Sophie Germain: mathématicienne extraordinaire - Scripps College

Sophie Germain’s Theorems 55
Claim 5. 1. For relatively prime integers y and z, the expressions y + z and
y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 are relatively prime.
Proof of Claim 5.1.
Suppose r is a prime which divides both y + z and
y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 . Then
y + z ≡ 0 (mod r)
y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 ≡ 0 (mod r).
Thus, z ≡ −y (mod r). Plugging z = −y into the second equation gives us
y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 ≡ py p−1 (mod r).
So py p−1 ≡ (mod r). Then either p ≡ 0 (mod r) or y p−1 ≡ 0 (mod r).
First suppose y p−1 ≡ 0 (mod r). Then y ≡ 0 (mod r). But z ≡ −y (mod r),
so z ≡ 0 (mod r). This implies that r divides z. This is a contradiction since
y and z are relatively prime. Thus r does not divide y p−1 .
Now suppose p ≡ 0 (mod r). Since p is prime, this means p = r. Then
y + z ≡ 0 (mod p)
y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 ≡ 0 (mod p).