Find all polynomial $P(x)$ $\in$ $\mathbb R[x]$ :$(P(x))^2-1 = P(x^2+1)$ Find all polynomial $P(x)$ $\in$ $\mathbb R[x]$ satisfying $deg$ $ P(x)$ = $4n+2$ ( with $n = 0;1;2;...)$ and : $(P(x))^2-1 = P(x^2+1)$
It is easy to see that $P(x)$ $\neq$ $Const$ because $deg$ $ P(x)$ = $4n+2$ ( with $n = 0;1;2;...)$
So, $P(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0$
$\Rightarrow$ $(a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0)^2-1=a_n{(x^2+1)}^n+a_{n-1}{(x^2+1)}^{n-1}+a_{n-2}{(x^2+1)}^{n-2}+...+a_1(x^2+1)+a_0$
Considering the coefficient of $x^{2n}$: ${a_n}^2=a_n$ $\Rightarrow$ $a_n=1$ $\Rightarrow$ $a_n$ $\in \mathbb Q$
Consider the coefficient of $x^{2n-1}$  : $2a_na_{n-1}=0$ $\Rightarrow$ $a_{n-1}=0$ $\in \mathbb Q$
Consider the coefficient of $x^{2n-2}$ in the $Left$ $Side$: $2a_na_{n-2}+ a_{n-1}^2 $
Consider the coefficient of $x^{2n-2}$ in the $Right$ $Side$: We see that $x^{2n-2}$ appears only in $a_n(x^2+1)^n +a_{n-1}(x^2+1)^{n-1}$ $\Rightarrow$ $a_{n-2}$ $\in$ $\mathbb Q$
In a similar way, we can see that $P(x) \in \mathbb Q[x]$
$P(x)$ is an even polynomial with $a_k$$ > 0$($k$ is odd).
But at this point I have no further idea.Looking forward to getting help from everyone. Thank you very much!I sincerely apologize for my mistake. I have corrected my post, please forgive my ignorance.
 A: Recall that any equality between polynomials in $x$ that holds for inifnitely many $x$ must in fact be an equality between the polynomials per se (and hence holds for all $x$, including complex numbers).
Lemma. Let $f(x)\in\Bbb R[x]$ be a polynomial. If $f(x)^2$ is  even (as a function $\Bbb R\to\Bbb R$), then $f(x)$ is an even polynomial or an odd polynomial.
Proof. As
$$(f(x)-f(-x))\cdot (f(x)+f(-x))=f(x)^2-f(-x)^2=0, $$
one of the factors on the left must be $=0$ for infinitely many $x$, hence must be the zero polynomial. $\square$
Assume  $P$ is a polynomial of degree $4n+2$ with
$$\tag1P(x^2+1)=P(x)^2-1 $$
for all $x\in\Bbb R$.
This allows us to apply the lemma, and as $\deg P$ is even, it must be the case that $P$ is even, so
$$P(x)=Q(x^2) $$
for some polynomial $Q$ of odd degree $2n+1$.
Then $(1)$ becomes
$$Q((x^2+1)^2)+1=Q(x^2)^2.$$
If we substitute $x^2\leftarrow y-1$, this becomes
$$ Q(y^2)+1=Q(y-1)^2,$$ which immediately follows only for $y\ge -1$, but per introductory remark must hold as an identity between polynomials.
Apply the lemma to $Q(y-1)$ and as the degree is odd, we conclude that $Q(y-1)$ is odd. In particular, $Q(-1)=0$.
Then $P(i)=Q(i^2)=0$, hence $P(0)=P(i^2+1)=P(i)^2-1=-1$.
In particular, the set
$$A:=\{\,x\in \Bbb R\mid P(x)=0\lor P(x)=-1\,\} $$
is non-empty. Moreover, $A$ is finite because $P$ is a non-constant polynomial. Let $a=\max A$ and $b=a^2+1$. Then
$$P(b)=P(a)^2-1=(0\text{ or }1)-1\in\{0,-1\} $$
so that also  $b\in A$.
But $$b=a^2+1=a+\frac34+\left(a-\frac12\right)^2>a,$$
contradiction.
We conclude that no such $P$ exists.
