Some Results On The Flynn-Poonen-Schaefer Conjecture

TL;DR

This paper investigates the Flynn-Poonen-Schaefer conjecture on rational cycles of quadratic polynomials, establishing conditions on the denominator of the parameter and confirming the conjecture for certain cases.

Contribution

It proves that rational cycles of length at least 3 require the denominator to be divisible by 16, and confirms the conjecture when the denominator has at most two prime factors.

Findings

Denominator of c divisible by 16 for cycles of length ≥ 3

Upper bound on rational periodic points based on prime factors

Conjecture holds if denominator has ≤ 2 prime factors

Abstract

For $c \in \mathbb{Q}$, consider the quadratic polynomial map $\varphi_c(x)=x^2-c$. Flynn, Poonen and Schaefer conjectured in 1997 that no rational cycle of $\varphi_c$ under iteration has length more than $3$. Here we discuss this conjecture using arithmetic and combinatorial means, leading to three main results. First, we show that if $\varphi_c$ admits a rational cycle of length $n \ge 3$, then the denominator of $c$ must be divisible by $16$. We then provide an upper bound on the number of periodic rational points of $\varphi_c$ in terms of the number of distinct prime factors of the denominator of $c$. Finally, we show that the Flynn-Poonen-Schaefer conjecture holds for $\varphi_c$ if that denominator has at most two distinct prime factors.

Equations

$$O_{\phi }(z)=({\phi }^k(z){)}_{k\ge 0},$$

$$\mathrm{Per}({\phi }_c)$$

$$\mathrm{Preper}({\phi }_c)$$

$$\mathrm{num}(\mathrm{Per}({\phi }_c))=d\cdot \mathrm{Per}({\phi }_c)\text{, }\mathrm{num}(\mathrm{Preper}({\phi }_c))=d\cdot \mathrm{Preper}({\phi }_c).$$

$$\prod _{1\le i<j\le n}(f(z_i)-f(z_j))=(-1{)}^{n-1}\prod _{1\le i<j\le n}(z_i-z_j).$$

$$\prod _{1\le i<j\le n}(f(z_i)-f(z_j))$$

$$=(-1{)}^{n-1}\prod _{1\le i<j\le n}(z_i-z_j).$$

$$f(u)-f(v)=(u-v)(u+v).$$

$$\prod_{i=0}^{r-1}\big{(}f^{i+1}(x)-f^{i+1}(y)\big{)}=\prod_{i=0}^{r-1}\big{(}f^{i}(x)-f^{i}(y)\big{)}.$$

$$f^{i+1}(x)-f^{i+1}(y)=(f^i(x)-f^i(y))(f^i(x)+f^i(y)).$$

$$\prod _{1\le i<j\le n}(f(x_i)-f(x_j))=\pm \prod _{1\le i<j\le n}(x_i-x_j).$$

$$\displaystyle\prod_{1\leq i\leq n}\big{(}X_{i}+X_{i+k}\big{)}$$

$$\displaystyle\prod_{1\leq i<j\leq n}\big{(}X_{i}+X_{j}\big{)}$$

$$\prod_{1\leq i<j\leq n}\big{(}X_{i}+X_{j}\big{)}=\pm d^{n(n-1)/2}.$$

$$n$$

$$m$$

$$r$$

$$\mathcal{C}-1/2\subset {\mathbb{Z}}_2.$$

$$f(t)$$

$$f(t-1/2)={\phi }_c(t)-1/2.$$

$$f(\mathcal{C}-1/2)=\mathcal{C}-1/2,$$

$$(x-1/2{)}^2+(x-1/2)-(c+1/4)=y-1/2.$$

$$\mathcal{C}-1/2=(z_1,\dots ,z_n)$$

$$\prod _{1\le i<j\le n}^n\frac{f(z_i)-f(z_j)}{z_i-z_j}=\pm 1.$$

$$\prod _{1\le i<j\le n}^n(z_i+z_j+1)=\pm 1.$$

$$|\mathrm{Preper}({\phi }_c)|\le 2^{s+2}+1.$$

$$|\mathrm{Per}({\phi }_c)|\le 2^s+2.$$

$$Z=(X^2-a)/d.$$

$$X\equiv \pm Y\mathrm{mod}{p}^r.$$

$$(X+Y)(X-Y)\equiv 0\mathrm{mod}{p}^r.$$

$$X^{\prime }-Y^{\prime }=(X^2-Y^2)/d.$$

$$(X+Y)(X-Y)\equiv 0\mathrm{mod}{2}^{r+1}.$$

$$X_1+Y_1=-(X_2+Y_2).$$

Full text

1. Introduction

Let $S$ be a set and $\varphi:S\to S$ a self map. For $z\in S$, the orbit of $z$ under $\varphi$ is the sequence of iterates

[TABLE]

where $\varphi^{k}$ is the $k^{th}$ iterate of $\varphi$ and $\varphi^{0}=\operatorname{Id}_{S}$. We say that $z$ is periodic under $\varphi$ if there is an integer $n\geq 1$ such that $\varphi^{n}(z)=z$, and then the least such $n$ is the period of $z$. In that case, we identify $O_{\varphi}(z)$ with the finite sequence $\mathcal{C}=(z,\varphi(z),\dots,\varphi^{n-1}(z))$, and we say that $\mathcal{C}$ is a cycle of length $n$. The element $z$ is said to be preperiodic under $\varphi$ if there is an integer $m\geq 1$ such that $\varphi^{m}(z)$ is periodic. For every rational fraction in $\mathbb{Q}(x)$ of degree $\geq 2$, its set of preperiodic points is finite, this being a particular case of a well known theorem of Northcott [7]. However, determining the cardinality of this set is very difficult in general. The following conjecture due to Flynn, Poonen and Schaefer [5] illustrates the difficulty in understanding, in general, the periodic points of polynomials, even those of degree $2$.

