# A peculiar conjecture including maximal prime factors

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.

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