The entire [non-linear] functions $f$ and $g$ do not have fixed points. Show that $f \circ g$ has infinitely many fixed points.
How do you prove this statement? Or if it is not true as stated, what additional hypothesis do we need to add to make it appropriate to call proving this statement an exercise?
This is from Problems in Mathematical Analysis by Biler and Witkowski, Exercise 7.73 on page 86 (Google Books link here). As stated, the exercise did not require $f$ and $g$ to be non-linear, but we clearly need that. And seeing the statement of a similar conjecture here that was made by Gross, I'm hoping that non-linearity is the only other necessary hypothesis.
$g(z) - z$ is a non-constant entire function with no zeros, so we can write $$ g(z) = z + \exp(G(z))$$ for some non-constant entire function $G$. Similarly, $$ f(z) = z + \exp(F(z))$$ Thus $$ f(g(z)) = g(z) + \exp(F(g(z))) = z + \exp(G(z)) + \exp(F(g(z))) $$ The question now is whether $\exp(G(z)) + \exp(F(g(z)))$ takes the value $0$ infinitely often. This would be the case if for some $z$ we had $G(z) = n \pi i + F(g(z))$ for infinitely many odd integers $n$.
Since $G(z) - F(g(z))$ is an entire function, according to Picard the only way that could fail is if $G(z) - F(g(z))$ is constant. If that constant is $c$, that would say $$f(g(z)) = z + \exp(F(g(z))+c) + \exp(F(g(z)))$$ But then $z = f(g(z)) - \exp(F(g(z))+c) - \exp(F(g(z)))$ only depends on $g(z)$, which implies $g$ is one-to-one. Since the only one-to-one entire functions are polynomials of degree $1$, we are done.