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?

  • $\begingroup$ @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. $\endgroup$
    – user236182
    Sep 4, 2015 at 16:39
  • $\begingroup$ @S.Panja-1729 It's enough to prove they're coprime to prove they're perfect squares. $\endgroup$
    – user236182
    Sep 4, 2015 at 16:40
  • $\begingroup$ Ok...He should add that statement , but he should write the actual question.. $\endgroup$
    – Empty
    Sep 4, 2015 at 16:42
  • $\begingroup$ @S.Panja-1729 He made his question clear at the end. $\endgroup$
    – user236182
    Sep 4, 2015 at 16:43
  • $\begingroup$ see math.stackexchange.com/questions/680972/… $\endgroup$
    – Will Jagy
    Sep 4, 2015 at 17:25

1 Answer 1


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.


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