# If $s,t \in \Bbb N_{>0}$ are co-prime of distinct parities, then $\gcd(s^2-t^2, 2st, s^2+t^2)=1$

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.

• @YuDing, the question title says they are distinct parities. – IntegrateThis Apr 21 at 1:10
• Note $p$ cannot divide $s^2$ and $t^2$ at the same time since $s, t$ coprime. So it must be $p|2$. And if $s, t$ are of different parity, $p$ cannot be $2$. – Yu Ding Apr 21 at 1:14
• ^^^ i.e. by prime $\,p\mid s^2-t^2$ we infer $\,p\mid s\iff p\mid t.\,$ But $\,s,t\,$ are coprime so $\,p\,$ divides neither $s$ nor $t$. $\ \$ – Bill Dubuque Apr 21 at 3:35

Let $$d$$ be the gcd you are after.
We have: $$d\,|\,\gcd(s^2-t^2,s^2+t^2)\implies d\,|\,2s^2\quad \text &\quad d\,|\,2t^2$$ Since $$d$$ is clearly odd (since, e.g., $$s^2+t^2$$ is odd) we see that $$d\,|\,\gcd(s^2,t^2)=1$$. Hence $$d=1$$.
• Why is $d$ odd? Thanks. – IntegrateThis Apr 21 at 1:13
• Since $s,t$ have different parities, $s^2+t^2$ is odd. – lulu Apr 21 at 1:14