Solution. Clearly, $ P_n(x)>P_{n-1}(x)\ \forall n>1, x\in\mathbb{R}$. If $ x\notin [-2,2]$, $ P_n(x)>x$. Therefore, all real solutons of $ P_n(x)=x$ are in $ [-2,2]$. Notice that $ P_n(x)=x$ has at most $ 2^n$ real roots. Let $ x=2\cos t$, Then it follows that $ P_n(x)=2\cos 2^nt$. It follows that $ P_n(x)=x\iff \cos 2^nt=\cos t\iff 2^nt=2\pi k+t\iff t=\frac{2\pi k}{2^n-1},$ where $ k=0,1,\hdots, 2^n-1$. And the conclusion follows. $ \Box$ Motivation. Consider the Chebyshev function $ T_n(x)$ and $ 2T_n\left(\frac{x}{2}\right)$.

Also we can use $x=y+\frac{1}{y}$

My solution: We have that a square plus 2=x. Subtracting 2 and square rooting we get another expression $=\pm \sqrt{2+x}$ If we keep on doing this we will get 2^n solutions. Now substitute x on the LHS for in the fractions an infinite number of times. To prove that all of these are real, we need to show that 2-the expression>0, but the expression is trivially <2

Can anyone write the solution in more elaborative way the above solution is pretty good but I am unable to track down the whole thing. Would be helpful for me. Thanks in advance.

$\textbf{\textcolor{red}{Claim:-}}$ all real roots of $P_{n}(x)=x$ lies in $[-2,2]$ $\textbf{\textcolor{blue}{Proof:-}}$ clearly through induction we can show that $p_{n}(x)>x$ $\forall$ $x>2$ and for $x<-2$ we have $P_{n}(x)>2$ $\forall$ $n$ hence we have al roots of $P_{n}(x)=x$ lies in [$-2,2]$ as claim follows $\square$ now we substitute $x=2\cos{\theta}$ , we have $P_{1}(x)=2\cos{2\theta}$ from again induction we have $P_{n}(x)=2\cos{2^{n}\theta}$ so we need to solve $\cos{2^{n}\theta}=\cos{\theta}\implies 2^{n}\theta= 2\pi m \pm \theta \implies \theta=\frac{2\pi m}{2^{n}-1}$ and $\theta=\frac{2\pi m}{2^{n}+1}$ where $m \in \mathbb{Z^{*}}$ hence we have all $2^{n}$ roots of $P_{n}(x)$ real and distinct $\blacksquare$