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 $Q(x)\ne 0$ and $\mathrm{deg}Q(x)=k<m$. 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. $\square$