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\title{On the number of residues of certain second-order linear recurrences}

\alttitle{Sur le nombre de résidus de certaines récurrences linéaires du second ordre}

%%%1
\author{\firstname{Federico} \lastname{Accossato}\CDRorcid{0009-0004-9861-0153}}

\address[1]{Department of Mathematical Sciences, Politecnico di Torino, 
Corso Duca degli Abruzzi 24, 10129 Torino, Italy}

\email{federico.accossato@polito.it}

%%%2
\author{\firstname{Carlo} \lastname{Sanna}\CDRorcid{0000-0002-2111-7596}\IsCorresp}

\address[1]{Department of Mathematical Sciences, Politecnico di Torino, 
Corso Duca degli Abruzzi 24, 10129 Torino, Italy}

\email{carlo.sanna@polito.it}



\thanks{The authors are members of GNSAGA of INdAM, and of CrypTO, the group of Cryptography and Number Theory of the Politecnico di Torino.}


\keywords{\kwd{Fractional parts of powers} \kwd{Lehmer sequences} \kwd{linear recurrences} \kwd{Pisot numbers} \kwd{primitive divisors} \kwd{residues}}

\altkeywords{\kwd{Parties fractionnaires des puissances} \kwd{suites de Lehmer} \kwd{récurrences linéaires} \kwd{nombres de Pisot} \kwd{diviseurs primitifs, résidus}}

\subjclass{11B37, 11B39, 11B50, 11K16}

\begin{DefTralics}
\newcommand{\coloneqq}{:=}
\end{DefTralics}

\begin{abstract}
 For every monic polynomial $f \in \mathbb{Z}[X]$ with $\deg(f) \geq 1$, let
 $\mathcal{L}(f)$ be the set of all linear recurrences with values in $\mathbb{Z}$ and characteristic polynomial $f$, and let
 \begin{equation*}
	 \mathcal{R}(f) \coloneqq \big\{\rho(\boldsymbol{x}; m) : \boldsymbol{x} \in \mathcal{L}(f), \, m \in \mathbb{Z}^+ \big\} ,
	\end{equation*}
 where $\rho(\boldsymbol{x}; m)$ is the number of distinct residues of $\boldsymbol{x}$ modulo $m$.

 Dubickas and Novikas proved that $\mathcal{R}(X^2 - X - 1) = \mathbb{Z}^+$.
 We generalize this result by showing that $\mathcal{R}(X^2 - a_1 X - 1) = \mathbb{Z}^+$ for every nonzero integer $a_1$.
 As a corollary, we deduce that for all integers $a_1 \geq 1$ and $k \geq 2$ there exists $\xi \in \mathbb{R}$ such that the sequence of fractional parts $\bigl(\mathrm{frac}(\xi \alpha^n)\big)_{n \geq 0}$, where $\alpha \coloneqq \big(a_1 + \sqrt{a_1^2 + 4}\,\big) / 2$, has exactly $k$ limit points.
 Our proofs are constructive and employ some results on the existence of special primitive divisors of certain Lehmer sequences.
\end{abstract}


\begin{altabstract}
 Pour chaque polynôme monique $f \in \mathbb{Z}[X]$ avec $\deg(f) \geq 1$, soit $\mathcal{L}(f)$ l'ensemble de toutes les récurrences linéaires avec des valeurs dans $\mathbb{Z}$ et un polynôme caractéristique $f$, et soit 
 \begin{equation*} 
	 \mathcal{R}(f) \coloneqq \big\{\rho(\boldsymbol{x}; m) : \boldsymbol{x} \in \mathcal{L}(f), \, m \in \mathbb{Z}^+ \big\} ,
 \end{equation*} 
où $\rho(\boldsymbol{x}; m)$ est le nombre de résidus distincts de $\boldsymbol{x}$ modulo $m$.

Dubickas et Novikas ont prouvé que $\mathcal{R}(X^2 - X - 1) = \mathbb{Z}^+$. Nous généralisons ce résultat en montrant que $\mathcal{R}(X^2 - a_1 X - 1) = \mathbb{Z}^+$ pour tout entier non nul $a_1$. Comme corollaire, nous déduisons que pour tous les entiers $a_1 \geq 1$ et $k \geq 2$, il existe $\xi \in \mathbb{R}$ tel que la séquence des parties fractionnaires $\bigl(\mathrm{frac}(\xi \alpha^n)\big)_{n \geq 0}$, où $\alpha \coloneqq \big(a_1 + \sqrt{a_1^2 + 4},\big) / 2$, a exactement $k$ points de limite. Nos preuves sont constructives et utilisent certains résultats sur l'existence de diviseurs primitifs spéciaux de certaines séquences de Lehmer.
\end{altabstract}





