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Let $k$ and $n$ be positive integers. Show that $(k+1)^2k^2(n+1)^4-2k(k+1)n(n+1)^2(2kn+k+1)+n^2(k+1)^2$ is a perfect square if and only if $k=n$.

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    @Ross, why must the third term have $(k+1)^2$? What if $k+1$ is a square?2011-07-17

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For $k=3$ and $n=792$, $ (k+1)^2k^2(n+1)^4-2\,k\,(k+1)\,n\,(n+1)^2(2\,k\,n+k+1)+n^2(k+1)^2=309\,396^2. $ This is the only perfect square for $1\le k\le2\,000$, $1\le n\le10\,000$ and $k\ne n$.

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    @DSM Thank you for discovering the typo, ans thanks also to t.b. for fixing it. I have been away from a computer for the last 15 days, so I coould not comment on this until today.2012-01-08
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Since Julian showed the "only if" part is not possible, here is a quick proof for the if part:

Suppose $n=k=x$. Then we have $(x+1)^2x^2(x+1)^4-2x(x+1)x(x+1)^2(2x^2+x+1)+x^2(x+1)^2.$ This then becomes

$(x+1)^{2}x^{2}\left((x+1)^{4}-2(x+1)(2x^{2}+x+1)+1\right).$ The inside term is exactly $x^{4}$ as $2(x+1)(2x^{2}+x+1)=4x^{3}+6x^{2}+4x+2.$ Hence $F(n,n)=n^{6}\left(n+1\right)^{2}.$