Conjecture 1.1**.**

Let $c\in\mathbb{Q}$. Consider the quadratic map $\varphi_{c}\colon\mathbb{Q}\to\mathbb{Q}$ defined by111The map $x\mapsto x^{2}+c$ is more common in the literature, but we slightly prefer to deal with $x\mapsto x^{2}-c$. $\varphi_{c}(x)=x^{2}-c$ for all $x\in\mathbb{Q}$. Then every periodic point of $\varphi_{c}$ in $\mathbb{Q}$ has period at most $3$.

See also [8] for a refined conjecture on the rational preperiodic points of quadratic maps over $\mathbb{Q}$. As the following classical example shows, rational points of period $3$ do occur for suitable $c\in\mathbb{Q}$.

Example 1.2**.**

Let $c=29/16$. Then the map $\varphi_{c}$ admits the cycle $\mathcal{C}=(-1/4,-7/4,5/4)$ of length $3$.

While Conjecture 1.1 has already been explored in several papers, it remains widely open at the time of writing. The main positive results concerning it are that period $4$ and period $5$ are indeed excluded, by Morton [6] and by Flynn, Poonen and Schaefer [5], respectively.

Theorem 1.3** (Morton).**

For every $c\in\mathbb{Q}$, there is no periodic point of $\varphi_{c}$ in $\mathbb{Q}$ of period $4$.

Theorem 1.4** (Flynn, Poonen and Schaefer).**

For every $c\in\mathbb{Q}$, there is no periodic point of $\varphi_{c}$ in $\mathbb{Q}$ of period $5$.

No period higher than $5$ has been excluded so far for the rational maps $\varphi_{c}$. However, Stoll showed that the exclusion of period $6$ would follow from the validity of the Birch and Swinnerton-Dyer conjecture [10].

Conjecture 1.1 is often studied using the height and $p$-adic Julia sets. Here we mainly use arithmetic and combinatorial means. Among our tools, we shall use the above two results and Theorem 2.11, a particular case of a theorem of Zieve [12] on polynomial iteration over the $p$-adic integers.

Given $0\not=c\in\mathbb{Q}$, let $s$ denote the number of distinct primes dividing the denominator of $c$. In [2], Call and Goldstine showed that the number of rational preperiodic points of $\varphi_{c}$ does not exceed the upper bound $2^{s+2}+1$. Among our present results, we show that any rational cycle of $\varphi_{c}$ has length at most $2^{s}+2$. We also show that the conjecture holds for $\varphi_{c}$ in case $s\leq 2$.

For convenience, in order to make this paper as self-contained as possible, we provide short proofs of some already known basic results.

1.1. Notation

Given $c\in\mathbb{Q}$, we denote by $\varphi_{c}\colon\mathbb{Q}\to\mathbb{Q}$ the quadratic map defined by $\varphi_{c}(x)=x^{2}-c$ for all $x\in\mathbb{Q}$. Most papers dealing with Conjecture 1.1 rather consider the map $x^{2}+c$. Our present choice allows statements with positive rather than negative values of $c$. For instance, with this choice, we show in [4] that if $\varphi_{c}$ admits a cycle of length at least $2$, then $c\geq 1$.

The sets of rational periodic and preperiodic points of $\varphi_{c}$ will be denoted by $\operatorname{Per}(\varphi_{c})$ and $\operatorname{Preper}(\varphi_{c})$, respectively:

[TABLE]

For a nonzero integer $d$, we shall denote by $\operatorname{supp}(d)$ the set of prime numbers $p$ dividing $d$. For instance, supp$(45)=\{3,5\}$. If $x\in\mathbb{Q}$ and $p$ is a prime number, the $p$-adic valuation $v_{p}(x)$ of $x$ is the unique $r\in\mathbb{Z}\cup\{\infty\}$ such that $x=p^{r}{x_{1}}/{x_{2}}$ with $x_{1},x_{2}\notin p\mathbb{Z}$ coprime integers. For $z\in\mathbb{Q}$, its numerator and denominator will be denoted by $\operatorname{num}(z)$ and $\operatorname{den}(z)$, respectively. They are the unique coprime integers such that $\operatorname{den}(z)\geq 1$ and $z=\operatorname{num}(z)/\operatorname{den}(z)$.

As usual, the cardinality of a finite set $E$ will be denoted by $|E|$.

2. Basic results over $\mathbb{Q}$

2.1. Constraints on denominators

The aim of this section is to show that if $\varphi_{c}$ has a periodic point of period at least $3$, then $\operatorname{den}(c)$ is divisible by $16$. The result below first appeared in [11].

Proposition 2.1**.**

Let $c\in\mathbb{Q}$. If $\operatorname{Per}(\varphi_{c})\not=\emptyset$, then $\operatorname{den}(c)=d^{2}$ for some $d\in\mathbb{N}$, and $\operatorname{den}(x)=d$ for all $x\in\operatorname{Preper}(\varphi_{c})$.

Proof.

Let $p$ be a prime dividing $\operatorname{den}(c)$, i.e. such that $v_{p}(c)<0$. Let $x\in\mathbb{Q}$.

Claim. If $v_{p}(x)\not=v_{p}(c)/2$, then the orbit of $x$ under $\varphi_{c}$ is infinite.

Indeed, consider the following two cases.

1. (1)

If $v_{p}(x)<v_{p}(c)/2$, then $v_{p}(\varphi_{c}(x))=v_{p}(x^{2}-c)=2v_{p}(x)$. Thus $v_{p}(\varphi_{c}(x))<v_{p}(c)<v_{p}(c)/2$ since $v_{p}(c)<0$. It follows that $v_{p}(\varphi_{c}^{n}(x))=2^{n}v_{p}(x)$ for all $n\geq 1$. 2. (2)

If $v_{p}(x)>v_{p}(c)/2$, then $v_{p}(\varphi_{c}(x))=v_{p}(x^{2}-c)=v_{p}(c)<v_{p}(c)/2$ and we are back in the preceding case. In particular, we have $v_{p}(\varphi_{c}^{n}(x))=2^{n-1}v_{p}(c)$ for all $n\geq 1$.

