The number is a perfect square if and only if $k=n$

Question

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 $k=n$.

Answer 1

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 $k\ne n$.

Answer 2

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