Is $f(\sqrt{x}) + f(-\sqrt{x})$ always rational? Suppose that $f$ is a rational function over $\mathbb{R}$ (or $\mathbb{C}$, or whatever field makes this easier to answer). Is $$\frac{f(\sqrt{x}) + f(-\sqrt{x})}{2}$$ always a rational function of $x$?
This question is inspired by a sequence question. Given a sequence $a_n$ with a rational generating function, is it true that, say, $a_{2n}$ also has a rational generating function? If $f$ is the generating function for $a_n$, then $(f(\sqrt{x}) + f(-\sqrt{x})) / 2$ is the generating function for $a_{2n}$.
As an example, take $f(x) = 1 / (1 - x)$, which corresponds to the all-ones sequence $a_n = 1$. Then $$\frac{f(\sqrt{x}) + f(-\sqrt{x})}{2} = \frac{1}{1 - x},$$ as we would expect.
This specific case might be "easy" to answer (not for me), but when I first heard about this someone mumbled something about Galois theory and field extensions that I didn't understand. I would like to know more about this if it's relevant, since this question generalizes to seemingly harder cases.

Here is a brutal-force solution: Write $f$ in the form
$$ f(x) = \frac{A_0(x^2) + x A_1(x^2)}{B_0(x^2) + x B_1(x^2)} $$
where $A_0, A_1, B_0, B_1$ are polynomials. Then
$$ \frac{f(\sqrt{x}) + f(-\sqrt{x})}{2}
= \frac{A_0(x)B_0(x) - x A_1(x)B_1(x)}{B_0(x)^2 - x B_1(x)^2}. $$
