Sophie Germain: mathématicienne extraordinaire - Scripps College

Sophie Germain’s Theorems 56
So
(y + z)(y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 ) ≡ 0 (mod p)
y p + z p ≡ 0 (mod p)
(−x) p ≡ 0 (mod p).
This implies that p divides x. This is a contradiction since we have assumed
that p does not divide x, y, or z. Thus r ≠ p.
So no prime r can divide p or y p−1 , then it cannot divide both y + z
and y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 .
Therefore y + z and
y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 are relatively prime. □
We return to our proof of Sophie Germain’s Theorem.
Since y + z and y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 are relatively
prime divisors of a pth power, they are both pth powers.
Similarly, we can factor (−y) p = x p + z p and (−z) p = x p + y p to give us
(−y) p = (x + z)(x p−1 − x p−2 z + x p−3 z 2 + · · · − xz p−2 + z p−1 )
(−z) p = (x + y)(x p−1 − x p−2 y + x p−3 y 2 + · · · − xy p−2 + y p−1 ), respectively,
where each of these factors is a pth power.
Then, there must be non-zero integers a, b, c, α, β, and γ, such that
y + z = a p y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 = α p and x = −aα
x + z = b p x p−1 − x p−2 z + x p−3 z 2 + · · · − xz p−2 + z p−1 = β p and y = −bβ
x + y = c p x p−1 − x p−2 y + x p−3 y 2 + · · · − xy p−2 + y p−1 = γ p and z = −cγ