Proof. We have $f(z_i)=z_{i+1}$ for all $1 \le i <n$ and $f(z_n)=z_1$ . Hence,

$$ \begin{align*} \prod_{1\leq i<j\leq n}(f(z_i)-f(z_j)) &= \prod_{1\leq i<j< n}(z_{i+1}-z_{j+1})\prod_{1\leq i< n}(z_{i+1}-z_{1}) \\ &= (-1)^{n-1}\prod_{1\leq i<j\le n}(z_{i}-z_{j}).\\[-3.6pc] \end{align*} $$

▪

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

(2.1) $$ \begin{align} f(u)-f(v) = (u-v)(u+v). \end{align} $$

Because $f^r(x)=x$ and $f^r(y)=y$ , we have

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

Now, it follows from (2.1) that

$$ \begin{align*}f^{i+1}(x)-f^{i+1}(y) = (f^{i}(x)-f^{i}(y))(f^{i}(x)+f^{i}(y)). \end{align*} $$

Because the right-hand side of (2.2) is nonzero, the formula in (i) follows.

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

$$ \begin{align*}\prod_{1\leq i<j\leq n}(f(x_i)-f(x_j)) = \, \pm \prod_{1\leq i<j\leq n} (x_i- x_j). \end{align*} $$

Using (2.1), and because the above terms are nonzero, the formula in (ii) follows.▪

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

Proof. The first point directly follows from equality (2.3). For the second one, if a prime p divides X and Y, then it divides d by the first point, a contradiction because $X,d$ are coprime. The last point follows from the first one and the hypothesis on the odd factors of d, which together imply $\operatorname {\mathrm {supp}}(X+Y) \subseteq \{2\}$ .▪

Proof. We have $\operatorname {\mathrm {den}}(c)=d^2$ for some $d \in \mathbb {N}$ , and $\operatorname {\mathrm {den}}(x)=d$ for all $x \in \operatorname {\mathrm {Preper}}(\varphi _c)$ . Assume $\operatorname {\mathrm {Per}}(\varphi _c)=\{x_1,\dots ,x_n\}$ . Let $X_i=\operatorname {\mathrm {num}}(x_i)$ for all i. Then, by Proposition 2.5, we have

$$ \begin{align*}\prod_{1\leq i<j\leq n} \big(X_i+X_j\big) = \pm d^{n(n-1)/2}.\end{align*} $$

Because 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 $1 \le i < j \le n$ . Of course, this is only possible if $n \le 2$ .▪

Proof. By Propositions 2.1 and 2.8, we have $\operatorname {\mathrm {den}}(c)=d^2$ 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 $\mathcal {C} \subseteq \operatorname {\mathrm {Per}}(\varphi _c)$ be a rational cycle of $\varphi _c$ of length $n \ge 3$ . For all $z \in \mathcal {C}$ , we have $\operatorname {\mathrm {den}}(z)=d$ , and hence $v_2(z)=-1$ by Proposition 2.1.

Recall that if $z_1,z_2 \in \mathbb {Q}$ satisfy $v_2(z)=v_2(z')=r$ for some $r \in \mathbb {Z}$ , then $v_2(z \pm z') \ge r+1$ .

In particular, for all $z \in \mathcal {C}$ , we have $v_2(z-1/2) \ge 0$ . 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

$$ \begin{align*}\mathcal{C}-1/2 \ \subset \ \mathbb{Z}_2. \end{align*} $$

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

$$ \begin{align*} f(t) & = \varphi_c(t+1/2)-1/2 \\ & = t^2+t-(c+1/4) \end{align*} $$

will do. Indeed, by construction, we have

$$ \begin{align*}f(t-1/2)=\varphi_c(t)-1/2. \end{align*} $$

Because $\varphi _c(\mathcal {C})=\mathcal {C}$ , it follows that

$$ \begin{align*}f(\mathcal{C}-1/2) = \mathcal{C}-1/2, \end{align*} $$

as desired. For the constant coefficient of f, we claim that $v_2(c+1/4) \ge 0$ . Indeed, let $x,y \in \mathcal {C}$ with $y = \varphi _c(x)$ . Thus, $f(x-1/2)=y-1/2$ , i.e.,

$$ \begin{align*}(x-1/2)^2+(x-1/2)-(c+1/4) = y-1/2. \end{align*} $$

Because $v_2(x-1/2), v_2(y-1/2) \ge 0$ , it follows that $v_2(c+1/4) \ge 0$ , as claimed. Therefore, $f(t) \in \mathbb {Z}_2[t]$ , as desired.

