Sophie Germain: mathématicienne extraordinaire - Scripps College

The First Attempts 35
4.1.2. Without loss of generality, we assume x 0 is even and we write
x 2 0 = 2st, (4.1)
y 2 0 = s 2 − t 2 , (4.2)
z 2 0 = s 2 + t 2 , (4.3)
with s and t relatively prime and of opposite parity, and with s > t > 0.
Equation (4.2) gives us y0 2 + t2 = s 2 , another Pythagorean triple with
integer solutions that can be similarly decomposed. We know that either y 0
or t must be even. Suppose that y 0 is even. Then x 0 and y 0 are both even so
they are not relatively prime. This is a contradiction, so y 0 cannot be even.
So there must be integers a and b such that
t = 2ab
y 0 = a 2 − b 2
s = a 2 + b 2
where a and b are relatively prime and of opposite parity, and with a > b > 0.
Then equation (4.1) can be rewritten as x 2 0 = 4ab(a2 + b 2 ).
Lemma 1. Relatively prime divisors of square are themselves squares. In fact,
relatively prime divisors of an nth power are themselves nth powers.
For a proof of this lemma some particularly good exercises can be found
in [5] or [22]. This proof uses the Fundamental Theorem of Arithmetic, that
every integer greater than 1 has a unique prime factorization. We will deal