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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.

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    Here, does $p^2(x)$ mean $p(x)^2$ or $p(p(x))$? (I'd think it's the latter, since $p(x)^2$ is usually written $p(x)^2$, but looking at the previous linked question makes it confusing.)2010-09-01

1 Answers 1

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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.

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    Wonderfully done!2010-09-01