I'm attempting to prove:
Define functions $f_m$ by the recursion relation such that $f_1(x) = 2x^2-1$ and $f_k(x) = f_1(f_{k-1}(x))$ . Then for all $ m > 0$, there exists no $x \in \mathbb Z$ such that $f_m = x$ other than the trivial $x = 1$.
I have no idea where to start, or even what even type of math could help me prove this right or wrong. (though after some research, I believe it fits under "iteration theory"). My only work is a brute-force method of computing roots for each function of the function, but that can't prove them all. Any direction towards the type of math this is, or a method to prove this would be extremely helpful.
Well, this post was edited several times while I wrote this, but I just assume that the corrected question is the one Hurkyl commented.
Equation $f_m(x)=x$ has no integer solution $x$ s.t $|x|>1$ because $|x|>1 \Rightarrow |2x^2-1|>|x|>1$.
Hence we just try calculating $f_m(x)$ with $x=0,-1$, and if you sensed that $x=\cos \theta \Rightarrow f_m(\cos \theta )=\cos (2^m \theta)$, you can easily get the proof you wanted.