Polynomials which satisfy $p^{2}(x)-1 = p(x^{2}+1)$ Can we find a polynomial $p(x) \in \mathbb{R}$ such that $\text{deg}\ p(x)>1$ and which satisfies $$p^{2}(x)-1=p(x^{2}+1)$$ for all $x \in \mathbb{R}$.
This question can be very well identified with my previous question.
 A: There is no solution. Here is the proof:
First note, that the highest coefficient of $p$ must be 1. In particular, $\lim_{x \to \infty} p(x) = \infty$.
Assume, that there exists a real value $c$ such that $p (c) = 0$. Then
$$ 0 = (p(c)^2 - 1)^2 - 1 = p(c^2 + 1)^2 - 1 = p((c^2 + 1)^2 + 1)$$
But $(c^2 + 1)^2 + 1 > c$, so any real root of $p$ would lead to another, higher root of $p$ which is impossible. So $p$ has no real root and thus $p(x) > 0$ for all $x$.
But then, $p(x)^2 - 1 = p(x^2 + 1) > 0$ and thus $p(x) > 1$.
Define a sequence $x_n$ by $x_0 = 0$ and $x_{n + 1}^2 + 1 = x_n$ with imaginary part of $x_n$ positive for $n > 0$. 
We first show, that $p(x_n) \neq 0$ for all $x_n$:
clearly, $p(x_0) \neq 0$ and $p(i)^2 - 1 = p(0) > 1$ and so $p(x_1) = p(i) \neq 0$.
Assume, that $p(x_{n+1}) = 0$ for some $n > 1$. Then $p(x_n) = -1$ and so $p(x_{n - 1}) = 0$. This is impossible.
From the functional equation, we obtain by differentiating
$$
p(x) p'(x) = x p'(x^2 + 1)
$$
We now show by induction, that $p'(x_n) = 0$ for all $n$.
For $n = 0$, we obtain $p(0) p'(0) = 0$, and thus $p'(0) = 0$.
Assume, that $p'(x_n) = 0$, then
$$
p(x_{n+1}) p'(x_{n+1}) = x_{n + 1} p'(x_{n+1}^2+1) = x_{n+1} p'(x_n) = 0
$$
and so, $p'(x_{n+1}) = 0$ as required.
But $x_n$ is a sequence converging to $\frac{1 + i \sqrt{3}}{2}$, and in particular consists of infinitely many distinct points, all of which are roots of $p'$ which is impossible.
