37th Hellenic Mathematical Olympiad 2020

Let: $\mathrm{deg}P(x)=m\ge 1$, $\mathrm{deg}Q(x)=n\ge 1$. By taking the degrees of the two members of the given equation we get the equality: $$\begin{array}{rl}3mn=1+m+3n& \iff (m-1)(3n-1)=2\\ & \iff m-1=1,3n-1=2\text{or }m-1=2,3n-1=1(\text{impossible in }{\mathbb{N}}^{\ast })\\ & \iff m=2,n=1.\end{array}$$ Let $P(x)=a{x}^2+bx+c$, $a,b,c\in \mathbb{R}$, $a\ne 0$, $Q(x)=dx+e$, $d,e\in \mathbb{R}$, $d\ne 0$. Then, from the given relation we have: $$a((Q(x){)}^3{)}^2+b(Q(x){)}^3+c=xP(x)(Q(x){)}^3,(1)$$ From which we conclude that $Q(x){|}_{c=0}\Rightarrow c=0$. Then from (1) we have: $$\begin{array}{rl}& a((Q(x){)}^3{)}^2+b(Q(x){)}^3-xP(x)(Q(x){)}^3=0\\ \iff & (Q(x){)}^3\left(a(Q(x){)}^3+b-xP(x)\right)\\ \iff & \underset{Q(x)\ne 0}{a(Q(x){)}^3}+b-xP(x)=0\\ & xP(x)=a(Q(x){)}^3+b\end{array}(2)$$ Then from relation (2) we have: $$\begin{array}{rl}& x(a{x}^2+bx)=a(dx+e{)}^3+b\\ \iff & a{x}^3+b{x}^2=a{d}^3{x}^3+3a{d}^{2}e{x}^2+3ad{e}^{2}x+a{e}^3+b\\ \iff & a=a{d}^3,b=3a{d}^{2}e,3ad{e}^2=0,a{e}^3+b=0\\ & \underset{a,d\ne 0}{\iff }d=1,e=0,b=0,a\in \mathbb{R}.\end{array}$$ Hence: $P(x)=a{x}^2$, $Q(x)=x$, $a\in \mathbb{R}$ and it is easy to verify that they satisfy our problem.