Can $f(g(x))$ be a polynomial?

Question

Let $f(x)$ and $g(x)$ be nonpolynomial real-entire functions.

Is it possible that $f(g(x))$ is equal to a polynomial ?

edit

Some comments :

I was thinking about iterations.

So for instance $f(f(x)) = $ some polynomial.

However such $f$ are usually (Always ?) not entire because of the fact that a non-linear polynomial has more than 1 fixpoint.

This lead me to consider adding the strong condition

$(f(g(x)) - g(f(x)))^2$ is not indentically $0$.

But I guess that is a followup question.

edit 2

Real-entire means entire and real-analytic.

Answer 1

There are various ways to show that this can not be true. E.g., by Casorati-Weierstrass, the image of $|z|>R$ under $g$ is dense in the plane for every $R>0$, so the image of the same domain under $f\circ g$ contains a dense subset of $f(\mathbb{C})$ which is itself dense in the plane, showing that $f \circ g$ has an essential singularity at $\infty$.

(And it does not matter that $f$ and $g$ are real.)