<s>[INST] Let $f(x)=x^2-2$ with $x\in [-2,2]$. Show that the equation $f^{n}(x)=x$ has $2^{n}$ real roots. [Where $f^{n}(x)=f(f^{n-1}(x))$] Let $f(x)=x^2-2$ with $x\in [-2,2]$. Show that the equation $f^{n}(x)=x$ has $2^{n}$ real roots. [Where $f^{n}(x)=f(f^{n-1}(x))$]
My solution:
Let $x=2\cos(\theta)$ for $\theta\in [0,\pi]$
$\implies$
$f(x)=4\cos^2 (\theta)-2=2\cos(2\theta)$
$f(f(x))=(f(x))^2-2=(2\cos^22\theta)^2-2=2(2\cos^{2}2\theta-1)=2\cos(2^2 \theta)$
$\implies$$f(f(x))=2\cos(4\theta)$
Similarly $f(f(f(x)))=2\cos(2^3\theta)$
$\vdots$
$\underbrace {f\circ f\circ\cdots \circ f}_\text{n times}(x)=2\cos(2^n \theta)$
$\implies$
From Question i.e, $f^{n}(x)=x$
$\implies$
$2\cos(2^n\theta)=2\cos\theta$
$\implies$
$2^n\theta=2m\pi \pm \theta$
$\implies$
$\theta=\dfrac{2m\pi}{2^{n}-1}$ $\quad$ or $\quad$ $\theta=\dfrac{2m\pi}{2^{n}+1}$ $\forall \; \theta \in [0,\pi]$
Checking Result for $n=1$
$\theta=\dfrac{2m\pi}{1}\;$ or $\; \theta=\dfrac{2m\pi}{3}$
$\implies$ $\theta = \dfrac{2\pi}{3}, 0$ i.e., $2$ solution.
Checking Result for $n=2$
$\theta=\dfrac{2m\pi}{3}\;$ or $\; \theta=\dfrac{2m\pi}{5}$
$\implies$ $0, \dfrac{2\pi}{3},\dfrac{2\pi}{5},\dfrac{4\pi}{5}$ i.e. $4$ solution.
$\implies$ There are $2^{n}$ distinct root of $f^{n}(x)=x\;$ equation.
Is my Solution Correct?
 [/INST] I think that your idea is very nice.
I would add proofs for the following claims :
Claim 1 : $f^{n}(2\cos\theta)=2\cos(2^n\theta)$ (your idea is very nice, but I would prove this claim by induction rigorously)
Claim 2 : The number of integers $m$ such that $0\leqslant\dfrac{2m\pi}{2^{n}-1}\leqslant\pi$ is $2^{n-1}$, and the number of integers $M$ such that $0\leqslant\dfrac{2M\pi}{2^{n}+1}\leqslant\pi$ is $2^{n-1}+1$.
Claim 3 : $\dfrac{2m\pi}{2^{n}-1}=\dfrac{2M\pi}{2^{n}+1}$ holds if and only if $m=M=0$.
(It follows from Claim 2 and 3 that $f^n(x)=x$ has $2^{n-1}+(2^{n-1}+1)-1=2^n$ real roots.)

Claim 1 : $f^{n}(2\cos\theta)=2\cos(2^n\theta)$.
Proof : If $n=1$, $f(2\cos\theta)=(2\cos\theta)^2-2=2(2\cos^2\theta-1)=2\cos(2\theta)$. Suppose that $f^{n}(2\cos\theta)=2\cos(2^n\theta)$. Then, $f^{n+1}(2\cos\theta)=(f^n(2\cos\theta))^2-2=2(2\cos^2(2^n\theta)-1)=2\cos(2^{n+1}\theta)$.$\ \square$

Claim 2 : The number of integers $m$ such that $0\leqslant\dfrac{2m\pi}{2^{n}-1}\leqslant\pi$ is $2^{n-1}$, and the number of integers $M$ such that $0\leqslant\dfrac{2M\pi}{2^{n}+1}\leqslant\pi$ is $2^{n-1}+1$.
Proof : Since $0\leqslant\dfrac{2m\pi}{2^{n}-1}\leqslant\pi\iff 0\leqslant m\leqslant 2^{n-1}-1$, the number of such integers $m$ is $2^{n-1}$. Since $0\leqslant\dfrac{2M\pi}{2^{n}+1}\leqslant\pi\iff 0\leqslant M\leqslant 2^{n-1}$, the number of such integers $M$ is $2^{n-1}+1$.$\ \square$

Claim 3 : $\dfrac{2m\pi}{2^{n}-1}=\dfrac{2M\pi}{2^{n}+1}$ holds if and only if $m=M=0$.
Proof : Since $m=\dfrac{(2^n-1)M}{2^n+1}$ with $\gcd(2^n-1,2^n+1)=\gcd(2^n-1,2)=1$, $M$ has to be a multiple of $2^n+1$. It follows from $0\leqslant M\leqslant 2^{n-1}$ that $M=m=0$.$\ \square$
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