Belarusian Mathematical Olympiad

Answer: $P(x)=ax(x-1)(x+1)$, where $a\in \mathbb{R}$.

Set $x=1$ and $x=-1$ in the initial identity $$(x-1)P(x+1)-(x+1)P(x-1)=4P(x).(1)$$ Thus we obtain $-2P(0)=4P(1)$ and $-2P(0)=4P(-1)$ respectively. Setting $x=0$ in (1), we obtain $-P(1)-P(-1)=4P(0)$, so, taking into account two previous equalities, we have $P(-1)=P(0)=P(1)=0$.

Therefore $0$, $1$, and $-1$ are zeroes of the polynomial $P(x)$, hence $P(x)=x(x-1)(x+1)Q(x)$, where $Q(x)$ is some polynomial. Setting this representation of $P(x)$ in (1), we obtain $$\begin{gathered} (x-1)(x+1)x(x+2)Q(x+1)-(x+1)(x-1)(x-2)xQ(x-1)= \\ =4x(x-1)(x+1)Q(x), \end{gathered}$$ whence $$(x+2)Q(x+1)-(x-2)Q(x-1)=4Q(x)\text{for all }x\in \mathbb{R}.(2)$$ Set $x=2$ in (2), then we obtain $Q(3)=Q(2)$. By induction on $k$, show that $Q(k)=Q(2)$ for all positive integers $k\ge 2$. Indeed, let $Q(k)=Q(2)$ for all $k=2,3,\dots ,m$, where $m\ge 3$. Then setting $x=m$ in (2), we obtain $(m+2)Q(m+1)-(m-2)Q(m-1)=4Q(m)$, so, taking into account our assumption, we obtain $(m+2)Q(m+1)=(m+2)Q(2)$, i.e. $Q(m+1)=Q(2)$. It follows that $Q(k)=Q(2)$ for all positive integers $k\ge 2$.

Since the polynomial $Q(x)-Q(2)$ has zero value at infinitely many points, we conclude that this polynomial is equal to zero identically, i.e. $Q(x)=Q(2)=a$ for some real number $a$.

Therefore, the polynomial satisfying (1) needs to have the form $P(x)=ax(x-1)(x+1)$. It is easy to verify that any polynomial of such form (for any real $a$) satisfies (1).