Sophie Germain: mathématicienne extraordinaire - Scripps College

The First Proof 49
are positive integers. Then
P 1 = A 2 + 5B 2
P1 2 = (A 2 + 5B 2 ) 2
= A 4 + 10A 2 5B 2 + 25B 4
Q 1 = 2B 2
Q 2 1 = (2B 2 )
= 4B 4
So P1 2 < P 0 = A(A 4 + 50A 2 B 2 + 125B 4 ) and we know 0 < P 1 < P 0 . Similarly,
Q 2 1 < Q 0 = 5B(A 4 + 10A 2 B 2 + 5B 4 ) so 0 < Q 1 < Q 0 .
This argument can be repeated indefinitely. That is, we can always find
positive integers P n and Q n which are smaller than P n−1 and Q n−1 , respectively.
By the principle of infinite descent, there can be no initial integers
P 0 and Q 0 . Thus there are no non-trivial integer solutions to the equation
X 5 + Y 5 + Z 5 = 0 where 10|Z. This concludes subcase A. of Case 2.
Proof of Case 2. B.
We are not going to prove Case 2. B. since the techniques
are very similar to those used in Case 2. A. A complete proof can be
seen in [6].
We would begin by supposing that there is a non-trivial integer solution
x, y, and z to the equation X 5 + Y 5 + Z 5 = 0. Without loss of generality, we
would assume z is divisible by 5, so x and y have opposite parity.