33rd Hellenic Mathematical Olympiad

Let $Q(x)=x^n+a_{n-1}{x}^{n-1}+\cdots +a_0$. Then the coefficient of the term of maximal degree of the polynomial $Q\left(\frac{(x+1{)}^2}{2}\right)-Q\left(\frac{(x-1{)}^2}{2}\right)$ can be found from the difference ${\left(\frac{(x+1{)}^2}{2}\right)}^n-{\left(\frac{(x-1{)}^2}{2}\right)}^n$. It is $$\frac{2n{x}^{2n-1}}{2^n}+\frac{2n{x}^{2n-2}}{2^n}=\frac{4n}{2^n}{x}^{2n-1}(1).$$ The corresponding coefficient in the left hand side is equal to $2$, and so $\frac{4n}{2^n}=2\iff 2^{n+1}=4n$. Since $2^{n+1}>4n$, for $n\ge 3$, we must have $n=1$ or $n=2$ and then from (1) the degree of $P$ is $2-1=1$ or $2\cdot 2-1=3$.

For $x=0$ in the given relation, we obtain $2P(0)=Q(1/2)-Q(1/2)=0$, and hence $P(0)=0$.

For $n=1$, then $P(x)=ax$ and since $P(1)=1$, we find $$P(x)=x\text{ and }Q(x)=x+a_0,a_0\in \mathbb{R}.$$

For $n=2$, then, if $Q(x)=x^2+bx+c$, we have: $$\begin{array}{rl}2P(x)& =Q\left(\frac{(x+1{)}^2}{2}\right)-Q\left(\frac{(x-1{)}^2}{2}\right)\\ & =\frac{1}{4}((x+1{)}^4-(x-1{)}^4)+\frac{b}{2}((x+1{)}^2-(x-1{)}^2)\\ & =\frac{1}{4}(8x^3+8x)+\frac{b}{2}(4x)\\ & =2x^3+2(1+b)x.\end{array}$$ Hence $P(x)=x^3+(1+b)x$. Since $P(1)=1$, we find $b=-1$, and finally $$P(x)=x^3\text{ and }Q(x)=x^2-x+c,c\in \mathbb{R}.$$