If $f : \mathbb{N} \to \mathbb{N}$ and $f(x)=f(2x^2)$ find all possible solutions

Question

If $f : \mathbb{N} \to \mathbb{N}$ and $f(x)=f(2x^2)$ find all possible solutions to the functional equation.

I think it's a constant since it doesn't really fit anything. But how do I show that?

Answer 1

You are going to be working a long time as there are continuum many solutions. You can choose $f(1)$ to be anything, then $f(2),f(8),f(128),$ and so on have to have the same value. There is nothing to constrain $f(3)$ so it can be anything you like, then $f(18),f(648)$ and so on have to match that. Every prime starts a new series, as does any product of odd primes. I haven't figured out if the chain that starts with $4$ ever meets the chain that starts with $2$. Yes, $f$ can be constant, but it doesn't need to be.