Prove $m-n$ and $2m+2n+1$ are coprime

Question

I have a small problem:

Let $m$ and $n$ be integers such that $2m^2+m = 3n^2+n$. Prove that $m-n$ and $2m+2n + 1$ are perfect square.

My work:

We have $$(m-n)(2m+2n+1) = 2(m^2-n^2) + m-n = n^2.$$

So, we need to prove that $m-n$ and $2m+2n+1$ are coprime. But I don't get further. Anyone can give me a hint?

@S.Panja-1729 No, read the whole question. He wants to prove they're coprime, since that is the only thing he needs to finish the problem. — user236182, Sep 04 '15 at 16:39
@S.Panja-1729 It's enough to prove they're coprime to prove they're perfect squares. — user236182, Sep 04 '15 at 16:40
Ok...He should add that statement , but he should write the actual question.. — Empty, Sep 04 '15 at 16:42
see http://math.stackexchange.com/questions/680972/if-m-n-in-mathbb-z-2-satisfies-3m2m-4n2n-then-m-n-is-a-perfect — Will Jagy, Sep 04 '15 at 17:25

score 4 · Accepted Answer · answered Sep 04 '15 at 16:34

4

Assume for contradiction $p\mid \gcd(m-n,2m+2n+1)$ for some prime $p$.

But then $p\mid n^2\iff p\mid n$, and so $p\mid m-n\implies p\mid m$.

However, $p\mid 2m+2n+1$ is then impossible.

answered Sep 04 '15 at 16:34

user236182

1 Answers1