In both cases, the $p$-adic valuation of $\varphi_{c}^{n}(x)$ tends to $-\infty$ for $n\to\infty$, whence the claim.

If now $x\in\operatorname{Preper}(\varphi_{c})$, then the claim implies $v_{p}(x)=v_{p}(c)/2$. Note that such points $x$ exist by hypothesis on $\varphi_{c}$. Hence $v_{p}(c)$ is even, and since this occurs for all primes $p$ dividing $\operatorname{den}(c)$, it follows that $\operatorname{den}(c)=d^{2}$ for some $d\in\mathbb{N}$, and that $\operatorname{den}(x)=d$. ∎

Consequently, since we are only interested in rational cycles of $\varphi_{c}$ here, we shall only consider those $c\in\mathbb{Q}$ such that $\operatorname{den}(c)=d^{2}$ for some $d\in\mathbb{N}$. Moreover, we shall frequently consider the set $\operatorname{num}(\operatorname{Per}(\varphi_{c}))$ of numerators of rational periodic points of $\varphi_{c}$.

Corollary 2.2**.**

Let $c\in\mathbb{Q}$. Assume $\operatorname{Per}(\varphi_{c})\not=\emptyset$. Let $d\in\mathbb{N}$ such that $\operatorname{den}(c)=d^{2}$. Then

[TABLE]

Proof.

Directly follows from the equality $\operatorname{den}(\operatorname{Preper}(\varphi_{c}))=\{d\}$ given by Proposition 2.1. ∎

2.2. Basic remarks on periodic points

In this section, we consider periodic points of any map $f\colon A\to A$ where $A$ is a domain.

Lemma 2.3**.**

Let $A$ be a commutative unitary ring and $f\colon A\to A$ a self map. Let $z_{1}\in A$ be a periodic point of $f$ of period $n$, and let $\{z_{1},\dots,z_{n}\}$ be the orbit of $z_{1}$. Then

[TABLE]

Proof.

We have $f(z_{i})=z_{i+1}$ for all $1\leq i<n$ and $f(z_{n})=z_{1}$. Hence

[TABLE]

∎

Proposition 2.4**.**

Let $A$ be a domain and $f\colon A\to A$ a map of the form $f(x)=x^{2}-c$ for some $c\in A$. Assume that $f$ admits a cycle in $A$.

$(i)$* Let $x,y\in A$ be two distinct periodic points of $f$, of period $m$ and $n$, respectively. Let $r=\emph{lcm}(m,n)$. Then $\displaystyle\prod_{i=0}^{r-1}(f^{i}(x)+f^{i}(y))\,=\,1.$*

$(ii)$* Assume $\operatorname{Per}(f)=\{x_{1},x_{2},\ldots,x_{N}\}$. Then $\displaystyle\prod_{1\leq i<j\leq N}(x_{i}+x_{j})\,=\,\pm 1.$*

Proof.

First observe that for all $u,v\in A$, we have

[TABLE]

$(i)$ Since $f^{r}(x)=x$ and $f^{r}(y)=y$, we have

[TABLE]

Now, it follows from (1) that

[TABLE]

Since the right-hand side of (2) is nonzero, the formula in $(i)$ follows.

Moreover, since $f$ permutes $\operatorname{Per}(f)$, we have

[TABLE]

Using (1), and since the above terms are nonzero, the formula in $(ii)$ follows. ∎

2.3. Sums of periodic points

Here are straightforward consequences of Proposition 2.4 for $\varphi_{c}$. The result below originally appeared in [3].

Proposition 2.5**.**

Let $c\in\mathbb{Q}$. Assume $\operatorname{Per}(\varphi_{c})=\{x_{1},x_{2},\ldots,x_{n}\}$ with $n\geq 1$. Let $d=\operatorname{den}(x_{1})$ and $X_{i}=\operatorname{num}(x_{i})$ for all $1\leq i\leq n$. Then, for all $1\leq k\leq n-1$, we have

[TABLE]

Proof.

By Proposition 2.1, we have $\operatorname{den}(x_{i})=d$ for all $i$. Now chase the denominator in the formulas of Proposition 2.4. ∎

These other consequences will play a crucial role in the sequel.

Corollary 2.6**.**

Let $c\in\mathbb{Q}$. Let $x,y$ be two distinct points in $\operatorname{Per}{(\varphi_{c})}$. Set $X=\operatorname{num}(x),Y=\operatorname{num}(y)$ and $d=\operatorname{den}(x)$. Then

1. (i)

$\operatorname{supp}(X+Y)\subseteq\operatorname{supp}(d)$. That is, any prime $p$ dividing $X+Y$ also divides $d$. 2. (ii)

$X$* and $Y$ are coprime.* 3. (iii)

If no odd prime factor of $d$ divides $X+Y$, then $X+Y=\pm 2^{t}$ for some $t\in\mathbb{N}$.

Proof.

The first point directly follows from equality (4). For the second one, if a prime $p$ divides $X$ and $Y$, then it divides $d$ by the first point, a contradiction since $X,d$ are coprime. The last point follows from the first one and the hypothesis on the odd factors of $d$, which together imply $\operatorname{supp}(X+Y)\subseteq\{2\}$. ∎

Example 2.7**.**

Consider the case $c=29/16$ of Example 1.2, where $d=4$ and $\varphi_{c}$ admits the cycle $\mathcal{C}=(-1/4,-7/4,5/4)$. Here $\operatorname{num}(\mathcal{C})=(-1,-7,5)$, with pairwise sums $-8,-2,4$, respectively. This illustrates all three statements of Corollary 2.6. Viewing $\mathcal{C}$ as a set, we have $\mathcal{C}\subseteq\operatorname{Per}(\varphi_{c})$. We claim $\mathcal{C}=\operatorname{Per}(\varphi_{c})$. For otherwise, let $x=X/4$ be yet another periodic point of $\varphi_{c}$. Then $X-1,X-7,X+5$ would also be powers of $2$ up to sign. The only possibility is $X=3$ as easily seen. But $3/4$ is only a preperiodic point, since under $\varphi_{c}$ we have $3/4\mapsto-5/4\mapsto-1/4\mapsto-7/4\mapsto 5/4\mapsto-1/4.$

2.4. Divisibility properties of $\operatorname{den}(c)$

Our bounds on cycle lengths of $\varphi_{c}$ involve the denominator of $c$. The following proposition and corollary already appear in [11].

