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Congruences, applications by Pascal Boyer

By Pascal Boyer

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M. Wright. An introduction to the theory of numbers. Oxford University Press, 1980. [3] K. Ireland and M. Rosen. A classical introduction to modern number theory. Springer-Verlag, 1982. [4] K. Ireland and M. Rosen. A classical introduction to modern number theory. Springer-Verlag, 1982. [5] D. Perrin. Cours d’alg`ebre. Ellipses, 1998. [6] P. Samuel. Th´eorie alg´ebrique des nombres. Hermann, 1967. -P. Serre. Cours d’arithm´etique. Puf, 1970. P. Smart. The algorithmic resolution of diophantine equations, volume 41.

Le cas n quelconque s’en d´eduit imm´ediatement en remarquant que si le r´esultat est vrai pour n et m, il l’est pour mn. 19 (i) Comme 1+p+· · ·+pp−1 ≡ 1 mod 2, ses facteurs premiers sont tous impairs ; en outre comme il est congru `a 1 + p mod p2 , il poss`ede au moins un diviseur premier non congru `a 1 modulo p2 . 2 (ii) Raisonnons par l’absurde : soit n tel que np ≡ p mod q ; on a alors np ≡ pp ≡ 1 mod q car pp − 1 = (p − 1)(1 + p + · · · + pp−1 . Comme q est un nombre premier ne divisant pas n, il est premier avec n et donc d’apr`es le petit th´eor`eme de Fermat, on a nq−1 ≡ 1 mod q.

On a alors uv = (y/7)2 de sorte que u = t2 et v = s2 avec = ±1 et s, t ≥ 0 premiers entre eux. On trouve alors les solutions x = ±3, 4 et y = 0, qui conviennent. 22 (i) Si x est pair, on a y 2 ≡ −1 mod 8. En ´ecrivant y impair sous la forme 2k + 1, on obtient y 2 = 1 + 4k(k + 1) ≡ 1 mod 8 contradiction. (ii) On a x3 +8 = (x+2)(x2 −2x+4) avec x impair de la forme 2k+1 ; (x2 −2x+4) = 4k 2 +3. On en d´eduit donc qu’il existe un p premier divisant x2 − 2x + 4 avec p ≡ 3 mod 4. Or si p premier divise y 2 + 1 alors p ≡ 1 mod 4, d’o` u la contradiction.

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