2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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 $R^2(0)\ne 0$.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.