How to find all polynomials P(x) such that $P(x^2-2)=P(x)^2 -2$? I am trying the fallowing exercise :
Solve $P(X^2 -2)=P(X)^2 -2$ with P a monic polynomial (non-constant)
My attempt :
Let P satisfying $P(X^2-2) = (P(X))^2-2$
Then $Q(X)=P(X^2-2) = (P(X))^2-2$ 
Therefore,  $$Q(X^2-2) = (P(X^2-2))^2-2 = (P(X)^2-2)^2-2 = Q^2-2$$
As X is a solution, by defining the sequence:
$(P_n)_{n \geq 1}$ with $P_1 = X$ and for all $n \geq 1, P_{n+1} = P_n^2-2$
We obtain a sequence of polynomials which are solutions. 
But I don't know how to prove it's the only one. If someone have an idea to prove it or an another method to solve the problem ? 
Thank you in advance for your time.

Lemma : If $P(x)^2$ is a polynomial in $x^2$, then so is either P(x) or P(x)/x.
By the lemma, there is a polynomial Q such that $P(x)=Q(x^2-2)$ or $P(x)=xQ(x^2-2)$.
Then $Q((x^2-2)^2-2)=Q(x^2-2)^2−2$ or $(x^2-2)Q((x^2-2)^2-2)=x^2Q(x^2-2)^2-2$
Substituting $x^2-2=y$ yields $Q(y^2-2)=Q(y)^2-2$ and $yQ(y^2-2)=(y+2)Q(y)^2+1$ 
Suppose that $yQ(y^2-2)=(y+2)Q(y)^2-2$. Setting $y=-2$ we obtain that $Q(2)=1$
Note that, if $a\neq 0$ and $Q(a)=1$ then also $aQ(a^2+1)=(a+2)-2$ and hence $Q(a^2+1)=1$.
We thus obtain an infinite sequence of points at which Q takes value 1.
Namely the sequence given by $a_{n+1}=a_{n^2}-2$. 
Therefore $Q≡1$. It follows that if $Q≢1$, then $P(x)=Q(x^2-2)$. 
Now we can easily list all solutions: these are the polynomials of the form $T(T(⋯(T(x))⋯))$, where $T(x)=x^2-2$.
NB: Lemma/proof,
Let $P(x)=a_nx^n+a_{n−1}x_{n−1}+⋯+a_0$, $a_n≠0$. 
The coefficient at $x^{2n−1}$ (of $P(x)^2$) is $2a_na_{n−1}$, from which we get $a_{n−1}=0$. Now the coefficient at $x_{2n−3}$ equals $2a_na_{n−3}$; hence $a_{n−3}=0$, and so on. Continuing in this manner we conclude that $a_{n−2k−1}=0$ for $k=0,1,2,…,$ i.e. $P(x)=a_nx^n+a_{n−2}x^{n−2}+a_{n−4}x_{n−4}+⋯$.