For the next step, we set

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

with $f(z_i)=z_{i+1}$ for $i \le n-1$ and $f(z_n)=z_1$ .

Step 2. By Theorem 2.11, applied to the polynomial $g=f$ and to its n-periodic point $\alpha =z_1$ , we have $n \in \{1,2,4\}$ . Because $n \ge 3$ 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.▪

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

(3.1) $$ \begin{align} Z=(X^2-a)/d. \end{align} $$

Because Z is an integer, it follows that $X^2 \equiv a \bmod d$ .▪

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

$$ \begin{align*}(X+Y)(X-Y) \equiv 0 \bmod p^r. \end{align*} $$

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, because X is coprime to d. Therefore, $p^r$ divides $X+Y$ or $X-Y$ , as desired.

Case 2. Assume $p=2$ . Then, $r \ge 2$ by hypothesis. Let $x=X/d, y=Y/d \in \operatorname {\mathrm {Preper}}(\varphi _c)$ . Let $x'=\varphi _c(x)=X'/d$ and $y'=\varphi _c(y)=Y'/d$ . Then, $X',Y'$ are odd because coprime to d. By (3.1), we have $X'=(X^2-a)/d$ and $Y'=(Y^2-a)/d$ . Hence,

$$ \begin{align*}X'-Y'=(X^2-Y^2)/d. \end{align*} $$

Because $2^r$ divides d and because $X'-Y'$ is even, it follows that

$$ \begin{align*}(X+Y)(X-Y) \equiv 0 \bmod 2^{r+1}. \end{align*} $$

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

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

Proof. We may assume $k_1 \le k_2$ . 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 $\varepsilon _2=1$ .▪

Proof. Assume for a contradiction that $X_1+Y_1 = X_2+Y_2 = \pm 2^k$ . Let $p \in \operatorname {\mathrm {supp}}(d)$ be odd, if any such factor exists. 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. Because p does not divide $X_i+Y_i$ for $1 \le i \le 2$ , i.e., $X_i \not \equiv -Y_i \bmod p$ , it follows that $X_i \equiv Y_i \bmod p$ . Because $X_1 \equiv \pm X_2 \bmod p$ and $X_1+Y_1 = X_2+Y_2$ , it follows that $X_1 \equiv X_2 \bmod p$ and the claim is proved, i.e.,

$$ \begin{align*}X_1 \equiv X_2 \equiv Y_1 \equiv Y_2 \equiv \bmod p. \end{align*} $$

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

$$ \begin{align*} \pm 2^{k+1} & = (X_1+Y_1)+(X_2+Y_2) \\ & = (X_1+X_2)+(Y_1+Y_2) \\ & = (X_1+Y_2)+(X_2+Y_1). \end{align*} $$

It now follows from Lemma 3.6 that at least two of $X_1, Y_1, X_2,Y_2$ are equal, contradicting the hypothesis that they are pairwise distinct. Hence, $X_1+Y_1 = -(X_2+Y_2)$ , as claimed.▪

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.

Proof. The first inequality is obvious. We now show $|\operatorname {\mathrm {Per}}(\varphi _c)| \le m+2$ .

Assume the contrary. Then, there are three distinct elements $X, Y, Z$ in $\operatorname {\mathrm {num}}(\operatorname {\mathrm {Per}}(\varphi _c))$ 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 $|\operatorname {\mathrm {Per}}(\varphi _c)| \ge m+3$ . The claim then implies that there are at least three distinct classes mod d each containing two distinct elements in $\operatorname {\mathrm {num}}(\operatorname {\mathrm {Per}}(\varphi _c))$ . That is, there are six distinct elements $X_1, Y_1$ , $X_2,Y_2$ and $X_3, Y_3$ in $\operatorname {\mathrm {num}}(\operatorname {\mathrm {Per}}(\varphi _c))$ such that $X_i\equiv Y_i \bmod d$ for $1 \le i \le 3$ . Again, Proposition 3.4 implies $X_i+Y_i=\pm 2$ for $1 \le i \le 3$ . This situation is excluded by Proposition 3.7, and the proof is complete.▪