\begin{document}

\maketitle


\section{Introduction}

Let $a_1, \dots, a_k$ be integers.
An integer sequence $\boldsymbol{x} = (x_n)_{n\geq 0}$ is a \emph{linear recurrence} with \emph{characteristic polynomial}
\begin{equation*}
	f = X^k - a_1 X^{k - 1} - a_2 X^{k - 2} - \cdots - a_k
\end{equation*}
if for all integers $n \geq k$ we have that
\begin{equation}\label{equ:recurrence}
	x_n = a_1 x_{n - 1} + a_2 x_{n - 2} + \cdots + a_k x_{n - k} .
\end{equation}
The terms $x_0, \dots, x_{k-1}$, which together with $f$ completely determine $\boldsymbol{x}$ via~\eqref{equ:recurrence}, are the \emph{initial values} of $\boldsymbol{x}$.
We let $\mathcal{L}(f)$ denote the set of all (integral) linear recurrences with characteristic polynomial $f$.
It is easily seen that each $\boldsymbol{x} \in \mathcal{L}(f)$ is ultimately periodic modulo $m$, for every integer $m \geq 1$, and in fact (purely) periodic if $\gcd(m, a_k) = 1$.
Indeed, properties of linear recurrences modulo $m$ have been studied extensively, including: which residues modulo $m$ appear in $\boldsymbol{x}$ and how frequently~\cite{MR369240,MR3298566,MR2024599,MR1119645,MR1117027,MR1393478}, and for which values of $m$ the linear recurrence $\boldsymbol{x}$ contains a complete system of residues modulo $m$~\cite{MR3068231,MR294238,MR4357629,MR951911,MR3696266}.


We let $\tau(\boldsymbol{x}; m)$ denote the (minimal) period of $\boldsymbol{x}$ modulo $m$, that is, the minimal integer $t \geq 1$ such that $x_{n + t} \equiv x_n \pmod m$ for all sufficiently large integers $n \geq 0$.
Moreover, we let $\rho(\boldsymbol{x}; m) \coloneqq \#\big\{x_n \bmod m : n \geq 0\big\}$ be the number of distinct residues of $\boldsymbol{x}$ modulo $m$, and we put
\begin{equation*}
	\mathcal{R}(f) \coloneqq \big\{\rho(\boldsymbol{x}; m) : \boldsymbol{x} \in \mathcal{L}(f), \, m \in \mathbb{Z}^+ \big\} .
\end{equation*}
Dubickas and Novikas~\cite{MR4125906}, motivated by some problems on fractional parts of powers of Pisot numbers~\cite{MR3397268}, proved that $\mathcal{R}(X^2 - X - 1) = \mathbb{Z}^+$ and stated that it ``may be very difficult in general'' to determine $\mathcal{R}(f)$.
Sanna~\cite{MR4350306} considered the special case in which $f$ is a quadratic polynomial with roots $\alpha, \beta$ such that $\alpha\beta = \pm 1$ and $\alpha/\beta$ is not a root of unity, and proved two results.
First, that $\mathcal{R}(f)$ contains all integers $n \geq 7$ with $n \neq 10$ and $4 \nmid n$.
Second, that $4D_0 \mathbb{Z}^+ \subseteq \mathcal{R}(f)$ if $\alpha\beta = 1$, and $8D_0 \mathbb{Z}^+ \subseteq \mathcal{R}(f)$ if $\alpha\beta = -1$; where $D_0$ is the squarefree part of the discriminant of $f$, and it is assumed that $D_0 \equiv 1 \pmod 4$ and $D_0 \geq 5$.

Our result is the following.

\begin{theorem}\label{thm:main}
	Let $a_1$ be a nonzero integer.
	Then $\mathcal{R}(X^2 - a_1 X - 1) = \mathbb{Z}^+$.
 In other words, for every $n \in \mathbb{Z}^+$ there exist $\boldsymbol{x} \in \mathcal{L}(f)$ and $m \in \mathbb{Z}^+$ such that $\rho(\boldsymbol{x}; m) = n$.
 Moreover, one can choose $\boldsymbol{x}$ and $m$ so that all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero if and only if $|a_1| = 1$ and $n \geq 4$, or $|a_1| \geq 2$.
\end{theorem}

The assumption that $a_1$ is nonzero is not a restriction, since it is easy to prove that $\mathcal{R}(X^2 - 1) = \{1, 2\}$.
We remark that the proof of Theorem~\ref{thm:main} is constructive.
It provides an algorithm that, given as input a nonzero integer $a_1$ and an integer $n \geq 1$, returns as output a modulo $m$ and the initial values $x_0, x_1$ of a linear recurrence $\boldsymbol{x} \in \mathcal{L}(X^2 - a_1 X - 1)$ such that $\rho(\boldsymbol{x}; m) = n$ and, if $|a_1| \geq 2$ or $n \geq 4$, all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero.

