SAUDI ARABIAN MATHEMATICAL COMPETITIONS

In terms of $G(x):=P(x)-\frac{c}{2}$, the given condition can be rewritten as $$(x+1)G(x-1)=(x-1)G(x)\text{ for all }x\in \mathbb{R}.$$ It follows immediately (with $x=\pm 1$) that $G(0)=0$ and $G(-1)=0$. Therefore, $G(x)=x(x+1)Q(x)$ for some $Q\in \mathbb{R}[x]$. Then $$Q(x-1)=Q(x)\text{ for all }x\in \mathbb{R}\setminus \{-1,0,1\}.$$ This is true iff $Q$ is a constant; i.e., $$P(x)=kx(x+1)+\frac{c}{2}\forall x\in \mathbb{R}$$ where $k$ is a real constant.

Remark. We have $(x+2)G(x)=xG(x+1)$ for all $x\in \mathbb{R}$. This is a special case of Problem 2 in the test for Level 4+ (where $a=-1,b=-2$ ).