Let $m$ and $n$ be positive integers. Show that $$4mn-m-n$$ can never be a square.

In my attempt I started by assuming for the sake of contradiction that $$4mn-m-n=k^2$$ for some $k \in \mathbb{N_0^+}$. Then I considered $k^2 \pmod3$ (I couldn't find a way for $\pmod4$ or $\pmod8$ to help) $$k^2 \equiv 0,1 \pmod3$$

This yield three possibilities for $m$ and $n$ $\pmod3$ either $$m,n \equiv 0 \pmod3$$ $$m \equiv 0,\;\;\;n \equiv 2 \pmod3 \; \; \;(WLOG)$$ $$m,n \equiv 2 \pmod3$$ Case $1$: $m,n \equiv 0 \pmod3$

Let $m=3m'$ and $n=3n'$ then $k^2=36m'n'-3m'-3n'$ Hence $3|k^2 \implies 3|k$ Let $k=3k'$ then $$9k'^2=36m'n'-3m'-3n'$$ $$\implies3k'^2=12m'n'-m'-n'$$ Here I'm stuck and don't know if anything I've done is helpful and don't feel inclined to pursue the other cases given what happened here. I was hoping to reduce this to some sort of infinite descent argument - that didn't happen. I now think that my approach was probably completely wrong. Any help would be greatly appreciated.

  • 2
    $\begingroup$ The tag says contest-math. Is it allowed to post these questions based on competition rules? $\endgroup$
    – Amzoti
    May 23, 2013 at 2:16
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    $\begingroup$ It is contest math from the 1980's and these questions are in the public domain. So yes it is. $\endgroup$
    – John Marty
    May 23, 2013 at 2:19
  • $\begingroup$ @Amzoti The tag contest-math is generally used for problems of the same type used in contests, rather than questions in ongoing contests. $\endgroup$ May 23, 2013 at 6:05
  • $\begingroup$ @Amzoti That's why I said "generally". I just wanted to make sure you were aware that the tag does not indicate it was part of an ongoing competition. $\endgroup$ May 23, 2013 at 6:07

1 Answer 1


Assume on the contrary that $4mn-m-n=k^2$. Then $(4m-1)(4n-1)=(2k)^2+1$. Since $4m-1 \equiv 3 \pmod{4}$ and $4m-1>0$, $4m-1$ has a prime factor $p \equiv 3 \pmod{4}$. Then $p \mid (2k)^2+1$, which gives a contradiction since $(\frac{-1}{p})=(-1)^{\frac{p-1}{2}}=-1$.


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