR LA THÃÂORIE DES
NOMBRES ; Seconde ÃÂ©dition. Paris, Courcier, 1808.
which you can view at

With all my best wishes.

Udai Venedem