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 '15 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 '15 at 16:40
  • $\begingroup$ Ok...He should add that statement , but he should write the actual question.. $\endgroup$ – Empty Sep 4 '15 at 16:42
  • $\begingroup$ @S.Panja-1729 He made his question clear at the end. $\endgroup$ – user236182 Sep 4 '15 at 16:43
  • $\begingroup$ see math.stackexchange.com/questions/680972/… $\endgroup$ – Will Jagy Sep 4 '15 at 17:25

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.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.