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Find all real polynomials with $P(0)=0$ and $P(x^2+1)=(P(x))^2+1$

I' like a solution for this problem and proof verification:

My proof:

Let $a_0=0$ and $a_{n+1}=a_n^2+1$, clearly $P(a_0)=a_0$ and if $P(a_n)=a_n$ then $P(a_{n+1})=P(a_n^2+1)=(P(a_n))^2+1=a_n^2+1=a_{n+1}$. So $P(x)=x$ for infinitely many values of $x$ (because $a_n$ is strictly increasing), hence $P$ is the identity polynomial, and it clearly works.

Asinomás
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  • "$P(x)=0$"? Is it for a particular fixed $x$? If that is supposed to be all $x$, the question is trivial. If not, please add quantifiers... – Clement C. Aug 20 '16 at 01:34
  • I think the problem should say "$P(0) = 0$ and .$P(x^2+1) = (P(x))^2+1$ for all real $x$." – JimmyK4542 Aug 20 '16 at 01:35
  • Oh yeah, my bad – Asinomás Aug 20 '16 at 01:37
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    for infinite values of x "For infinitely many values of x" would sound better. – dxiv Aug 20 '16 at 01:41
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    To be more detailed you could also prove that your sequence indeed contains infinitely many distinct terms – clark Aug 20 '16 at 01:43
  • @clark right thanks. – Asinomás Aug 20 '16 at 01:50
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    The proof is good and is the natural one in this case. If restated in more general terms of commuting polynomials $P(x)$ and $Q(x) = x^2+1$, maybe this answer to When functions commute under composition could be relevant. – dxiv Aug 20 '16 at 02:02
  • After checking a few cases, I am convinced that, without the assumption that $P(0)=0$, the polynomial solutions to $P\left(X^2+1\right)=\big(P(X)\big)^2+1$ in characteristic not equal to $2$ are the constant polynomials $P(X)=\frac{1\pm\sqrt{-3}}{2}$ if $\sqrt{-3}$ exists, the identity polynomial $P(X)=X$, and the polynomials $P(X)=Q_n(X)$ for $n=1,2,\ldots$, where $Q_1(X)=X^2+1$ and $Q_n(X)=Q_1\big(Q_{n-1}(X)\big)$ for $n=2,3,\ldots$. – Batominovski Aug 21 '16 at 20:41
  • In characteristic $2$, there are weird solutions. Over $\mathbb{F}_4=\mathbb{F}_2[\alpha]$ with $\alpha^2+\alpha+1=0$, $P(X)=X^2+X+\alpha$ and $P(X)=X^3+\alpha,X^2+\alpha X$ are solutions. – Batominovski Aug 21 '16 at 20:50

1 Answers1

2

Let $P(x)=a_1x+a_2x^2+…..+a_nx^n$ be a polynomial such that $$P(x^2+1)=(P(x))^2+1$$ and $P(0)=0$ (it is immediate that $P(1)=1$). We have $$P(x^2+1)=\sum a_k(x^2+1)^k=\sum a_k\sum\binom{k}{ l}(x^2)^{k-l}$$ $$\\((P(x))^2+1=\sum a_k^2x^{2k}+2\sum a_ka_jx^{k+j}+1$$ It follows that the polynomial $$R(x)= \sum a_k\sum\binom{k}{ l}(x^2)^{k-l}-\left(\sum a_k^2x^{2k}+2\sum a_ka_jx^{k+j}+1\right)$$ must be identically zero because it has more roots than its degree so all the coefficients must be zero.

It can be seen that even for the second degree this is not the case and that the only possibility is $$\color{red}{P(x)=x}$$

$$***$$

We give another solution because the verification about coefficients in the precedent one is not immediate

$$***$$

SECOND SOLUTION.-It is clear that $P(x)=x$ goes well. Put $$P(x)=x+Q(x)$$ so that
$$ P(x^2+1)=(P(x))^2+1\iff Q(x^2+1)=2xQ(x)+(Q(x))^2\qquad (*) $$ We have from $(*)$, $$ (Q(x))^2+2xQ(x)-Q(x^2+1)=0\iff Q(x)=-x\pm\sqrt{x^2+Q(x^2+1)}\qquad (**)$$

Note that if $Q(x_0)=0$ then $Q(x_0^2+1)=0$

Because of $P(0)=0\Rightarrow P(1)=1\Rightarrow P(2)=2\Rightarrow P(5)=5$, one has $Q(0)=Q(1)=Q(2)=Q(5)=0$.

It follows from $(**)$ we can deduce an infinity of roots for $Q(x)$ (for example $26,677$ and $458330$ the three first deduced from $Q(5)=0$).

Consequently $Q(x)=0$ for all $x$ and $$P(x)=x$$ is the only solution.

Piquito
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  • I agree that "It is easily seen" when $P$ is quadratic, however in general if we write out $R(x)$ we get:

    $$0 = x^{2n}(a_n-a_n^2) - 2a_{n}a_{n-1}x^{2n-1} + (a_{n-1}-a_{n-1}^2 + n a_n - 2a_na_{n-2})x^{2n-2} + \ldots$$

    How do you show that $a_n=a_{n-1}=\ldots=a_{2} = 0$ is the only solution to this equation system?

     – Winther Aug 20 '16 at 13:29