Proof. We have $|\operatorname {\mathrm {Per}}(\varphi _c)|\,\le \, m+2$ by the above theorem, and $m \le 2^s$ by Corollary 3.3.▪

Proof. Let us start with 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 $\operatorname {\mathrm {num}}(\mathcal {C})$ , say

$$ \begin{align*}d^{-1}\varphi_{a} \colon \operatorname{\mathrm{num}}(\mathcal{C}) \to \operatorname{\mathrm{num}}(\mathcal{C}). \end{align*} $$

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

(3.2) $$ \begin{align} X+Y \not\equiv 0 \bmod q. \end{align} $$

Because d is invertible mod q, the map $d^{-1}\varphi _{a}$ induces a map

(3.3) $$ \begin{align} f \colon \mathbb{Z}/q\mathbb{Z} \to \mathbb{Z}/q\mathbb{Z}, \end{align} $$

where $f(x)=d^{-1}(x^2-a)$ for all $x \in \mathbb {Z}/q\mathbb {Z}$ . Thus, we may view $\pi _q(\operatorname {\mathrm {num}}(\mathcal {C}))$ 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 (3.2) 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 $0$ .

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 $\pi _3(\operatorname {\mathrm {num}}(\mathcal {C}))$ consists of at most one $0$ and all other elements equal to some $u \in \{\pm 1\}$ . Because $n \ge 3$ , this n-sequence contains two cyclically consecutive occurrences of u. Therefore, $f(u) = u$ . Hence, $\pi _3(\operatorname {\mathrm {num}}(\mathcal {C}))$ contains u as its unique element repeated n times.

(ii) Assume $d \not \equiv 0 \bmod q$ where $q=5$ . Because $n \ge 4$ and the n-sequence $\pi _5(\operatorname {\mathrm {num}}(\mathcal {C}))$ contains at most one $0$ , it must contain three cyclically consecutive nonzero elements $u_1,u_2,u_3 \in \mathbb {Z}/5\mathbb {Z} \setminus \{0\}$ . Because that set contains at most two pairwise nonopposite elements, it follows that $u_i=u_j$ for some $1 \le i < j \le 3$ . 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 $\pi _5(\operatorname {\mathrm {num}}(\mathcal {C}))$ 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 $\pi _5(\operatorname {\mathrm {num}}(\mathcal {C}))$ consists of the sequence $u_1,u_2$ repeated $n/2$ times. This concludes the proof.▪

Proof. By Corollary 3.11, we have $|\mathcal {C}| \,\le \, 2^s+2$ . If $|\mathcal {C}| = 2^s+2$ , then, by Remark 3.10, there exist two pairs $\{X_1, Y_1\}, \{X_2, Y_2\}$ in $\operatorname {\mathrm {num}}(\mathcal {C})$ such that $X_1+Y_1=2$ and $X_2+Y_2=-2$ . Because $d \not \equiv 0 \bmod 3$ , 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 $X_1+Y_1=-(X_2+Y_2)$ .▪

Proof. Let $\mathcal {C}$ be a rational cycle of $\varphi _c$ of length $n \ge 3$ . Then, d is even, and hence $s \geq 1$ .

$\bullet $ If $s=1$ , then d is a power of $2$ . By Corollary 3.11 and Proposition 3.15, we have $ |\operatorname {\mathrm {Per}}(\varphi _c)| \, \le 4$ and $|\mathcal {C}| \le 3$ . See also Reference Eliahou and Fares[4].

$\bullet $ Assume now $s=2$ . Then, $d=2^{r_1}p^{r_2}$ where p is an odd prime and $r_1 \ge 2$ . By Theorem 3.9, we have $|\mathcal {C}| \le |\operatorname {\mathrm {Per}}(\varphi _c)| \le 6$ . 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 Proposition 3.15, and we are done. Assume now $p=3$ , so that $d=2^{r_1}3^{r_2}$ . Let m denote the number of classes of $\operatorname {\mathrm {num}}(\mathcal {C})$ mod $q=5$ . It follows from Lemma 3.13 that $m \le 2$ . Because the order of every element in $(\mathbb {Z}/ 5\mathbb {Z})^*$ belongs to $\{1, 2, 4\}$ , it follows from Zieve’s Theorem 2.12 that $|\mathcal {C}|$ is a power of $2$ . Hence, $|\mathcal {C}| \in \{1, 2, 4\}$ , and we are done.▪