A conjecture about Pythagorean triples I noticed for the integer solutions of $a^2 + b^2 = c^2$, there don't seem to be cases where both a and b are odd numbers. In fact, I saw this property pop up on a nice question, which required you to prove it. So I have tried proving it, but I have failed so far. A reductio ad absurdum method seems to me the most likely way of proving this. 
So far, I have shown that if both a and b were odd, then $a^2 + b^2 = 4\cdot y^2$ (where y is some integer). This is because an odd squared is still odd and 2 odds make an even, and if $c^2$ is even, then it must have a factor of 4.
However this isn't too great and my brain seems to be rather uncreative today. Any tips on how I should proceed?

The square of an odd number is always a multiple of four plus one: $(2n+1)^2=4(n^2+n)+1$.
So the sum of two odd squares cannot be a multiple of four.
