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 $P(x)=a{x}^n-b{x}^{n-1}+P^{\prime }(x)$ with $\mathrm{deg}{P}^{\prime }(x)\le n-2$. 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 $a{x}^n-b{x}^{n-1}+P^{\prime }(x)-{\beta }_i=a(x-{\delta }_1)(x-{\delta }_2)\cdots (x-{\delta }_n)$. Since $n\ge 2$, it follows by Vi\'eta's formulas that the sum ${\delta }_1+{\delta }_2+\cdots +{\delta }_n$ 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 $\frac{b}{na}$.

$\square$