For every $t \in \mathbb{R}$, let $\floorpart(t)$ be the \emph{floor function} of $t$, that is, the greatest integer not exceeding $t$, and let $\fracpart(t) \coloneqq t - \floorpart(t)$ be the \emph{fractional part} of $t$.
As a corollary of their aforementioned result, Dubickas and Novikas~\cite{MR4125906} proved that for every integer $k \geq 2$ there exists $\xi \in \mathbb{R}$ such that the sequence $\big(\!\fracpart\!\big(\xi ((1 + \sqrt{5})/2)^n\big)\big)_{n \geq 0}$ has exactly $k$ limit points.

From Theorem~\ref{thm:main}, we deduce the following corollary.

\begin{corollary}\label{cor:fractional-parts}
 Let $a_1 \geq 1$ be an integer, and let $\alpha \coloneqq \big(a_1 + \sqrt{a_{\smash{1}}^2 + 4}\,\big) / 2$.
 Then, for every integer $k \geq 2$, and for every integer $k \geq 1$ if $a_2 \geq 2$, there exists $\xi \in \mathbb{Q}(\alpha)$ such that the sequence $\big(\!\fracpart(\xi \alpha^n)\big)_{n \geq 0}$ has exactly $k$ limit points.
\end{corollary}

We remark that sequence of fractional parts $\big(\!\fracpart(\xi \alpha^n)\big)_{n \geq 0}$, where $\alpha > 1$ is a real algebraic number and $\xi \in \mathbb{R}$, has been studied by several authors~\cite{MR2184199,MR2201605,MR2240352,MR0017744,MR0002326,MR2256803,MR3397268}.

The paper is structured as follows.
Section~\ref{sec:preliminaries} contains several preliminary lemmas.
More precisely, Section~\ref{sec:lehmer} is devoted to prove the existence of some special primitive divisors of specific Lehmer sequences, Section~\ref{sec:orders} contains some elementary but useful results on certain multiplicative orders, and Section~\ref{sec:second-order-recurrences} provides the main lemmas for the construction of the desired sequences in $\mathcal{L}(f)$.
Then, Sections~\ref{sec:proof-main} and~\ref{sec:proof-fractional-parts} are devoted to the proofs of Theorem~\ref{thm:main} and Corollary~\ref{cor:fractional-parts}, respectively.

\section{Preliminaries}\label{sec:preliminaries}

Hereafter, let $a_1$ be a nonzero integer, let $f \coloneqq X^2 - a_1 X - 1$, let $D \coloneqq a_1^2 + 4$ be the discriminant of $f$, let $K \coloneqq \mathbb{Q}\big(\!\sqrt{D}\big)$ be the splitting field of $f$, and let $\alpha, \beta \in K$ be the roots of $f$.
Note that $D$ is not a square in $\mathbb{Q}$, so that $K$ is a quadratic number field.
Moreover, note that $\alpha$ and $\beta$ are distinct real numbers, $\alpha\beta=-1$, $\alpha + \beta = a_1$, $(\alpha-\beta)^2 = D$, and $\alpha / \beta$ is not a root of unity.

We begin with the following lemma.

\begin{lemma}\label{lem:reversal}
 Let $\boldsymbol{x} \in \mathcal{L}(f)$, let $m \geq 1$ be an integer, and let $t \coloneqq \tau(\boldsymbol{x}; m)$.
 Then there exists $\boldsymbol{y} \in \mathcal{L}(X^2 + a_1 X - 1)$ such that $y_n \equiv x_{(-(n + 1) \bmod t)}\! \pmod m$ for every integer $n \geq 0$, where $(-(n+1) \bmod t)$ is the unique integer $r \in {[0,t)}$ such that $r \equiv -(n+1) \pmod t$.
 In~particular, we have that $\tau(\boldsymbol{x}; m) = \tau(\boldsymbol{y}; m)$, $\rho(\boldsymbol{x}; m) = \rho(\boldsymbol{y}; m)$, and that all residues of $\boldsymbol{x}$ modulo $m$ are nonzero if and only if all residues of $\boldsymbol{y}$ modulo $m$ are nonzero.
\end{lemma}
\begin{proof}
	Let us extend $\boldsymbol{x}$ to negative indices by defining $x_n \coloneqq -a_1 x_{n + 1} + x_{n + 2}$ for every integer $n \leq -1$.
	Then, we have that $x_n = -a_1 x_{n + 1} + x_{n + 2}$ and $x_{n + t} \equiv x_n \pmod m$ for every integer $n$.
	Let $\boldsymbol{y} \in \mathcal{L}(X^2 + a_1 X - 1)$ with $y_0 = x_{t - 1}$ and $y_1 = x_{t - 2}$.
	Since $y_{n + 2} = -a_1 y_{n + 1} + y_n$ for every integer $n \geq 0$, it follows easily by induction that $y_n = x_{t - (n + 1)}$ for every integer $n \geq 0$.
	Hence, by the periodicity of $\boldsymbol{x}$, we get that $y_n \equiv x_{(-(n + 1) \bmod t)}\! \pmod m$ for every integer $n \geq 0$, as desired.
\end{proof}

