Phorphyrion wrote: Find all pairs of polynomials $p$, $q$ with complex coefficients so that \[p(x)\cdot q(x)=p(q(x)).\] If $P\equiv 0$, equation is true whatever is $Q(x)$ and : Solution $\boxed{\text{S1 : }P(x)=0\text{ and }Q(x)=R(x)\quad\forall x}$, which indeed fits, whatever is $R(x)\in\mathbb C[x]$ If $P\not\equiv 0$ and $Q\equiv 0$ equation is $P(0)=0$ and : Solution $\boxed{\text{S2 : }P(x)=xR(x)\text{ and }Q(x)=0\quad\forall x}$, which indeed fits, whatever is $R(x)\in\mathbb C[x]$ with $R\not\equiv 0$ If $PQ\not\equiv 0$, let $p,q$ degrees of $p(x),q(x)$ : we have $p+q=pq$, which is $(p-1)(q-1)=1$ and so $p=q=0$ or $p=q=2$ 1) $p=q=0$ Solution : $\boxed{\text{S3 : }P(x)=c\text{ and }Q(x)=1\quad\forall x}$, which indeed fits, whatever is $c\in\mathbb C$ with $c\ne 0$ 2) $p=q=2$ $P(x)Q(x)=P(Q(x))$ implies $P(0)=0$ (setting $ x$ to any root of $Q(x)$) $P(x)$ solution implies $cP(x)$ solution, so WLOG $P(x)$ monic. Plugging $P(x)=x^2+bx$ in equation, we get $x^2+bx=Q(x)+b$ And solution $\boxed{\text{S4 : }P(x)=ax^2+abx\text{ and }Q(x)=x^2+bx-b\quad\forall x}$, which indeed fits, whatever are $a,b\in\mathbb C$ with $a\ne 0$

Isn't this just a simple degree calculation, or I might be missing something. And just by analyzing the case of $Q(x)=0$ and $P(x)=0$?