Saudi Arabia Mathematical Competitions 2012

Let $S$ be the set of all $a$ with $P(a)=a$. For each $a\in S$, let $T_a$ be the set of solutions to $P(P(x))=P(x)=a$ with $x\ne a$. Since $P(P(x))=P(x)$ and $P(x)\ne x$ has an odd number of solutions, the union of all $T_a$ has an odd number of elements. Therefore one of the $T_a$ has an odd number of elements; choose this $a$ to be our $c$ and let $T_a=\{b_1,b_2,\dots ,b_k\}$ where $k$ is odd. We know that $P(x)-a$ has an even number of distinct real roots: $a,b_1,b_2,\dots ,b_k$. Because $P(x)-a$ has odd degree and its nonreal roots come in conjugate pairs, the combined multiplicity of its real roots must be odd. Then if neither $(x-a{)}^2$ nor $(x-b_i{)}^2$ for any $i$ divide $P(x)-a$, the combined multiplicity of its real roots would be even, a contradiction. So either $(x-a{)}^2$ or $(x-b_i{)}^2$ for some $i$ divide $P(x)-a$. This shows that our choice of $c$ works, and the proof is complete.