Find all $f(x)$ such that $f(x)+f(1-x)=f(x^2)$ is satisfied
This is a question I made up, and my solution is as follows:
Let $x=1$ (or $x=0$), then $f(0)+f(1)=f(1)$ which means that $f(0)=0$. Substitute $-x$ for $x$. We get that $$f(x)+f(1-x)=f(-x)+f(1+x)=f(x^2)$$Note that substituting $x+1$ into the original functional equation yields $f(x+1)+f(-x)=f((x+1)^2)$, meaning that $$f(x^2)=f((x+1)^2)$$Thus we need to find a function that is the same when the input is the square of an integer. Two such functions are $$\frac{a\lfloor\sqrt{x}\rfloor}{\sqrt{x}}\text{ and }\frac{a\lceil\sqrt{x}\rceil}{\sqrt{x}}$$Where $a$ is any real number.
I am not sure how to prove that these are (not) the only functions that solve the functional equation. (The case $f(x)=0$ is included in the two functions when $a=0$). So what are the other functions, if there are any?
Edit: The functions in my solution are only true when $x$ is an integer (meaning that, according to @BrunoB's comment, the functions aren't solutions), so this means that $x\in\mathbb{Z}$. I would like to see a solution for $x\in\mathbb{R}$ or maybe even for $x\in\mathbb{C}$.