Determine all polynomials $P(x)$ with real coefficients such that

$$
(x+1) P(x-1)-(x-1) P(x)
$$

is a constant polynomial.
The answer is $P(x)$ being any constant polynomial and $P(x) \equiv k x^{2}+k x+c$ for any (nonzero) constant $k$ and constant $c$.

Let $\Lambda$ be the expression $(x+1) P(x-1)-(x-1) P(x)$, i.e. the expression in the problem statement.

Substituting $x=-1$ into $\Lambda$ yields $2 P(-1)$ and substituting $x=1$ into $\Lambda$ yields $2 P(0)$. Since $(x+1) P(x-1)-(x-1) P(x)$ is a constant polynomial, $2 P(-1)=2 P(0)$. Hence, $P(-1)=P(0)$.

Let $c=P(-1)=P(0)$ and $Q(x)=P(x)-c$. Then $Q(-1)=Q(0)=0$. Hence, $0,-1$ are roots of $Q(x)$. Consequently, $Q(x)=x(x+1) R(x)$ for some polynomial $R$. Then $P(x)-c=x(x+1) R(x)$, or equivalently, $P(x)=x(x+1) R(x)+c$.

Substituting this into $\Lambda$ yields

$$
(x+1)((x-1) x R(x-1)+c)-(x-1)(x(x+1) R(x)+c)
$$

This is a constant polynomial and simplifies to

$$
x(x-1)(x+1)(R(x-1)-R(x))+2 c .
$$

## Sun $\mathbb{E}^{\circ}$

## Life Financial

Since this expression is a constant, so is $x(x-1)(x+1)(R(x-1)-R(x))$. Therefore, $R(x-1)-R(x)=0$ as a polynomial. Therefore, $R(x)=R(x-1)$ for all $x \in \mathbb{R}$. Then $R(x)$ is a polynomial that takes on certain values for infinitely many values of $x$. Let $k$ be such a value. Then $R(x)-k$ has infinitely many roots, which can occur if and only if $R(x)-k=0$. Therefore, $R(x)$ is identical to a constant $k$. Hence, $Q(x)=k x(x+1)$ for some constant $k$. Therefore, $P(x)=k x(x+1)+c=k x^{2}+k x+c$.

Finally, we verify that all such $P(x)=k x(x+1)+c$ work. Substituting this into $\Lambda$ yields

$$
\begin{aligned}
& (x+1)(k x(x-1)+c)-(x-1)(k x(x+1)+c) \\
= & k x(x+1)(x-1)+c(x+1)-k x(x+1)(x-1)-c(x-1)=2 c .
\end{aligned}
$$

Hence, $P(x)=k x(x+1)+c=k x^{2}+k x+c$ is a solution to the given equation for any constant $k$. Note that this solution also holds for $k=0$. Hence, constant polynomials are also solutions to this equation.