Hellenic Mathematical Olympiad

Let $Q(x) = x^n + a_{n-1}x^{n-1} + \dots + a_0$. 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
$$\frac{2n x^{2n-1}}{2^n} + \frac{2n x^{2n-2}}{2^n} = \frac{4n}{2^n} x^{2n-1} \quad (1).$$
The corresponding coefficient in the left hand side is equal to $2$, and so $\frac{4n}{2^n} = 2 \Leftrightarrow 2^{n+1} = 4n$. 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 $P(x) = a x$ and since $P(1) = 1$, we find
$$P(x) = x \text{ and } Q(x) = x + a_0, \quad a_0 \in \mathbb{R}.$$

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