By rewriting the original relation we get that: $(x+1)(P(x)-P(x-1))=(x-1)(P(x+1)-P(x))$ Now, if $P(x)-P(x-1)=0$ for some $x \ge 2$ then we can deduce that $P(x+1)-P(x)=0$ for infinitely many values, so $P(x)$ is constant which clearly satisfies the above relation. Now suppose $P(x+1)-P(x)\neq 0$. By telescoping products on the above relation we get that: $P(n)-P(n-1)=cn(n-1)$ for any natural $n>2$ and for some constant $c$, and because $P(x)$ is a polynomial we get that $P(x)=cx(x-1)$ And now, by using telescoping sums on $P(n)-P(n-1)$ we get that $P(x)=ax(x-1)(x+1)+b$ for some constants $a$ and $b$, which clearly satifies the relation.