Let $p(z)$ be a given polynomial of degree $3$.
How to find nonconstant entire functions $f,g$ such that $f(f(z)) = p(g(z))$ or prove they do not exist ?
I considered the related equation $f(f(z)) = g^2(z)$.
In particular finding nontrivial solutions , so not $C x^{2n}$.
We get $\sqrt f(f(z)) = g(z)$ , so $f(f(z))$ must have all its roots with even multiplicity. Therefore we rewrite $f(z) = F^2(z)$ and $F(F^2(z)) = g(z)$.
We then arrive at $F^2(z) = F^{-1}(g(z))$ so $g(z) = F(G(z)) = \sqrt f(G(z))$.
Then $f(f(z)) = f(G(z))$. I got confused after this , so instead of continuing this approach I switched to using derivatives :
$D f(f(z)) = D g^2(z)$.
$f'(f(z)) f'(z) = 2 g(z) g'(z)$
From here on I tried to consider arguments like " if $f'(z)$ divides $g'(z)$ " and the alike , but without succes.
I have the impression Im close to a solution here.
edit : As Thomas Andrews point out there is a trivial solution : $f(z)=g(z)=p(z)$. But of course one wonders about the existance of others.
Given any entire function $h$, you can define $f(z)=p(h(z))$ and $g(z)=h(p(h(z)))$ then $f\circ f = (p\circ h)\circ(p\circ h) = p\circ(h\circ p\circ h) = p\circ g$.
That's a general trick that will work for finding solutions to $f\circ f = p\circ g$ in any semigroup (an algebra with one associative binary operation.)
Not sure if there are any other solutions in the specific case of the algebra of entire functions.