Proposition 2.8**.**

Let $c\in\mathbb{Q}$. If $\operatorname{den}(c)$ is odd, then $|\operatorname{Per}(\varphi_{c})|\leq 2$.

Proof.

We have $\operatorname{den}(c)=d^{2}$ for some $d\in\mathbb{N}$, and $\operatorname{den}(x)=d$ for all $x\in\operatorname{Preper}(\varphi_{c})$. Assume $\operatorname{Per}(\varphi_{c})=\{x_{1},\dots,x_{n}\}$. Let $X_{i}=\operatorname{num}(x_{i})$ for all $i$. Then by equality (4) in Proposition 2.5, we have

[TABLE]

Since $d$ is odd by assumption, each factor $X_{i}+X_{j}$ is odd as well, whence $X_{i}\not\equiv X_{j}\bmod 2$ for all $1\leq i<j\leq n$. Of course, this is only possible if $n\leq 2$. ∎

Remark 2.9**.**

If $c\in\mathbb{Z}$, then $\operatorname{den}(c)=1$ and the above result implies that $\varphi_{c}$ admits at most two periodic points.

Corollary 2.10** ([11]).**

Let $c\in\mathbb{Q}$. If $\varphi_{c}$ admits a rational cycle of length at least $3$, then $\operatorname{den}(c)$ is even.

2.5. Involving $p$-adic numbers

We shall now improve Corollary 2.10 by showing that under the same hypotheses, $\operatorname{den}(c)$ must in fact be divisible by $16$. For that, we shall need Morton’s Theorem 1.3 excluding period $4$, as well as a result below due to Zieve concerning periodic points of polynomials over the $p$-adic integers.

As usual, $\mathbb{Z}_{p}$ and $\mathbb{Q}_{p}$ will denote the rings of $p$-adic integers and numbers, respectively. A result in [1] contains a generalization of the above proposition. It says that any polynomial $g(x)=x^{p}+\alpha$ with $\alpha\in\mathbb{Z}_{p}$, either admits $p$ fixed points in $\mathbb{Q}_{p}$ or else a cycle of length exactly $p$ in $\mathbb{Q}_{p}$. For $z\in\mathbb{Q}_{p}$, we denote by $v_{p}(z)$ the $p$-adic valuation of $z$.

Here is a particular case of a theorem of Zieve [12] that we shall use to improve Corollary 2.10. See also [9, Theorem 2.21 p. 62].

Theorem 2.11**.**

Let $p$ be a prime number and let $g$ be a polynomial in $\mathbb{Z}_{p}[t]$ of degree at least $2$. Let $\alpha$ be a periodic point of $g$ in $\mathbb{Z}_{p}$ and let

[TABLE]

Then $n\in\{m,mr,mrp^{e}\}$ for some integer $e\geq 1$ such that $p^{e-1}\leq 2/(p-1).$

We may now sharpen Corollary 2.10.

Theorem 2.12**.**

Let $c\in\mathbb{Q}$. If $\varphi_{c}$ admits a rational cycle of length $n\geq 3$, then $\operatorname{den}(c)$ is divisible by $16$.

Proof.

By Propositions 2.1 and 2.8, we have $\operatorname{den}(c)=d^{2}$ for some even positive integer $d$. Assume for a contradiction that $d$ is not divisible by $4$. Hence $v_{2}(d)=1$ and $v_{2}(c)=-2$. Let $\mathcal{C}\subseteq\operatorname{Per}(\varphi_{c})$ be a rational cycle of $\varphi_{c}$ of length $n\geq 3$. For all $z\in\mathcal{C}$, we have $\operatorname{den}(z)=d$ and hence $v_{2}(z)=-1$ by Proposition 2.1.

Recall that, if $z_{1},z_{2}\in\mathbb{Q}$ satisfy $v_{2}(z)=v_{2}(z^{\prime})=r$ for some $r\in\mathbb{Z}$, then $v_{2}(z\pm z^{\prime})\geq r+1$.

In particular, for all $z\in\mathcal{C}$, we have $v_{2}(z-1/2)\geq 0$. Therefore the translate $\mathcal{C}-1/2$ of $\mathcal{C}$ may be viewed as a subset of the local ring $\mathbb{Z}_{(2)}\subset\mathbb{Q}$, and hence of the ring $\mathbb{Z}_{2}$ of $2$-adic integers. That is, we have

[TABLE]

Step 1. In view of applying Theorem 2.11, we seek a polynomial in $\mathbb{Z}_{2}[t]$ admitting $\mathcal{C}-1/2$ as a cycle. The polynomial

[TABLE]

will do. Indeed, by construction we have

[TABLE]

Since $\varphi_{c}(\mathcal{C})=\mathcal{C}$, it follows that

[TABLE]

as desired. For the constant coefficient of $f$, we claim that $v_{2}(c+1/4)\geq 0$. Indeed, let $x,y\in\mathcal{C}$ with $y=\varphi_{c}(x)$. Thus $f(x-1/2)=y-1/2$, i.e.

[TABLE]

Since $v_{2}(x-1/2),v_{2}(y-1/2)\geq 0$, it follows that $v_{2}(c+1/4)\geq 0$, as claimed. Therefore $f(t)\in\mathbb{Z}_{2}[t]$, as desired.

