@KingGeorge Prove that if m^2 | n^2, then m | n

Question

anonymous · Answer

I tried doing k = (n/m)^2 but then I got stuck. Maybe I was approaching it wrong?

kinggeorge · Answer

It seems like that should work. You just have to note that $k$ is a perfect square, and so when you take the square root, you get an integer.

anonymous · Answer

How do you know that it is a perfect square?

kinggeorge · Answer

Alternatively, let $p$ be a prime dividing $m$. Then $p^2|n^2$. Thus, $p|n^2$, so $p$ must divide $n$. That means that $m|n$.

kinggeorge · Answer

And that was my mistake. We don't know offhand that it's a perfect square.

anonymous · Answer

Wait, yes, we do. Since m^2 | n^2, then m <= n, then our expression n/m must not be a fraction.

anonymous · Answer

Which means that (n/m)^2 must be a perfect square

anonymous · Answer

Since m^2 | n^2, then m <= n, then our expression n/m must be an integer

kinggeorge · Answer

Nice. So it looks like we have two ways to prove this now :)

anonymous · Answer

You don't see any faults in my logic do you?

kinggeorge · Answer

It looks fine to me. We know that $n^2=m^2\cdot k$ for some integer $k$, so $k=n^2/m^2$. But using properties of squares, we get that $k=(n/m)^2$, so it must be that $n/m$ is an integer $k_0$, which then implies that $m|n$.