Sophie Germain: mathématicienne extraordinaire - Scripps College

Sophie Germain’s Theorems 58
Since x ≡ 0 (mod q), we can add multiples of x to get
py p−1 ≡ px p−1 − px p−2 y + px p−3 y 2 + · · · − pxy p−2 + py p−1 (mod q)
≡ p(x p−1 − x p−2 y + x p−3 y 2 + · · · − xy p−2 + y p−1 ) (mod q)
≡
pγ p (mod p).
We know γ ≢ 0 (mod p), since this would imply that z is divisible by p.
Then there must be an integer, g, the inverse of γ, such that gγ ≡ 1 (mod q).
Thus we have,
α p ≡ py p−1 (mod q)
≡
pγ p (mod q).
So
α p g p ≡ pγ p g p (mod q)
(αg) p ≡ p(γg) p (mod q)
(αg) p ≡ p (mod q).
This violates condition 1.
We have found an integer n = αg such that
n p ≡ p (mod q).
We have shown that if p has an auxiliary prime q such that condition 2
is satisified and if there are non-trivial integer solutions x, y, and z to the
equation X p + Y p + Z p such that p ∤ xyz, then condition 1 is not satisfied.
Thus, if p has an auxiliary prime q such that conditions 1 and 2 are satis-