For the next step, we set

[TABLE]

with $f(z_{i})=z_{i+1}$ for $i\leq n-1$ and $f(z_{n})=z_{1}$.

Step 2. We now apply Theorem 2.11 to the polynomial $g=f$ and to its $n$-periodic point $\alpha=z_{1}$. We need to compute the corresponding numbers $m$ and $r$ in that theorem, where $m$ is the period of $z_{1}$ in $\mathbb{Z}/2\mathbb{Z}$.

We claim that $m=1$. By Lemma 2.3, for the cycle $(z_{1},z_{2},\dots,z_{n})$ of $f$, we have

[TABLE]

Since $f(x)-f(y)=(x-y)(x+y+1)$ for all $x,y$, this yields

[TABLE]

Therefore $v_{2}(z_{i}+z_{j}+1)=0$ for all $1\leq i<j\leq n$, which in turn implies $v_{2}(z_{i}-z_{j})\geq 1$ for all $i<j$. Consequently, the cycle $(z_{1},z_{2},\dots,z_{n})$ collapses to the cycle $(z_{1})$ of length $1$ in $\mathbb{Z}/2\mathbb{Z}$. This settles the claim.

Since $m=1$, we have $(f^{m})^{\prime}(t)=f^{\prime}(t)=2t+1$ in $\mathbb{Z}_{2}[t]$, whence $f^{\prime}(z_{1})=1$ in $\mathbb{Z}/2\mathbb{Z}$. Therefore $r=1$ by definition.

By Theorem 2.11, it follows that $n\in\{1,2^{e}\}$ for some integer $e\geq 1$ such that $2^{e-1}\leq 2/1$. Hence $e\leq 2$ and so $n\in\{1,2,4\}$. Since $n\geq 3$ by assumption, it follows that $n=4$. But period 4 for $\varphi_{c}$ is excluded by Morton’s Theorem 1.3. This contradiction concludes the proof of the theorem. ∎

Remark 2.13**.**

Theorem 2.12 is best possible, as witnessed by Example 1.2 where period 3 occurs for $\varphi_{c}$ with $c=29/16$.

3. An upper bound on $|\operatorname{Per}(\varphi_{c})|$

Let $c\in\mathbb{Q}$. Throughout this section, we assume $\operatorname{den}(c)=d^{2}$ with $d\in 4\mathbb{N}$. Recall that this is satisfied whenever $\varphi_{c}$ admits a rational cycle $\mathcal{C}$ of length $n\geq 3$, as shown by Proposition 2.1 and Theorem 2.12.

Let $s=|\operatorname{supp}(d)|$. The following upper bound on $|\operatorname{Preper}(\varphi_{c})|$ was shown in [2]:

[TABLE]

Our aim in this section is to obtain an analogous upper bound on $|\operatorname{Per}(\varphi_{c})|$, namely

[TABLE]

The proof will follow from a string of modular constraints on the numerators of periodic points of $\varphi_{c}$ developped in this section.

3.1. Constraints on numerators

We start with an easy observation.

Lemma 3.1**.**

Let $c=a/d^{2}\in\mathbb{Q}$ with $a,d$ coprime integers. Let $x\in\operatorname{Preper}(\varphi_{c})$. Let $X=\operatorname{num}(x)$. Then $X^{2}\equiv a\bmod d$.

Proof.

We have $x=X/d$ by Proposition 2.1. Let $z=\varphi_{c}(x)$. Then $z\in\operatorname{Preper}(\varphi_{c})$, whence $z=Z/d$ where $Z=\operatorname{num}(z)$. Now $z=x^{2}-c=(X^{2}-a)/d^{2}$, whence

[TABLE]

Since $Z$ is an integer, it follows that $X^{2}\equiv a\bmod d.$ ∎

Here is a straightforward consequence.

Proposition 3.2**.**

Let $c\in\mathbb{Q}$ such that $\operatorname{den}(c)=d^{2}$ with $d\in 4\mathbb{N}$. Let $X,Y\in\operatorname{num}(\operatorname{Preper}(\varphi_{c}))$. Let $p\in\operatorname{supp}(d)$ and $r=v_{p}(d)$ the $p$-valuation of $d$. Then

[TABLE]

In particular, $\operatorname{num}(\operatorname{Preper}(\varphi_{c}))$ reduces to at most two opposite classes mod $p^{r}$.

Proof.

It follows from Lemma 3.1 that $X^{2}\equiv Y^{2}\bmod d.$ Hence

[TABLE]

Case 1. Assume $p$ is odd. Then $p$ cannot divide both $X+Y$ and $X-Y$, for otherwise it would divide $X$ which is impossible since $X$ is coprime to $d$. Therefore $p^{r}$ divides $X+Y$ or $X-Y$, as desired.

Case 2. Assume $p=2$. Then $r\geq 2$ by hypothesis. Let $x^{\prime}=\varphi_{c}(x)=X^{\prime}/d$ and $y^{\prime}=\varphi_{c}(y)=Y^{\prime}/d$. Then $X^{\prime},Y^{\prime}$ are odd since coprime to $d$. By (6), we have $X^{\prime}=(X^{2}-a)/d$ and $Y^{\prime}=(Y^{2}-a)/d$. Hence

[TABLE]

Since $2^{r}$ divides $d$ and since $X^{\prime}-Y^{\prime}$ is even, it follows that

[TABLE]

Now $4$ cannot divide both $X+Y$ and $X-Y$ since $X,Y$ are odd. Therefore $X+Y\equiv 0\bmod 2^{r}$ or $X-Y\equiv 0\bmod 2^{r}$, as desired. ∎

Corollary 3.3**.**

Let $c\in\mathbb{Q}$ such that $\operatorname{den}(c)=d^{2}$ with $d\in 4\mathbb{N}$. Let $s=|\operatorname{supp}(d)|$. Then $\operatorname{num}(\operatorname{Preper}(\varphi_{c}))$ reduces to at most $2^{s}$ classes mod $d$.

