1
$\begingroup$

Let $\operatorname{gpf}(n)$ be the greatest primefactor of n. By experimenting I found the following conjecture.

Given $m,n\in\mathbb N_{>1}$. Then $m+n=\operatorname{gpf}(m)\cdot\operatorname{gpf}(n)$ implies that $m+n$ is a perfect square.

The conjecture is verified for $m,n<5000$. I have no idea how to prove this - if it is thrue, that is. I am thankful for proves, hints and counter-examples.

$\endgroup$
  • 7
    $\begingroup$ If the equality holds, then $\operatorname{gpf}(m) \mid n$ and $\operatorname{gpf}(n) \mid m$. Thus $\operatorname{gpf}(m) = \operatorname{gpf}(n)$. $\endgroup$ – Daniel Fischer Sep 23 at 11:57

Your Answer

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

Browse other questions tagged or ask your own question.