Saudi Arabian Mathematical Competitions

Denote $m$ as degree of $P(x)$ then by comparing the leading coefficients $a$ of two sides, we get $a^{n} = a$. We consider two cases based on the parity of $n$.

1. If $n$ is even then $a = 1$. Put $P(x) = x^{m} + Q(x)$ with $Q \in \mathbb{R}[x]$ and if $Q(x) \equiv 0$, we obtain $P(x) = x^{m}$ as a solution. Otherwise, suppose that $Q(x) \neq 0$ and $\operatorname{deg} Q(x) = k < m$. So
$$\left(x^{m} + Q(x)\right)\left(x^{2m} + Q\left(x^{2}\right)\right) \cdots \left(x^{mn} + Q\left(x^{n}\right)\right) - x^{\frac{mn(n+1)}{2}} = Q\left(x^{\frac{n(n+1)}{2}}\right).$$
The degree of LHS is $\frac{mn(n+1)}{2} - (m - k)$, while the degree of RHS is $\frac{kn(n+1)}{2}$. Note that
$$\frac{mn(n+1)}{2} - (m - k) - \frac{kn(n+1)}{2} = \frac{(m - k)(n + 2)(n - 1)}{2} > 0$$
which implies that the degree of both sides cannot be equal. Thus $Q(x) \equiv 0$.

2. If $n$ is odd then $a = \pm 1$ and the case $a = 1$ can be solved similarly as above. For the case $a = -1$, put $P(x) = -\left(x^{m} + Q(x)\right)$ then process similarly, we get $P(x) = -x^{m}$ as another solution.

Hence, $P(x) = x^{m}$ for some $m \in \mathbb{Z}^{+}$ when $n$ is even and $P(x) = \pm x^{m}$ when $n$ is odd. $\square$