Proof.

Set $\operatorname{supp}(d)=\{p_{1},\dots,p_{s}\}$ and $d=p_{1}^{r_{1}}\dots p_{s}^{r_{s}}$. By Proposition 3.2, the set $\operatorname{num}(\operatorname{Preper}(\varphi_{c}))$ covers at most 2 distinct classes mod $p_{i}^{r_{i}}$ for all $1\leq i\leq s$. Therefore, by the Chinese Remainder Theorem, this set covers at most $2^{s}$ distinct classes mod $d$. ∎

The particular case in Proposition 3.2 where $X,Y\in\operatorname{num}(\operatorname{Per}(\varphi_{c}))$ and $X\equiv+Y\bmod p^{r}$ for all $p\in\operatorname{supp}(d)$, i.e. where $X\equiv Y\bmod d$, has a somewhat surprising consequence and will be used more than once in the sequel.

Proposition 3.4**.**

Let $c\in\mathbb{Q}$ such that $\operatorname{den}(c)=d^{2}$ with $d\in 4\mathbb{N}$. Let $X,Y\in\operatorname{num}(\operatorname{Per}(\varphi_{c}))$ be distinct. If $X\equiv Y\bmod d$, then $X+Y=\pm 2$.

Proof.

As $X,Y$ are coprime to $d$, they are odd. We claim that $\operatorname{supp}(X+Y)=\{2\}$. Indeed, let $p$ be any prime factor of $X+Y$. Then $p$ divides $d$ by Corollary 2.6. Hence $p$ divides $X-Y$ since $d$ divides $X-Y$ by hypothesis. Therefore $p$ divides $2X$, whence $p=2$ since $X$ is odd. It follows that $X+Y=\pm 2^{t}$ for some integer $t\geq 1$. Since $d\in 4\mathbb{N}$ and $d$ divides $X-Y$, it follows that $4$ divides $X-Y$. Hence $4$ cannot also divide $X+Y$ since $X,Y$ are odd. Therefore $t=1$, i.e. $X+Y=\pm 2$ as desired. ∎

Example 3.5**.**

Consider the case $c=29/16$ of Example 1.2, where $\varphi_{c}$ admits the cycle $\mathcal{C}=(-1/4,-7/4,5/4)$. In $\operatorname{num}(\mathcal{C})=(-1,-7,5)$, only $-7$ and $5$ belong to the same class mod $4$, and their sum is $-2$ as expected.

3.2. From $\mathbb{Z}/d\mathbb{Z}$ to $\mathbb{Z}$

Our objective now is to derive from Proposition 3.2 the upper bound $|\operatorname{Per}(\varphi_{c})|\leq 2^{s}+2$ announced earlier. For that, we shall need the following two auxiliary results.

Lemma 3.6**.**

Let $k\in\mathbb{N}$. Up to order, there are only two ways to express $2^{k}$ as $2^{k}=\varepsilon_{1}2^{k_{1}}+\varepsilon_{2}2^{k_{2}}$ with $\varepsilon_{1},\varepsilon_{2}=\pm 1$ and $k_{1},k_{2}\in\mathbb{N}$.

Proof.

We may assume $k_{1}\leq k_{2}$. There are two cases.

1. (1)

If $k_{1}=k_{2}$, then $2^{k_{1}}(\varepsilon_{1}+\varepsilon_{2})=2^{k}$, implying $k_{1}=k_{2}=k-1$ and $\varepsilon_{1}=\varepsilon_{2}=1$. 2. (2)

If $k_{1}<k_{2}$, then $2^{k_{1}}(\varepsilon_{1}+\varepsilon_{2}2^{k_{2}-k_{1}})=2^{k}$, implying $k=k_{1}=k_{2}-1$, $\varepsilon_{1}=-1$ and $\varepsilon_{2}=1$. ∎

Proposition 3.7**.**

Let $c\in\mathbb{Q}$ such that $\operatorname{den}(c)=d^{2}$ with $d\in 4\mathbb{N}$. If there are distinct pairs $\{X_{1},Y_{1}\},\{X_{2},Y_{2}\}\subseteq\operatorname{num}(\operatorname{Per}(\varphi_{c}))$ such that $X_{1}+Y_{1}=\pm(X_{2}+Y_{2})=\pm 2^{k}$ for some $k\in\mathbb{N}$, then

[TABLE]

Proof.

Assume for a contradiction that $X_{1}+Y_{1}=X_{2}+Y_{2}=\pm 2^{k}$. Let $p\in\operatorname{supp}(d)$ be odd. We claim that $X_{1},X_{2},Y_{1},Y_{2}$ all belong to the same nonzero class mod $p$. Indeed, we know by Proposition 3.2 that $X_{1},X_{2},Y_{1},Y_{2}$ belong to at most two opposite classes mod $p$. Since $p$ does not divide $X_{i}+Y_{i}$ for $1\leq i\leq 2$, i.e. $X_{i}\not\equiv-Y_{i}\bmod p$, it follows that $X_{i}\equiv Y_{i}\bmod p$. Since $X_{1}\equiv\pm X_{2}\bmod p$ and $X_{1}+Y_{1}=X_{2}+Y_{2}$, it follows that $X_{1}\equiv X_{2}\bmod p$ and the claim is proved. Therefore no sum of two elements in $\{X_{1},Y_{1},X_{2},Y_{2}\}$ is divisible by $p$. Hence, by the third point of Corollary 2.6, any sum of two distinct elements in $\{X_{1},Y_{1},X_{2},Y_{2}\}$ is equal up to sign to a power of $2$. Moreover, we have

[TABLE]

It now follows from Lemma 3.6 that at least two of $X_{1},Y_{1},X_{2},Y_{2}$ are equal. This contradiction concludes the proof. ∎

Notation 3.8**.**

For any $h\in\mathbb{Z}$, we shall denote by $\pi_{h}\colon\mathbb{Z}\to\mathbb{Z}/h\mathbb{Z}$ the canonical quotient map mod $h$.

