IMO Team Selection Test 3

Write $\gamma_i$ with $1 \le i \le n^3$ for the distinct roots of $Q(x)$. For these $\gamma_i$, we have that $P(P(P(\gamma_i))) - P(\gamma_i) = 0$, so $P(\gamma_i)$ is a zero of the polynomial $P(P(x)) - x$. The degree of $P(P(x)) - x$ is $n^2$, so this polynomial has at most $n^2$ distinct roots. Denote these by $\beta_i$. For each root $\beta_i$, the equation $P(x) = \beta_i$ has at most $n$ solutions. Therefore in total we have at most $n^2 \cdot n$ numbers $\gamma_i$ such that $P(\gamma_i)$ is a root of $P(P(x)) - x$. As by assumption this maximum is attained, there are exactly $n^2$ roots $\beta_i$ of $P(P(x)) - x$ and for each $\beta_i$ there are exactly $n$ solutions of $P(x) = \beta_i$.

We write $P(x) = a x^n - b x^{n-1} + P'(x)$ with $\deg P'(x) \le n-2$. Note that for every $\beta_i$, we get a subset $\delta_1, \delta_2, \dots, \delta_n$ of the roots $\gamma_1, \dots, \gamma_{n^3}$ such that $P(x) - \beta_i = a(x - \delta_1)(x - \delta_2) \cdots (x - \delta_n)$. So $a x^n - b x^{n-1} + P'(x) - \beta_i = a(x - \delta_1)(x - \delta_2) \cdots (x - \delta_n)$. Since $n \ge 2$, it follows by Vi\'eta's formulas that the sum $\delta_1 + \delta_2 + \dots + \delta_n$ of the roots of $P(x) - \beta_i$ equals $b/a$. We conclude that we have $n^2$ groups of $n$ roots of $Q(x)$ which all have the same arithmetic mean $\frac{b}{n a}$.

$\square$