Find all polynomials $P(x)$ with odd degree such that $P(x^2 - 2) = P^2(x)-2$ The problem says :

Find all polynomials $P(x)$ with odd degree such that
  $$P(x^2 - 2) = P^2(x)-2$$

I tried a lot if ways (using high school mathematics) but the only solution I have so far is $P(x) = x$. Can anyone solve this problem using only high school mathematics ?
PS: I have reduced the solution set to  the subset of all polynomials with leading coefficient 1.

Let $K(x) = x^2-2$ and for any $n \geq 1$ let $K^n(x)$ be repeated application of $K$ (so $K^1(x)=K(x)$ and $K^{n+1}(x) = K(K^n(x))$). Then for fixed $x$ the sequence $K^n(x)$ is bounded if and only if $\lvert x \rvert \leq 2$. Since $P(K^n(x))=K^n(P(x))$ it follows that $\lvert P(x) \rvert \leq 2$ for all $\lvert x \rvert \leq 2$. As noted, the leading coefficient of $P$ is $1$.
It is well known that a polynomial $T$ of degree $n$ with leading coefficient $2^{n-1}$ maps $[-1,1]$ into $[-1,1]$ if and only if $T=T_n$ is a Chebyshev polynomial of the first kind.
Conclude that $P(2x)/2 = T_n$ where $\deg(P)=n$. So $P$ is characterized by $$P(x+x^{-1})=x^n+x^{-n}.$$ It is easy to check that such a polynomial indeed has the required property.
