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.
 A: This extended comment has the sole purpose of showing a simple Mathematica code to print the $P(x)$ polynomials up to the tenth degree that satisfy the relation $P(x^2-2) - (P(x))^2 + 2 = 0$.
P[x_] = Sum[ToExpression[StringJoin["a", ToString[n]]] x^n, {n, 0, 10}];
coeff = CoefficientList[P[x^2 - 2] - P[x]^2 + 2, x];
zeros = ConstantArray[0, Length[coeff]];
sol = Solve[coeff == zeros];
poly = Table[P[x] /. sol[[n]], {n, Length[sol]}];
TableForm[Sort[poly]]


It's clear that, for $n \ge 1$, behind all this there is a sequence function, in fact:
Q[x_] = FindSequenceFunction[Sort[poly][[3 ;; All]], n];
TraditionalForm[Q[x]]
TableForm[Table[Expand[Q[x]], {n, 10}]]


From this simple numerical experiment I deduce that the relation:
$$P(x^2-2) - (P(x))^2 + 2 = 0$$
is satisfied by:
$$
P(x) = -1 
\; \; \; \vee \; \; \; 
P(x) = \left(\frac{x-\sqrt{x^2-4}}{2}\right)^n + \left(\frac{x+\sqrt{x^2-4}}{2}\right)^n
$$
where is assumed $n \in \mathbb{Z}$.
A: 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.
A: For every $n\ge 0$ there exists a unique polynomial of degree $n$ such that 
$$P_n(t+1/t) = t^n + 1/t^n$$
For instance, $P_0(x) = 2$, $P_1(x) =x$, $P_2(x) = x^2-2$, and so on. It is easy to see that 
$$P_m\circ P_n(x) = P_{mn}(x)$$
so any two $P_m$'s commute. In particular, $P_n(P_2(x))= P_2(P_n(x))$. 
Let now $n>0$ and $P$ of degree $n$. We can write 
$$P(x) = \alpha \cdot P_n(x) + Q(x)$$
where $\alpha \ne 0$, and $Q$ is of degree at most $n-1$. Assume that we have
$$P(x^2-2) = P(x)^2-2$$
We get $$\alpha P_n(x^2 - 2) + Q(x^2 -2) = (\alpha P_n(x)+ Q(x))^2 - 2$$
or, since $P_n$ commutes with $x^2-2$
$$\alpha (P_n(x)^2 -2) + Q(x^2-2) = \alpha^2 P_n(x)^2+ 2 \alpha P_n(x) Q(x) + Q(x)^2 -2$$
The leading terms have to match, so $\alpha = 1$.  We now get 
$$ Q(x^2 -2) = 2  P_n(x) Q(x) + Q(x)^2$$
If $Q\ne 0$ then LHS has degree less than RHS, contradiction. 
Therefore, for any $n>0$ the only polynomial of degree $n$ commuting with $x^2-2$ is $P_n(x)$. Note that there are two constant polynomials that commute, $2$ and $-1$. 
