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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.

Dane Bouchie
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    What you wrote doesn't make much sense: did you mean something like "Define functions $f_m$ by the recursion relation $f_1(x) = f(x)$ and $f_k(x) = f(f_{k-1}(x))$. Prove, for all $m > 0$, there is no $x \in \mathbf{Z}$ such that $f_m(x) = x$, except for $x=1$."? –  Dec 18 '14 at 05:27
  • What if $f$? You say that $f_k(x)$ is $f(f_{k-1})$, but never say what $f$ is. If you mean $f_1(f_{k-1})$ then obviously what you want to prove is false. – Andrés E. Caicedo Dec 18 '14 at 05:51
  • "Then for all m>0 and x∈ℤ other than the trivial x=1" is not even a sentence. – Andrés E. Caicedo Dec 18 '14 at 05:53
  • What is " each function of the function"? What do you mean when you say that you cannot prove all the roots? – Andrés E. Caicedo Dec 18 '14 at 05:55
  • You still haven't said what $f$ is, and writing $f_m=x$ makes no sense. – Andrés E. Caicedo Dec 18 '14 at 05:56

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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.

cjackal
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