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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.

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  • $\begingroup$ @YuDing, the question title says they are distinct parities. $\endgroup$ Apr 21, 2019 at 1:10
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    $\begingroup$ 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 $2$. $\endgroup$
    – Yuval
    Apr 21, 2019 at 1:14
  • $\begingroup$ ^^^ 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 $t$. $\ \ $ $\endgroup$ Apr 21, 2019 at 3:35

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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 $d=1$.

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  • $\begingroup$ Why is $d$ odd? Thanks. $\endgroup$ Apr 21, 2019 at 1:13
  • $\begingroup$ Since $s,t$ have different parities, $s^2+t^2$ is odd. $\endgroup$
    – lulu
    Apr 21, 2019 at 1:14
  • $\begingroup$ Ok, thanks for your time, I understand now. $\endgroup$ Apr 21, 2019 at 1:16

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