SAUDI ARABIAN MATHEMATICAL COMPETITIONS

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

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

Now, let $b\ne 0$, 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 $$-bP(0)=0\Rightarrow P(0)=0.$$ Substituting $x=a,2a,3a,\dots ,ka$ (in succession) into the given condition, we also have $$\begin{array}{rl}& aP(0)=(a-b)P(a)\Rightarrow P(a)=0\\ & 2aP(a)=(2a-b)P(2a)\Rightarrow P(2a)=0\\ & 3aP(2a)=(3a-b)P(3a)\Rightarrow P(3a)=0\\ & \dots \\ & (k-1)aP((k-2)a)=((k-1)a-b)P((k-1)a)\Rightarrow P((k-1)a)=0.\end{array}$$ Thus, $0,a,2a,\dots ,(k-1)a$ 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: $$\begin{array}{rl}& x(x-a)\cdots (x-ka)Q(x-a)=x(x-a)\cdots (x-ka)Q(x)\\ & \iff Q(x-a)=Q(x)\forall x\notin \{0,a,2a,\dots ,ka\}\\ & \iff Q(x)\equiv c\\ & \iff P(x)=cx(x-a)(x-2a)\cdots (x-(k-1)a)\forall x\in \mathbb{R}\end{array}$$ 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 $P(x)\equiv cx(x-a)(x-2a)\cdots (x-(k-1)a)$. - If $\frac{b}{a}=k\notin \mathbb{N}$, then $P(x)\equiv 0$.