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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 the one i have asked here: http://math.stackexchange.com/questions/2346/continuous-function-satisfying-fkxfxk

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Note that the constant $p(x) = \frac{1 \pm \sqrt 5}{2}$ is a solution, but Wolfram|Alpha did not help me find any other quadratic solutions. –  Joshua Shane Liberman Sep 1 '10 at 12:26
    
@JSL: As i pointed out i want the deg of polynomial to be >1 –  anonymous Sep 1 '10 at 12:39
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There are no solutions of degree 1..6. As the degree goes up, the system of equations gets 'worse' (in that the method of indeterminate equations leaves a larger remainder that cannot be eliminated). I would conjecture that there are no solutions other than the one Joshua gave. –  Jacques Carette Sep 1 '10 at 12:52
    
No solutions up to degree 8. Stopping now. –  Jacques Carette Sep 1 '10 at 12:54
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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.) –  ShreevatsaR Sep 1 '10 at 13:35

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up vote 13 down vote accepted

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! –  anonymous Sep 1 '10 at 14:00

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