Sophie Germain: mathématicienne extraordinaire - Scripps College

The First Proof 45
Returning to the proof of Fermat’s Last Theorem for n = 5, we note
here that we are setting up an infinite descent argument. We have taken
the equation x 5 + y 5 + z 5 = 0 and have assumed that there exist non-trivial
integer solutions. From these solutions, we have found positive integers P 0
and Q 0 . We are going to show that P 0 and Q 0 are positive integers greater
than P 1 and Q 1 , respectively, which in turn are positive integers. That is,
we will demonstrate that P 0 > P 1 > 0 and Q 0 > Q 1 > 0. By the principle
of mathematical induction, we can continue finding P i and Q i such that
P i > P i+1 > · · · > 0 and Q i > Q i+1 > · · · > 0 for all i ∈ Z + . This will show
that no such initial solution could exist, finishing our proof.
Recall that we have P0 2 − 5Q2 0 = (P √ √
0 + Q 0 5)(P0 − Q 0 5).
Claim 4. 2. The integers P 0 = q 2 + 25r 2 and Q 0 = 10r 2 are relatively prime. In
particular, P 0 is not divisible by either 2 or 5.
Proof of Claim 4.2.
Suppose that m|P 0 and m|Q 0 for some prime m. Then
m|q 2 + 5 2 r 2 and m|10r 2 . So m = 2, m = 5, m|q or m|r. Recall that q and r
are relatively prime and have opposite parity and, in particular, 5 ∤ q. Since
q 2 + 5 2 r 2 is odd, thus m ≠ 2. Since 5 ∤ q, we know 5 ∤ q 2 + 5 2 r 2 , so m ≠ 5.
Thus m|q or m|r. But q and r are relatively prime, so if m|q, then m ∤ 10r 2
and if m|r, then m ∤ q 2 + 5 2 r 2 . So our supposition is wrong. Thus there
are no primes which divide P 0 and Q 0 , so they are relatively prime. In
particular, since Q 0 = 10r 2 is divisible by 2 and 5, P 0 is not divisible by
either 2 or 5. □
√
Continuing our proof of n = 5, we want to show that P 0 + Q 0 5 is a
fifth power.