Sophie Germain: mathématicienne extraordinaire - Scripps College

Sophie Germain’s Theorems 60
For Sophie Germain primes p, we know q = 2p + 1 so q − 1 = 2p. In
particular,
n q−1 ≡ n 2p (mod q)
≡ (n p ) 2 (mod q)
≡
1 (mod q).
We can then write (n p ) 2 − 1 ≡ 0 (mod q) which we factor to get
(n p − 1)(n p + 1) ≡ 0 (mod q). We now see that n p ≡ ±1 (mod q) for all integers
n such that n ≢ 0 (mod q).
We need to show that p ≢ ±1 (mod q)
Claim 5. 2. For a Sophie Germain prime pair p and q = 2p + 1, we know that
p ≢ ±1 (mod q).
Proof of Claim 5.2.
Assume p ≡ ±1 (mod q). Then p ± 1 ≡ 0 (mod q), so
q divides p ± 1. But we defined p and q such that q = 2p + 1 = p + p + 1, so
q > p + 1 > p − 1. Thus q does not divide p ± 1 and p ≢ ±1 (mod q). □
We return to our proof of the Sophie Germain Primes Theorem. We
know n p ≡ ±1 (mod q) ≢ p (mod q) for all integers n ≢ 0 (mod q).
Now suppose n ≡ 0 (mod q). Then n p ≡ 0 (mod q) ≢ p (mod q).
Thus condition 1 holds for all integers n if p is a Sophie Germain Prime.
Let us consider the integer solutions x, y, and z. We showed above that