31st Hellenic Mathematical Olympiad

The given equation is written as $$(x-2)(x-4)P(x)=x(x+2)P(x-2),\text{ for all }x\in \mathbb{R},(1)$$ and so for $x=0,-2$ and $4$ we obtain: $P(0)=P(-2)=P(2)=0$. Therefore the polynomial $P(x)$ takes the form: $$P(x)=x(x+2)(x-2)Q(x),(2)$$ where $Q(x)$ is a polynomial with real coefficients. From (2), equation (1) takes the form: $$(x-2)(x-4{)}^{2}x(x+2)Q(x)=x(x+2)(x-2)x(x-4)Q(x-2),(3)$$ for all $x\in \mathbb{R}$. Equivalently, for all $x\in \mathbb{R}$ we have $$x(x+2)(x-2)(x-4)[(x-2)Q(x)-xQ(x-2)]=0,(4)$$ Since the polynomial $x(x-2)(x+2)(x-4)$ is not the zero polynomial, from relation (4) we get: $$(x-2)Q(x)-xQ(x-2)=0,\text{ for all }x\in \mathbb{R}.(5)$$ From (5) for $x=0$ we get $Q(0)=0$, and hence $Q(x)=xR(x)$, where $R(x)$ is a polynomial with real coefficients. From (5), relation (4) becomes: $$\begin{array}{rl}& (x-2)xR(x)-x(x-2)R(x-2)=0\\ \iff & (x-2)[R(x)-R(x-2)]=0,\end{array}$$ From which, since $x(x-2)\ne 0$ for all $x\ne 0$, we have that: $$R(x)=R(x-2),\text{ for all }x\in \mathbb{R}.$$ Last relation gives $R(x)=R(x+2k)$, $k\in \mathbb{Z}$, for all $x\in \mathbb{R}$, and hence the polynomial $R(x)$ takes the same value for infinite values of $x\in \mathbb{R}$, for example $c=R(0)=R(2k)$, $k\in \mathbb{Z}$. Hence $R(x)=c$, for all $x\in \mathbb{R}$, and hence $$P(x)=x(x+2)(x-2)Q(x)=x(x+2)(x-2)xR(x)=c{x}^2(x^2-4).$$