0
$\begingroup$

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.

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

1 Answer 1

Reset to default
2
$\begingroup$

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 $d=1$.

$\endgroup$
3
  • $\begingroup$ Why is $d$ odd? Thanks. $\endgroup$
    – IntegrateThis
    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. $\endgroup$
    – IntegrateThis
    Apr 21, 2019 at 1:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.