Sophie Germain: mathématicienne extraordinaire - Scripps College

Sophie Germain’s Theorems 57
We have already assumed that x is divisible by q. Then
2x = (x + z) + (x + y) − (y + z) = b p + c p − a p ≡ 0 (mod q).
By condition 2, q|abc. Since q is prime, then q must divide exactly one of a,
b, or c.
Suppose q divides b. Then
−bβ ≡ 0 (mod q)
y ≡ 0 (mod q).
This implies that q|y, which contradicts our assumption that x, y, and z are
relatively prime. So b ≢ 0 (mod q).
Similarly, we find that c ≢ 0 (mod q), so q ∤ c.
Thus q|a. Then we have
a ≡ 0 (mod q)
a p ≡ 0 (mod q)
y + z ≡ 0 (mod q)
y ≡ −z (mod q)
Recall from the proof of Claim 5.1 that z = −y gives us
y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 ≡ py p−1 (mod q)
so, α p ≡ py p−1 (mod q).