Suppose that $s>t$ where $s,t$ are positive integers of distinct parities. I want to show that if $s,t$ are co-prime of distinct parities, then $\gcd(s^2-t^2, 2st, s^2+t^2)=1$
Thoughts: Suppose that $gcd(s,t)=1$ and one is odd, while the other is even, and let $ d = gcd(s^2-t^2, 2st, s^2+t^2)$. If $p$ is a prime dividing $d$, then $p|2s^2, p|2t^2, p|2st$. So that either $p|2$, or $p|st$ (and consequently $p$ divides either $s$ or $t$ or both). This is about as far as I have gotten, hints appreciated.