National Math Olympiad – Final Round, 2015

Let $p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_0$, where $a_n \neq 0$. The leading term of the polynomial $p(p(x))$ equals $a_n(a_nx^n)^n = a_n^{n+1}x^{n^2}$, and the leading term of the polynomial $p(x)^3$ equals $(a_nx^n)^3 = a_n^3x^{3n}$, they are both of odd degree. Since the polynomial $p(x)^3 - p(p(x))$ has only non-negative values it must be of even degree, hence the leading terms must cancel each other out. Therefore $a_n^{n+1}x^{n^2} = a_n^3x^{3n}$, i.e. $n^2 = 3n$ and $a_n^{n+1} = a_n^3$. It follows $n=3$ and $a_n=1$ since $a_n \neq 0$, i.e. the polynomial $p$ is of degree 3 with a leading coefficient 1. Since the coefficient of the polynomial $p$ at $x^2$ is equal to the polynomial $p(x) = x^3+ax+b$. We use this in the given inequality and simplifying it to get $ax^3+a^2x+ab+b \le 0$ for all $x \in \mathbb{R}$. This is possible only if $a=0$, since the polynomial of odd degree always has positive values at some points. The inequality thus simplifies to $b \le 0$. The polynomials that satisfy the conditions of the problem are exactly those of the form $p(x) = x^3+b$, where $b \le 0$.