Quite standard. Plugging $x=n$ gives that $n$ is a root of $P$. Let $P(x)=(x-n)^tQ(x)$ for maximal $t$. Now the equation rewrites as $(x+n+1)^tQ(x^2+x-n^2)=(x-n)^tQ(x)^2+Q(x)$, so plugging $x=n$ again gives $(2n+1)^tQ(n)=Q(n)$, hence $n$ is root of $Q$ as well and thus $P \equiv 0$.

Very nice!

trinhquockhanh wrote: What book is it?

It's my solution, I compiled it in a PDF for better storage.