7

What can be $P(0)$, when $P(x^2+1)=(P(x))^2+1$ and $P(x)$ is polynomial?

Let $P(0)=0$, then $P(1)=1$, $P(2)=2$, $P(5)=5$, $P(26)=26$, $P(677)=677$ ... and so on. Then $P(x)=x$, because all the points on $y=P(x)$ are $y=x$.

If $P(0)=2$, then $P(x)=(x^2+1)^2+1$ for the same reason.

But when $P(0)=3$, we have that $\lim_{x→∞}\log_xP(x)$ does not converge into an integer. So I think $P(x)$ cannot be a polynomial.

Then what are the values of $P(0)$ that makes $P(x)$ polynomial?

asdf1048576
  • 87
  • If you differentiate both sides, you get $P(0)=0$, otherwise, $P'(0)=P'''(0)=0$. Maybe the same idea can be applied to show that $P^{2n+1}(0)=0$, but I have not enough time to expand this. – Konstantinos Gaitanas May 15 '19 at 10:41
  • For p(0)=5: $x^8 + 4x^6 + 8x^4 + 8*x^2 + 5$ – joro May 15 '19 at 10:55
  • 1
    (I should have accounted $P(x)=x$ as $0$-th iterate of $x^2+1$. Actually in the linked short proof, it appears separately as it's first shown that $P$ is either odd or even, and then the odd and even case are treated separately.) – YCor May 15 '19 at 11:36
  • 1
    Apparently, some comments were deleted when this question was migrated. See https://meta.mathoverflow.net/questions/4219/migrating-a-question-removes-some-comments – Gerry Myerson May 16 '19 at 03:05

0 Answers0