In light of Lemma~\ref{lem:reversal}, hereafter we assume that $a_1 \geq 1$.

\begin{remark}
 Since Dubickas and Novikas already proved the case $a_1 = 1$ of Theorem~\ref{thm:main} and Corollary~\ref{cor:fractional-parts}, we could assume that $a_1 \geq 2$.
 However, we choose to include the case $a_1 = 1$ in our proofs in order to state some of the intermediary results with greater generality.
\end{remark}

\subsection{Lehmer sequences}\label{sec:lehmer}

Let $\gamma, \delta$ be complex numbers such that $\gamma\delta$ and $(\gamma + \delta)^2$ are nonzero coprime integers and $\gamma / \delta$ is not a root of unity.
The \emph{Lehmer sequence} $(\ell_n)$ associated to $\gamma, \delta$ is defined by
\begin{equation}\label{equ:lehmer-definition}
	\ell_n \coloneqq 
	\begin{cases}
		(\gamma^n - \delta^n) / (\gamma - \delta) & \text{ if $n$ is odd}; \\
		(\gamma^n - \delta^n) / (\gamma^2 - \delta^2) & \text{ if $n$ is even};
	\end{cases}
\end{equation}
for all integers $n \geq 0$.
The conditions on $\gamma, \delta$ ensure that each $\ell_n$ is an integer.
A prime number $p$ is a \emph{primitive divisor} of $\ell_n$ if $p \mid \ell_n$ but $p \nmid (\gamma^2 - \delta^2)^2 \ell_1 \cdots \ell_{n-1}$ (cf.~\cite{MR1863855}).
Note that $\ell_1 = \ell_2 = 1$, so that $\ell_n$ can have primitive divisors only if $n \geq 3$.

The sequence of \emph{cyclotomic numbers} $(\phi_n)$ associated to $\gamma, \delta$ is defined by
\begin{equation}\label{equ:cyclotomic-definition}
	\phi_n \coloneqq \prod_{\substack{1 \,\leq\, k \,\leq\, n \\[1pt] \gcd(n,k) \,=\, 1}} \left(\gamma - \mathrm{e}^{\frac{2\pi\mathbf{i}k}{n}} \delta\right) ,
\end{equation}
for all integers $n \geq 1$.
It can be proved that $\phi_n \in \mathbb{Z}$ for every integer $n \geq 3$~\cite[p.~84]{MR1863855}.

The next lemma relates the cyclotomic numbers with the primitive divisors of the Lehmer sequence.
For every prime number $p$, let $\nu_p$ denote the $p$-adic valuation.
For every integer $n \geq 4$, let $P_3(n)$ denote the greatest prime factor of $n /\! \gcd(n, 3)$.

\begin{lemma}\label{lem:product-of-primitive-divisors}
	Let $n \geq 5$ be an integer with $n \neq 6$.
	Then
	\begin{equation*}
		|\phi_n| \, \Big/ \!\! \prod_{p \emph{ prim.\ div.\ } \ell_n} p^{\nu_p(\ell_n)}
		\in
		\begin{cases}
			\{1, P_3(n)\} & \emph{ if } n \neq 12 ; \\
			\{1, 2, 3, 6\} & \emph{ if } n = 12 ; \\
		\end{cases}
	\end{equation*}
	where the product runs over all the primitive divisors $p$ of $\ell_n$.
\end{lemma}
\begin{proof}
	From~\eqref{equ:cyclotomic-definition} it follows easily that
	\begin{equation}\label{equ:binomial}
		\gamma^m - \delta^m = \prod_{d \,\mid\, m} \phi_d ,
	\end{equation}
	for all integers $m \geq 1$.
	In turn, applying to~\eqref{equ:binomial} the M\"obius inversion formula and taking into account~\eqref{equ:lehmer-definition}, we get that
	\begin{equation}\label{equ:cyclotomic-as-lehmer-product}
		\phi_m = \prod_{d \,\mid\, m} \left(\gamma^d - \delta^d\right)^{\mu(m / d)} = \prod_{d \,\mid\, m} \ell_d^{\,\mu(m / d)} ,
	\end{equation}
	for all integers $m \geq 3$, where $\mu$ is the M\"obius function and where we also employed the well-known fact that $\sum_{d \,\mid\, m} \mu(d) = 0$ for every integer $m \geq 2$.

