SELECTION and TRAINING SESSION

(Solution by A. Asanau, D. Voynov.) Replace $x$ by $1-x$ in $$p(x^2)=p(x)q(1-x)+p(1-x)q(x),(1)$$ then we have $p((x-1{)}^2)=p(1-x)q(x)+p(x)q(1-x)=p(x^2)$. It follows that the polynomial $p(x^2)$ is periodic, i.e., $p(x)$ is a constant polynomial, $p(x)=a$ for some $a\in \mathbb{R}$. If $a=0$, then (1) is valid for any arbitrary polynomial $q(x)$. If $a\ne 0$, then (1) becomes $q(1-x)+q(x)=1$ for all $x$. Replacing $x$ by $x+0.5$ we can rewrite the last equality as $$q(0.5-x)-0.5+q(0.5+x)-0.5=0.(2)$$ Let $h(x)=q(0.5+x)-0.5$. Then (2) shows that $h(-x)=-h(x)$, so we may write $h(x)=xr(x^2)$ for some polynomial $r(x)$. Hence $q(0.5+x)=xr(x^2)+0.5$. Replace $x$ by $x-0.5$ here, we obtain $q(x)=(x-0.5)r((x-0.5{)}^2)+0.5$ which obviously satisfies the condition.