2. find all polynomials $P$ with real coefficients so that the following equation holds for all $x \in \mathbb{R}$:

$$
(x-2) P(x+2)+(x+2) P(x-2)=2 x P(x)
$$
Solution (Viviane):

Substituting $x=0$ and $x=2$ gives $P(-2)=P(0)=P(2)$, so we can write $P(x)$ as $Q(x)(x-2)(x)(x+2)+a$. Substituting into the original equation gives

$$
(x-4) Q(x-2)+(x+4) Q(x+2)=2 x Q(x)
$$

If we find $s<4$ so that $Q(s)=Q(s+2)=k$, then the following follows

$$
Q(s-2)=\frac{2 s k-(s+4) k}{s-4}=k
$$

Inductively, $Q(s-2 n)=k$ then follows, so $Q$ is constant. Inserting $x=-4$ into the above equation yields $Q(-6)=Q(-4)$. Since $-6<4$, $Q$ is therefore constant. Substituting provides that

$$
P(x)=b(x-2)(x)(x+2)+a
$$

are actually solutions.

Marking Scheme:

- $P(x)=Q(x) x(x-2)(x+2)+a 2$ points
- $\mathrm{Q}$ constant or somehow limit the degree of $P$ sensibly: 5 points in total
- finish: 2 points