	Let $p$ be a prime number.
	If $p$ is a primitive divisor of $\ell_n$, then $p$ does not divide $\ell_k$ for every positive integer $k < n$.
	Hence, from~\eqref{equ:cyclotomic-as-lehmer-product} we have that $\nu_p(\phi_n) = \nu_p(\ell_n)$.

	If $p$ is not a primitive divisor of $\ell_n$ and $p \mid \phi_n$, then it is known that $\nu_p(\phi_n) = 1$ and that either $p = P_3(n)$, if $n \neq 12$, or $p \in \{2, 3\}$, if $n = 12$; see the ``only if'' part of the proof of~\cite[Theorem~2.4]{MR1863855}, or see the paragraphs before, and the comment after,~\cite[Lemma~6]{MR0491445}.
\end{proof}

We need a lower bound for the absolute values of cyclotomic numbers.

\begin{lemma}\label{lem:lower-bound-cyclotomic}
	Suppose that $\gamma, \delta$ are real numbers with $\gamma\delta > 0$.
	Then
	\begin{equation*}
		|\phi_n| > |4 (\gamma - \delta)^2 \gamma \delta|^{\varphi(n) / 4}
	\end{equation*}
	for every integer $n \geq 3$, where $\varphi$ is the Euler function.
\end{lemma}
\begin{proof}
	This result is due to Ward~\cite[p.~233]{MR0071446}.
\end{proof}

Hereafter, we put $\gamma \coloneqq \alpha$ and $\delta \coloneqq -\beta$.
Note that this choice does indeed satisfy the conditions of Lehmer sequence.
Moreover, we have that $\gamma\delta = 1$, $\gamma - \delta = a_1$, $(\gamma + \delta)^2 = D$, and $(\gamma^2 - \delta^2)^2 = a_1^2 D$.
Furthermore, the first values of $(\ell_n)$ are
\begin{align}\label{equ:small-lehmer}
 \ell_1 = 1, \quad \ell_2 = 1, \quad \ell_3 = a_1^2 + 3, \quad \ell_4 = a_1^2 + 2, \quad
 \ell_5 = (a_1^2 + 1)(a_1^2 + 4) + 1, \quad \ell_6 = (a_1^2 + 1)\ell_3 .
\end{align}
In particular, note that, since $\ell_3$ or $\ell_4$ is even, if $n \geq 5$ then every primitive divisor of $\ell_n$ (if it exists) is odd.

The problem of determining which terms of a Lehmer sequence have a primitive divisor has a very long history.
The first complete classification was given by Bilu, Hanrot, and Voutier~\cite{MR1863855} (see also~\cite[Remark~3.1]{MR4350306} for some missing values in~\cite[Table~4]{MR1863855}).
We make use of the following two results.

\begin{lemma}\label{lem:existence-of-primitive-divisor}
	Let $n \geq 3$ be an integer.
	If $a_1 = 1$ and $n \notin \{3, 6, 10, 12\}$, or $a_1 = 3$ and $n \neq 3$, or $a_1 \notin \{1, 3\}$, then $\ell_n$ has at least one odd primitive divisor $p$.
\end{lemma}
\begin{proof}
	If $n \geq 13$ then it is known that $\ell_n$ has at least one odd primitive divisor~\cite[Lemma~8]{MR0491445}.
	(In~fact, this is true for all Lehmer sequences with real $\gamma, \delta$.)
	Hereafter, assume that $n \leq 12$.

	If $a_1 = 1$ and $n \notin \{3, 6, 10, 12\}$, then one can check that $\ell_n$ has an odd primitive divisor.
	If $a_1 = 3$ and $n \neq 3$, then one can check that $\ell_n$ has an odd primitive divisor.