Theorem 3.9**.**

Let $c\in\mathbb{Q}$ such that $\operatorname{den}(c)=d^{2}$ with $d\in 4\mathbb{N}$. Let $m=|\pi_{d}(\operatorname{num}(\operatorname{Per}(\varphi_{c})))|$. Then

[TABLE]

Proof.

The first inequality is obvious. We now show $|\operatorname{Per}(\varphi_{c})|\leq m+2$.

Claim. Each class mod $d$ contains at most 2 elements of $\operatorname{num}(\operatorname{Per}(\varphi_{c}))$.

Assume the contrary. Then there are three distinct elements $X,Y,Z$ in $\operatorname{num}(\operatorname{Per}(\varphi_{c}))$ such that $X\equiv Y\equiv Z\bmod d$. By Proposition 3.4, all three sums $X+Y$, $X+Z$ and $Y+Z$ belong to $\{\pm 2\}$. Hence two of them coincide, e.g. $X+Y=X+Z$. Therefore $Y=Z$, a contradiction. This proves the claim.

Now, assume for a contradiction that $|\operatorname{Per}(\varphi_{c})|\geq m+3$. The claim then implies that there are at least 3 distinct classes mod $d$ each containing two distinct elements in $\operatorname{num}(\operatorname{Per}(\varphi_{c}))$. That is, there are six distinct elements $X_{1},Y_{1}$, $X_{2},Y_{2}$ and $X_{3},Y_{3}$ in $\operatorname{num}(\operatorname{Per}(\varphi_{c}))$ such that $X_{i}\equiv Y_{i}\bmod d$ for $1\leq i\leq 3$. Again, Proposition 3.4 implies $X_{i}+Y_{i}=\pm 2$ for $1\leq i\leq 3$. This situation is excluded by Proposition 3.7, and the proof is complete. ∎

Remark 3.10**.**

The above proof shows that if $|\operatorname{Per}(\varphi_{c})|=m+2$, then there are exactly two classes mod $d$ containing more than one element of $\operatorname{num}(\operatorname{Per}(\varphi_{c}))$, and both classes contain exactly two such elements. Denoting $\{X_{1},Y_{1}\},\{X_{2},Y_{2}\}\subset\operatorname{num}(\operatorname{Per}(\varphi_{c}))$ these two special pairs, the proof further shows that $X_{1}+Y_{1}=\pm 2=-(X_{2}+Y_{2})$.

Corollary 3.11**.**

Let $c\in\mathbb{Q}$ such that $\operatorname{den}(c)=d^{2}$ with $d\in 4\mathbb{N}$. Let $s=|\operatorname{supp}(d)|$. Then

[TABLE]

Proof.

We have $|\operatorname{Per}(\varphi_{c})|\,\leq\,m+2$ by the above theorem, and $m\leq 2^{s}$ by Corollary 3.3. ∎

3.3. Numerator dynamics

Let $c=a/d^{2}\in\mathbb{Q}$ with $a,d$ coprime integers. Closely related to the map $\varphi_{c}$ is the map $d^{-1}\varphi_{a}\colon\mathbb{Q}\to\mathbb{Q}$. By definition, this map satisfies

[TABLE]

for all $x\in\mathbb{Q}$. As was already implicit earlier, we now show that cycles of $\varphi_{c}$ in $\mathbb{Q}$ give rise, by taking numerators, to cycles of $d^{-1}\varphi_{a}$ in $\mathbb{Z}$.

Lemma 3.12**.**

Let $c=a/d^{2}\in\mathbb{Q}$ with $a,d$ coprime integers. Let $\mathcal{C}\subset\mathbb{Q}$ be a cycle of $\varphi_{c}$. Then $\operatorname{num}(\mathcal{C})\subset\mathbb{Z}$ is a cycle of $d^{-1}\varphi_{a}$ of length $|\mathcal{C}|$.

Proof.

Recall that $\operatorname{den}(\mathcal{C})=\{d\}$ by Proposition 2.1. Let $x\in\mathcal{C}$ and $y=\varphi_{c}(x)$. Let $X=\operatorname{num}(x),Y=\operatorname{num}(y)$. Then $x=X/d,Y=y/d$. We have $y=x^{2}-c=(X^{2}-a)/d^{2}$. Hence $Y=(X^{2}-a)/d=d^{-1}\varphi_{a}(X)$. In particular, we have the formula

[TABLE]

for all $X\in\operatorname{num}(\mathcal{C})$. ∎

3.4. The cases $d\not\equiv 0$ mod 3 or mod 5

Lemma 3.13**.**

Let $c\in\mathbb{Q}$ and $\mathcal{C}\subseteq\operatorname{Per}(\varphi_{c})$ a cycle of positive length $n$.

(i)* If $d\not\equiv 0\bmod 3$ and $n\geq 3$, then $\operatorname{num}(\mathcal{C})$ reduces mod $3$ to exactly one nonzero element.*

(ii)* If $d\not\equiv 0\bmod 5$ and $n\geq 4$, then $\operatorname{num}(\mathcal{C})$ reduces mod $5$ to exactly one or two nonzero elements mod $5$.*

Proof.

First some preliminaries. Of course $\varphi_{c}$ induces a cyclic permutation of $\mathcal{C}$. By Proposition 2.1, we have $c=a/d^{2}$ with $a,d$ coprime integers. By Lemma 3.12, the rational map $d^{-1}\varphi_{a}$ induces a cyclic permutation of $\operatorname{num}(\mathcal{C})$, say

[TABLE]

Let $X,Y\in\operatorname{num}(\mathcal{C})$ be distinct. Then $\operatorname{supp}(X+Y)\subseteq\operatorname{supp}(d)$ by Corollary 2.6. In particular, let $q$ be any prime number such that $d\not\equiv 0\bmod q$. Then

[TABLE]

Since $d$ is invertible mod $q$, we may consider the reduced map

[TABLE]

where $f(x)=d^{-1}(x^{2}-a)$ for all $x\in\mathbb{Z}/q\mathbb{Z}$. Thus, we may view $\pi_{q}(\operatorname{num}(\mathcal{C}))$ as a sequence of length $n$ in $\mathbb{Z}/q\mathbb{Z}$, where each element is cyclically mapped to the next by $f$. Note that (8) implies that this $n$-sequence does not contain opposite elements $u,-u$ of $\mathbb{Z}/q\mathbb{Z}$, and in particular contains at most one occurrence of [math].

