Team Selection Test for IMO 2009

Let $P(x)=1+p{n}^2+{\prod }_{i=1}^{2p-2}Q(x^i)$. For $p=2$, $n=1$ and $Q(x)=2x+1$ work as $P(-1)=0$.

We will show that for odd primes no suitable $n$ and $Q(x)$ exist.

Since all $Q(a^i)$ have the same parity for $1\le i\le 2p-2$ for an integer $a$; if $P(a)=0$, then $p\equiv 3(\mathrm{mod}4)$.

We also have $a^i\equiv a^{i+p-1}(\mathrm{mod}p)$ for $1\le i\le p-1$ for an integer $a$. Therefore, ${\prod }_{i=1}^{2p-2}Q(a^i)\equiv ({\prod }_{i=1}^{p-1}Q(a^i){)}^2(\mathrm{mod}p)$, and $P(a)\equiv 1+({\prod }_{i=1}^{p-1}Q(a^i){)}^2\ne 0(\mathrm{mod}p)$ for $p\equiv 3(\mathrm{mod}4)$.