	Hereafter, assume that $a_1 \notin \{1, 3\}$.
	If $n \geq 5$ and $n \neq 6$, then by Lemma~\ref{lem:lower-bound-cyclotomic}, recalling that $\gamma - \delta = a_1$, $\gamma \delta = 1$, we get that $|\phi_n| > 2^{\varphi(n)} > n$.
	In turn, by Lemma~\ref{lem:product-of-primitive-divisors}, this implies that $\ell_n$ has an odd primitive divisor.

	It remains to prove that if $n \in \{3, 4, 6\}$ then $\ell_n$ has an odd primitive divisor.
	Since $a_1 \notin \{1, 3\}$, by reasoning modulo $8$ and modulo $9$, it follows easily that $a_1^2 + 2$ is not a power of $2$, and that neither $a_1^2 + 1$ nor $a_1^2 + 3$ is equal to $2^s 3^t$ for some integers $s, t \geq 0$.
	Hence, there exists a prime number $p \geq 5$ dividing $a_1^2 + 1$.
	We claim that $p$ is an odd primitive divisor of $\ell_6$.
	In fact, since $p \geq 5$, by~\eqref{equ:small-lehmer} and recalling that $D = a_1^2 + 4$, we get that $p \mid \ell_6$ and $p \nmid a_1 D \ell_1 \cdots \ell_5$.
	The proof that $\ell_3$ and $\ell_4$ have an odd primitive divisor proceeds similarly.
\end{proof}

\begin{remark}
	Lemma~\ref{lem:existence-of-primitive-divisor} is optimal, that is, one can verify that if $a_1 = 1$ and $n \in \{3, 6, 10, 12\}$, or if $a_1 = 3$ and $n = 3$, then $\ell_n$ does not have an odd primitive divisor.
\end{remark}

For every integer $n \geq 1$, we say that a prime number $p$ is a \emph{high primitive divisor} of $\ell_n$ if $p$ is an odd primitive divisor of $\ell_n$ such that $p^{\nu_p(\ell_n)} > n$.

\begin{lemma}\label{lem:existence-of-primitive-divisor-high-power}
	Let $n \geq 3$ be an integer.
	If $a_1 = 1$ and $n \notin \{3, 4, 6, 8, 10, 12, 14, 18, 24\}$, or $a_1 = 2$ and $n \notin \{4, 6, 12\}$, or $a_1 = 3$ and $n \notin \{3, 6\}$, or $a_1 \geq 4$, then $\ell_n$ has at least one high primitive divisor.
\end{lemma}
\begin{proof}
 Suppose that $n \geq 5$ and $n \neq 6$.
	We claim that if $(2a_1)^{\varphi(n) / 2} > n^2$ then $\ell_n$ has a high primitive divisor.
	Indeed, recalling that $\gamma - \delta = a_1$ and $\gamma \delta = 1$, by Lemma~\ref{lem:lower-bound-cyclotomic} we have that $|\phi_n| > (2a_1)^{\varphi(n) / 2}$.
	Hence, by Lemma~\ref{lem:product-of-primitive-divisors}, we get that $(2a_1)^{\varphi(n) / 2} > n^2$ implies that
	\begin{equation}\label{equ:big-product}
		\prod_{q \text{ prim.\ div.\ } \ell_n} q^{\nu_q(\ell_n)} > n .
	\end{equation}
	In particular, from~\eqref{equ:big-product} it follows that $\ell_n$ has at least one primitive divisor.
	Let $p$ be the greatest primitive divisor of $\ell_n$.
	It is known that each primitive divisor $q$ of $\ell_n$ satisfies $q \equiv \pm 1 \pmod n$~\cite[p.~427]{MR0491445}.
	Hence, either $p \geq n + 1$, or $p = n - 1$ and $p$ is the unique primitive divisor of $\ell_n$.
	In both cases, taking into account~\eqref{equ:big-product}, we get that $p^{\nu_p(\ell_n)} > n$.
	Hence, we have that $p$ is a high primitive divisor of $\ell_n$ (note that $p$ is odd since $n \geq 5$).

	If $n \geq 2 \cdot 10^9$ then $\varphi(n) > n / \log n$~\cite[Lemma~4.1]{MR0071446}.
	Hence, we get that
	\begin{equation*}
		(2a_1)^{\varphi(n) / 2} > 2^{n / (2\log n)} > n^2 ,
	\end{equation*}
	which implies that $\ell_n$ has a high primitive divisor.

	If $180 \leq n < 2 \cdot 10^9$, then $\varphi(n) > n / 6$~\cite[Lemma~4.2]{MR0071446}.
	Hence, we get that
	\begin{equation*}
		(2a_1)^{\varphi(n) / 2} > 2^{n / 12} > n^2 ,
	\end{equation*}
	which implies that $\ell_n$ has a high primitive divisor.

