6-th Czech-Slovak Match

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