We are now ready to prove statements (i) and (ii).

(i) Assume $d\not\equiv 0\bmod q$ where $q=3$. By the above, the $n$-sequence $\pi_{3}(\operatorname{num}(\mathcal{C}))$ consists of at most one [math] and all other elements equal to some $u\in\{\pm 1\}$. Since $n\geq 3$, this $n$-sequence contains two cyclically consecutive occurrences of $u$. Therefore $f(u)=u$. Hence $\pi_{3}(\operatorname{num}(\mathcal{C}))$ contains $u$ as it unique element repeated $n$ times.

(ii) Assume $d\not\equiv 0\bmod q$ where $q=5$. Since $n\geq 4$ and the $n$-sequence $\pi_{5}(\operatorname{num}(\mathcal{C}))$ contains at most one [math], it must contains three cyclically consecutive nonzero elements $u_{1},u_{2},u_{3}\in\mathbb{Z}/5\mathbb{Z}\setminus\{0\}$. Since that set contains at most two pairwise non-opposite elements, it follows that $u_{i}=u_{j}$ for some $1\leq i<j\leq 3$. Now $u_{1}\mapsto u_{2}\mapsto u_{3}$ by $f$. Therefore, if either $u_{1}=u_{2}$ or $u_{2}=u_{3}$, it follows that the whole sequence $\pi_{5}(\operatorname{num}(\mathcal{C}))$ consists of the one single element $u_{2}$ repeated $n$ times. On the other hand, if $u_{1}\not=u_{2}$, then $u_{1}=u_{3}$. In this case, the $n$-sequence $\pi_{5}(\operatorname{num}(\mathcal{C}))$ consists of the sequence $u_{1},u_{2}$ repeated $n/2$ times. This concludes the proof. ∎

Example 3.14**.**

Consider the case $c=a/d^{2}=29/16$ of Example 1.2, where $\varphi_{c}$ admits the cycle $\mathcal{C}=(-1/4,-7/4,5/4)$. Then $\operatorname{num}(\mathcal{C})=(-1,-7,5)$, a cycle of length $3$ of the map $d^{-1}\varphi_{a}=4^{-1}\varphi_{29}$. That cycle reduces mod $3$ to $(-1,-1,-1)$, as expected with statement (i) of the lemma. Statement (ii) does not apply since $n=3$, and it would fail anyway since $\operatorname{num}(\mathcal{C})$ reduces mod $5$ to the sequence $(-1,-2,0)$.

3.5. Main consequences

Proposition 3.15**.**

Let $c=a/d^{2}\in\mathbb{Q}$ with $a,d$ coprime integers and with $d\in 4\mathbb{N}$. Assume $d\not\equiv 0\bmod 3$. Let $s=|\operatorname{supp}(d)|$. For every rational cycle $\mathcal{C}$ of $\varphi_{c}$, we have

[TABLE]

Proof.

By Corollary 3.11, we have $|\mathcal{C}|\,\leq\,2^{s}+2$. If $|\mathcal{C}|=2^{s}+2$ then, by Remark 3.10, there exist two pairs $\{X_{1},Y_{1}\},\{X_{2},Y_{2}\}$ in $\operatorname{num}(\mathcal{C})$ such that $X_{1}+Y_{1}=2$ and $X_{2}+Y_{2}=-2$. Since $d\not\equiv 0\bmod 3$, Lemma 3.13 implies that $X_{1},X_{2},Y_{1},Y_{2}$ reduce to the same nonzero element $u$ mod $3$. This contradicts the equality $X_{1}+Y_{1}=-(X_{2}+Y_{2})$. ∎

Theorem 3.16**.**

If $\operatorname{den}(c)$ admits at most two distinct prime factors, then $\varphi_{c}$ satisfies the Flynn-Poonen-Schaefer conjecture.

Proof.

Let $\mathcal{C}$ be a rational cycle of $\varphi_{c}$ of length $n\geq 3$. Then $d$ is even and hence $s\geq 1$.

$\bullet$ If $s=1$, then $d$ is a power of $2$. By Corollary 3.15, we have $|\operatorname{Per}(\varphi_{c})|\,\leq 2^{1}+1=3$ and $|\mathcal{C}|\leq 3$. See also [3].

$\bullet$ Assume now $s=2$. Then $d=2^{2r_{1}}p^{r_{2}}$ where $p$ is an odd prime. By Theorem 3.9, we have $|\mathcal{C}|\leq|\operatorname{Per}(\varphi_{c})|\leq 6$. By Theorems 1.3 and 1.4, we have $|\mathcal{C}|\not=4,5$. It remains to show $|\mathcal{C}|\not=6$. We distinguish two cases. If $p\not=3$, then $|\mathcal{C}|\leq 2^{2}+1=5$ by Corollary 3.15 and we are done. Assume now $p=3$, so that $d=2^{2r_{1}}3^{r_{2}}$. Let $m$ denote the number of classes of $\operatorname{num}(\mathcal{C})$ mod $q=5$. It follows from Lemma 3.13 that $m\leq 2$. Since the order of every element in $(\mathbb{Z}/5\mathbb{Z})^{*}$ belongs to $\{1,2,4\}$, it follows from Zieve’s Theorem 2.11 that $|\mathcal{C}|$ is a power of $2$. Hence $|\mathcal{C}|\in\{1,2,4\}$ and we are done. ∎