TITLE: Solve $f(x)f(2x^2) = f(2x^3+x)$
QUESTION [10 upvotes]: The full question is to solve $f(x)f(2x^2) = f(2x^3+x)$, prove there are at most one solution per degree of $f(x)$ and then find all the solutions. (Here $f(x)$ is assumed to be polynomial.)
So far I have proven that $(x^2+1)^n$ works for even degree polynomials, and I'm pretty near certain there are no odd solutions.
Is the following a valid argument for there only being at most 1 solution per degree?
Let $K$ be the degree of $f(x)$ then $f(x)f(2x^2) = f(2x^3+x)$ produces $3k+1$ linearly independent equations with only $k+1$ independent variables. Therefore for any degree $K$ polynomial there can be at most 1 solution.
Finally, is there any elegant way to prove that no odd solutions exist?

REPLY [16 votes]: As observed in other answers, the leading coefficient of $f$, and the constant term both have to equal $1$ (if $f$ is not identically $0$). Now assume that $f$ has complex root with absolute value greater than 1. Let $z$ one of them with greatest absolute value. Now $2z^3 + z$ is also a root, by triangle inequality
$$
|2z^3 + z| = |z||2z^2 + 1| \geq |z|(|2z^2| - 1) > |z|(2 - 1) = |z|,
$$
contradicting the assumption. So all the roots have absolute value $\leq 1$. But the product of the absolute values of roots is $1$ (the constant term is $1$), so all of them are of absolute value one.
Again using the previous inequality with $|z| = 1$ instead of $|z| > 1$, we get that for all the roots
$$
|2z^2 + 1| = |2z^2| - 1,
$$
which is quite easily only possible if $z^2 = -1$ so $z = \pm i$. So only possible roots are $\pm i$ and if $i$ is, so is $2i^3 + i = -i$ and vice versa.
Lastly, to ensure that multiplicities of $i$ and $-i$ are equal: if $f$ has roots, it has factor $z^2 + 1$ by our observation, and factoring that out we see that the resulting polynomial also satisfies the functional equation. So indeed, only solutions are constant polynomials $0$ and $1$ and $(x^2+1)^k$ for positive integer $k$.