Saudi Arabian Mathematical Competitions

Let $P \in \mathbb{R}[x]$ satisfy the given condition.

First, we notice that if $b=0$ then $x P(x-a) = x P(x) \Rightarrow P(x-a) = P(x)$ (for all $x \neq 0$), hence $P(x) \equiv$ a constant (since $a \neq 0$), and we can recheck that any constant polynomial satisfies the given condition.

Now, let $b \neq 0$, and put $\frac{b}{a} = k$. We need only consider 2 cases.

1. $k \in \mathbb{N}$ : Substituting $x=0$ into the given condition, we have
$$-b P(0) = 0 \Rightarrow P(0) = 0.$$
Substituting $x=a, 2a, 3a, \ldots, k a$ (in succession) into the given condition, we also have
$$\begin{aligned} & a P(0) = (a-b) P(a) \Rightarrow P(a) = 0 \\ & 2a P(a) = (2a-b) P(2a) \Rightarrow P(2a) = 0 \\ & 3a P(2a) = (3a-b) P(3a) \Rightarrow P(3a) = 0 \\ & \ldots \\ & (k-1)a P((k-2)a) = ((k-1)a-b) P((k-1)a) \Rightarrow P((k-1)a) = 0. \end{aligned}$$
Thus, $0, a, 2a, \ldots, (k-1)a$ are roots of $P(x) = 0$; therefore, by Bezout's theorem, $P(x)$ can be written in the form
$$P(x) = x(x-a)(x-2a) \cdots (x-(k-1)a) Q(x)$$
for some $Q \in \mathbb{R}[x]$.

The given condition is then equivalent to:
$$\begin{aligned} & x(x-a) \cdots (x-ka) Q(x-a) = x(x-a) \cdots (x-ka) Q(x) \\ & \Leftrightarrow Q(x-a) = Q(x) \quad \forall x \notin \{0, a, 2a, \ldots, ka\} \\ & \Leftrightarrow Q(x) \equiv c \\ & \Leftrightarrow P(x) = c x(x-a)(x-2a) \cdots (x-(k-1)a) \quad \forall x \in \mathbb{R} \end{aligned}$$
for any real constant $c$.

2. $k \notin \mathbb{N}$ : In the same way, it is easy to check that all numbers $ma$, with $m \in \mathbb{N}$, are roots of $P(x) = 0$; which implies that $P(x) \equiv 0$; and we can re-check that the zero polynomial satisfies the given condition.

In summary:
- If $b=0$, then $P(x) \equiv c$.
- If $\frac{b}{a} = k \in \mathbb{N}$, then $P(x) \equiv c x(x-a)(x-2a) \cdots (x-(k-1)a)$.
- If $\frac{b}{a} = k \notin \mathbb{N}$, then $P(x) \equiv 0$.