6-th Czech-Slovak Match

By the Little Fermat theorem $n^3 \equiv n \pmod{3}$ for every $n \in \mathbb{Z}$. Then also $n^4 \equiv n^2 \pmod{3}$. Since $P(x)$ is a polynomial with all integer coefficients, it follows that
$$P(n^3) \equiv P(n) \pmod{3}, \quad P(n^4) \equiv P(n^2) \pmod{3}$$
Polynomial $Q(x)$ has integer coefficients therefore, from the last two congruences, we get
$$Q(n) \equiv (P(n)P(n^2))^2 + 1 \pmod{3}$$
But we can easily see that for every integer $m$ we have $m^2 + 1 \not\equiv 0 \pmod{3}$. So for every integer number $n$, we have $Q(n) \not\equiv 0 \pmod{3}$, from which it follows that polynomial $Q(x)$ can not have integer roots. The proof is finished. $\square$