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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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If it helps, you can reduce the polynomial modulo $k$ and $n$: $$F(k,n)\equiv n^2 \pmod k$$ $$F(k,n) \equiv (k^2+k)^2 \pmod n.$$ You can also reduce it modulo some other things, like $k^2+k$... – anon Jul 17 '11 at 3:28
Two of the terms have $(k+1)^2$ so the third must as well. Where does that come from? – Ross Millikan Jul 17 '11 at 4:07
@Ross, why must the third term have $(k+1)^2$? What if $k+1$ is a square? – Gerry Myerson Jul 17 '11 at 6:00

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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+1, this explains why things just were not working! – Eric Naslund Jul 17 '11 at 15:44
Actually, this seems quite correct. The counterexample given above is bogus, why do 9 people give it a positive rating without even checking? – user1111929 Dec 30 '11 at 16:22
For $k=3,n=729$, the result is not a perfect square, but rather 74674187856 = 2^4 * 3^2 * 331 * 487 * 3217. – r.e.s. Dec 30 '11 at 18:06
@r.e.s.: 729 is a typo for 792. k=3,n=792 gives 95725884816, which is 309396^2. – DSM Dec 30 '11 at 19:07
I fixed the typo discovered by DSM (thanks to user1111929's answer). A verification with wolframalpha is here – t.b. Dec 31 '11 at 4:16

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 $$F(n,n)=n^{6}\left(n+1\right)^{2}.$$

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