	If $5 \leq n \leq 179$ with $n \neq 6$, and $|a_1| \geq 7$, then
	\begin{equation*}
	(2a_1)^{\varphi(n) / 2} \geq 14^{\varphi(n) / 2} > n^2 ,
	\end{equation*}
	which implies that $\ell_n$ has a high primitive divisor.

	If $5 \leq n \leq 179$ with $n \neq 6$, and $a_1 \leq 6$, then with the aid of a computer one can check that $\ell_n$ has a high primitive divisor, unless $a_1 = 1$ and $n \in \{8, 10, 12, 14, 18, 24\}$, or $a_1 = 2$ and $n = 12$.
	Note that this verification is done more efficiently by factorizing $\ell_n$ only in the cases in which the inequality $(2a_1)^{\varphi(n) / 2} > n^2$ does not hold.

	If $a_1 = 2$ or $a_1 = 3$, then $\ell_3$ or $\ell_4$ has a high primitive divisor, respectively.
	Hereafter, assume that $n \in \{3, 4, 6\}$ and $a_1 \geq 4$.
	It remains to prove that $\ell_n$ has a high primitive divisor.
	This is done similarly to the end of the proof of Lemma~\ref{lem:existence-of-primitive-divisor}.
	Since $a_1 \geq 4$, by reasoning modulo $8$ and modulo $9$, it follows easily that $a_1^2 + 2$ is not equal to $2^s$ or $2^s 3$, for some integer $s \geq 0$, and that neither $a_1^2 + 1$ nor $a_1^2 + 3$ is equal to $2^s 3^t$ or $2^s 3^t 5$, for some integers $s, t \geq 0$.
	Hence, there exists a prime number $p \geq 5$ dividing $a_1^2 + 1$ and such that $p^{\nu_p(a_1^2 + 1)} > 6$.
	We claim that $p$ is a high primitive divisor of $\ell_6$.
	In fact, since $p \geq 5$, by~\eqref{equ:small-lehmer} and recalling that $D = a_1^2 + 4$, we get that $p \mid \ell_6$, $p \nmid a_1 D \ell_1 \cdots \ell_5$, and $p^{\nu_p(\ell_6)} > 6$.
	The proof that $\ell_3$ and $\ell_4$ have a high primitive divisor proceeds similarly.
\end{proof}

\begin{remark}
	Lemma~\ref{lem:existence-of-primitive-divisor-high-power} is optimal, that is,
	one can verify that if $a_1 = 1$ and $n \in \{3, 4, 6, 8, 10, 12, 14, 18, 24\}$, or $a_1 = 2$ and $n \in \{4, 6, 12\}$, or $a_1 = 3$ and $n \in \{3, 6\}$, then $\ell_n$ has no high primitive divisor.
\end{remark}

For every odd prime number $p$, let $\big(\frac{D}{p}\big)$ denote the Legendre symbol.
Note that $\big(\frac{D}{p}\big) = 1$ if and only if $f$ splits modulo $p$.

\begin{lemma}\label{lem:legendre-D-p-equal-1}
	Let $n \geq 3$ be an odd integer and let $p$ be an odd prime factor of $\ell_n$.
	Then we have that $\big(\frac{D}{p}\big) = 1$.
\end{lemma}
\begin{proof}
	Since $n$ is odd, we have that $v_n \coloneqq (\gamma^n + \delta^n) / (\gamma + \delta)$ is a symmetric expression of the algebraic integers $\alpha, \beta$ (recall that $\gamma = \alpha$ and $\delta = -\beta$).
	Hence, we get that $v_n$ is an integer.
	Furthermore, from $p \mid \ell_n$ and the identity
	\begin{equation*}
		D v_n^2 - a_1^2 \ell_n^2 = (\gamma + \delta)^2 v_n^2 - (\gamma - \delta)^2 \ell_n^2 = 4(\gamma\delta)^2 = 4,
	\end{equation*}
	we get that $D v_n^2 \equiv 2^2 \pmod p$.
	Since $p$ is odd, it follows that $p \nmid v_n$ and so $D \equiv (2 v_n^{-1})^2 \pmod p$.
	Thus $\big(\frac{D}{p}\big) = 1$, as desired.
\end{proof}

\subsection{Multiplicative orders}\label{sec:orders}

For every ideal $\mathfrak{i}$ of $\mathcal{O}_K$ and for every $\theta \in \mathcal{O}_K$, let $\ord_{\mathfrak{i}}(\theta)$ denote the multiplicative order of $\theta$ modulo $\mathfrak{i}$ (if it exists).

