Sophie Germain: mathématicienne extraordinaire - Scripps College

The First Proof 42
We take a detour here to prove that 5 2 2r and 5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 are
both fifth powers.
Claim 4. 1. The integers 5 2 2r and 5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 are relatively prime.
Proof of Claim 4.1.
Recall that q and r are relatively prime and have opposite
parity. Suppose that m is a prime such that m divides 5 2 · 2r and m
divides |5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 . Since m|5 2 · 2r, then either m = 5, m = 2,
or m|r. We know that 5 ∤ q, so 5 ∤ 5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 , so m ≠ 5. Since
q and r are relatively prime, if m|r then m ∤ q so m ∤ 5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 .
Thus it must be that m = 2. So 2|5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 . Since q and r have
opposite parity, 5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 must be odd, so 2 does not divide it.
Thus we have a contradiction. That is, our supposition that m is a prime
common divisor was wrong, and there are no primes which divide 2 · 5 2 r
and 5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 . Thus they are relatively prime. □
We return to our proof of n = 5. We know that the integers 5 2 2r and
5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 are relatively prime, and
z 5 = (5 2 2r)(5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 ). (4.6)
By Lemma 1, we can conclude that 5 2 2r and 5 3 r 4 + 2 · 5 2 r 2 q 2 + q 4 are
both fifth powers.
Let us consider the second factor further. Completing the square gives