Let $R(x) = \frac{F(x)}{G(x)}$ be a rational function with $F(x), G(x) \in \mathbb{Z}[x]$ and

$F(x), G(x)$ have no common root modulo $p$ for all primes $p$.

Consider the rational function $Q(x)=\underbrace{R(R(\ldots(R(x))))}_{\text{$n$ times}}$ where $n \in \mathbb{N}$.

Prove that if there is an integer $k$ such that $Q(k)=k$, then $R(R(k))=k$.

My thought,

in case $G(x) = 1$, $R(x)$ will be polynomial.

Let $a_1$ be the value such that $Q(a_1)=a_1$ and $R(a_i)=a_{i+1}, \forall i= 1, 2, \ldots, n-1$.

We have $a_2-a_1 \mid a_3-a_2 \mid \ldots \mid a_1-a_n \mid a_2-a_1$, so $\mid a_{i+1}-a_i \mid$ is constant.

Then $\mid R(a_i)-R(a_j)\mid = \mid a_i-a_j \mid, \forall i, j$

Since $\mid a_{i+1}-a_i \mid = c, \forall i$ which will be true when orbit = $1$ or $2$,

so $R(R(k))=k$ or $R(k)=k$, we can conclude that $R(R(k))=k$

Please suggest how to proceed. Thank you.

  • $\begingroup$ A few questions: 1. What is the $P(a_i)$ mentioned in your attempts? 2. Should there be some condition on n? Clearly, $n\geqslant 3$ for the question to be nontrivial but if $n$ is odd, $Q(k)=k$ and $R^2(k)=k$ would imply $R(k)=k$. (This would be a strictly stronger implication.) $\endgroup$ – daruma Jul 30 '18 at 8:30
  • $\begingroup$ @daruma. Thank you, typo edited. There is no other condition on $n$. $\endgroup$ – Dan Jul 30 '18 at 9:27
  • $\begingroup$ If $Q(k)=R^n(k)=k$ and $R^2(k)=k$, then we would get that $R(k)=k$ if $n$ is odd. Are you sure the question is correct? $\endgroup$ – daruma Jul 30 '18 at 9:30
  • $\begingroup$ On a different point, why is $|P(a_i)-P(a_j)|=|a_i-a_j|$? In general, polynomials are not isometries of $\mathbb{R}$ $\endgroup$ – daruma Jul 30 '18 at 9:32
  • $\begingroup$ @daruma. The question is correct. $\mid R(a_i)-R(a_j)\mid = \mid a_i-a_j \mid $ because $\mid a_{i+1}-a_i \mid$ is constant. Sorry, I don't understand "isometry". $\endgroup$ – Dan Jul 30 '18 at 12:54

Unless there is something I don't understand, the assertion is false. An easy example is $F(x)=1$ and $G(x)=1-x$. Then $R(0)=1$, $R(1)=\infty$ and $R(\infty)=0$. So for $n=3$, we have $Q(0)=R^3(0)=0$, but $R^2(0)\ne 0$.

  • 4
    $\begingroup$ And of course you could have avoided the embarrassment of an infinity by starting with $k=2$ instead. The period of the transformation is three, so the counterexample kills the conjecture. $\endgroup$ – Lubin Jul 30 '18 at 17:44
  • $\begingroup$ @Lubin, Thank you all. I got it now. $\endgroup$ – Dan Jul 31 '18 at 14:11

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