69th Belarusian Mathematical Olympiad

1. It is straightforward to verify the following Lemma. Let $x^2+bx+c$ be a trinomial with integer coefficients which is not divisible by $3$ for any integer $x$. Then this trinomial is congruent modulo $3$ to one of the next three trinomials: $x^2+1$, $x^2+x-1$ and $x^2-x-1$. Denote $x^2+1=p_1(x)$, $x^2+x-1=p_2(x)$ and $x^2-x-1=p_3(x)$. The straightforward verification shows that $x^6+x^4+x^2+1\equiv p_1(x)\cdot p_2(x)\cdot p_3(x)(\mathrm{mod}3)$. Therefore, we can change the polynomial $x^6+x^4+x^2+1$ in the condition by the polynomial $p_1(x)p_2(x)p_3(x)$.

Let $p(x)=a{x}^2+bx+c$. We distinguish two cases.

Case 1: $p(a)=0$. Then there exists $a_0$ such that $a{a}_0\equiv 1(\mathrm{mod}3)$. Applying the lemma to $a_{0}p(x)$ we get $a_{0}p(x)\equiv p_i(x)(\mathrm{mod}3)$, $1\le i\le 3$. That is, there exists $h_0(x)$ such that $a_{0}p(x)=p_i(x)-3h_0(x)$. Multiply this equality by $p_{i+1}(x)p_{i+2}(x)$ (where $3+1=2+2=1$ and $3+2=2$) to obtain $a_{0}p(x)p_{i+1}(x)p_{i+2}(x)=p_1(x)p_2(x)p_3(x)-3h_0(x)p_{i+1}(x)p_{i+2}(x)$. It remains to set $f(x)=a_0{p}_{i+1}(x)p_{i+2}(x)$ and $h(x)=h_0(x)p_{i+1}(x)p_{i+2}(x)$.

Case 2: $p|a$. In this case $bx+c$ is not divisible by $3$ for all integer $x$, whence $3|b$ and $3|c$. Thus there exists integer $c_0$ such that $c{c}_0\equiv 1(\mathrm{mod}3)$. So we can set $h(x)\equiv 0$ and $f(x)=c_0(x^6+x^4+x^2+1)$.