\begin{lemma}\label{lem:prim-div-order-alpha2}
	Let $n,v \geq 1$ be integers, let $p$ be a prime number not dividing $a_1 D$, and let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ lying over $p$.
	Then $p^v \mid \ell_n$ if and only if $\ord_{\mathfrak{p}^v}(\alpha^2) \mid n$.
	In particular, we have that $p$ is a primitive divisor of $\ell_n$ if and only if $n = \ord_{\mathfrak{p}}(\alpha^2)$.
\end{lemma}
\begin{proof}
	Since $p \nmid a_1 D$, we have that $\alpha + \beta$ and $\alpha - \beta$ are invertible modulo $\mathfrak{p}^v$.
	Hence, using~\eqref{equ:lehmer-definition}, we get that $p^v \mid \ell_n$ is equivalent to $\alpha^n \equiv (-\beta)^n \pmod{\mathfrak{p}^v}$, which in turn is equivalent to $(\alpha^2)^n \equiv 1 \pmod{\mathfrak{p}^v}$, since $\alpha / (-\beta) = \alpha^2$.
	Thus the claim follows.
\end{proof}

\begin{lemma}\label{lem:sqrt-1}
	Let $p$ be an odd prime number, let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ lying over $p$, and let $v \geq 1$ be an integer.
	Then the equation $X^2 \equiv 1 \pmod {\mathfrak{p}^v}$ has exactly two solutions modulo $\mathfrak{p}^v$, namely $1$ and $-1$.
\end{lemma}
\begin{proof}
	The equation $X^2 \equiv 1 \pmod {\mathfrak{p}^v}$ is equivalent to $\nu_{\mathfrak{p}}(X - 1) + \nu_{\mathfrak{p}}(X + 1) \geq v$,
	where $\nu_{\mathfrak{p}}$ is the $\mathfrak{p}$-adic valuation over $\mathcal{O}_K$.
	Moreover, we have that $\nu_{\mathfrak{p}}(X - 1)$ and $\nu_{\mathfrak{p}}(X + 1)$ cannot both be positive, since $2 \notin \mathfrak{p}$.
	Hence, either $X \equiv 1 \pmod {\mathfrak{p}^v}$ or $X \equiv -1 \pmod {\mathfrak{p}^v}$.
\end{proof}

\begin{lemma}\label{lem:orders}
	Let $p$ be an odd prime number, let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ lying over $p$, and let $v \geq 1$ be an integer.
	Put $a \coloneqq \ord_{\mathfrak{p}^v}(\alpha)$, $b \coloneqq \ord_{\mathfrak{p}^v}(\beta)$, and $c \coloneqq \ord_{\mathfrak{p}^v}(\alpha^2)$.
	Then the following statements~hold.
	\begin{enumerate}
		\item\label{ite:orders:1} If $c$ is odd then $a = c$ and $b = 2c$, or $a = 2c$ and $b = c$.
		\item\label{ite:orders:2} If $c$ is even then $a = b = 2c$.
		\item\label{ite:orders:3} $\lcm(a, b) = 2c$.
	\end{enumerate}
\end{lemma}
\begin{proof}
	Hereafter, all congruences are modulo $\mathfrak{p}^v$.
	Since $\alpha^{2c} \equiv 1$, we have that $a \mid 2c$ and, by Lemma~\ref{lem:sqrt-1}, that $\alpha^c \equiv \pm 1$.
	Moreover, from $\alpha^a \equiv 1$, we have that $\alpha^{2a} \equiv 1$, and so $c \mid a$.
	Hence, we get that $a = c$ if $\alpha^c \equiv 1$, and $a = 2c$ if $\alpha^c \equiv -1$.

	Using the fact that $\beta^2 = \alpha^{-2}$, we get that $c = \ord_{\mathfrak{p}^v}(\beta^2)$.
	Hence, by a reasoning similar to before, we have that $\beta^c \equiv \pm 1$, while $b = c$ if $\beta^c \equiv 1$, and $b = 2c$ if $\beta^c \equiv -1$.

	Suppose that $c$ is odd.
	Since $\alpha\beta = -1$, we have that $\alpha^c \beta^c \equiv (-1)^c \equiv -1$.
	Hence, either $\alpha^c \equiv 1$ and $\beta^c \equiv -1$, or $\alpha^c \equiv -1$ and $\beta^c \equiv 1$.
	By the previous considerations, it follows that either $a = c$ and $b = 2c$, or $a = 2c$ and $b = c$.
	This proves~\eqref{ite:orders:1}.

	Suppose that $c$ is even.
	Then, by the minimality of $c$, we get that $\alpha^c \equiv \beta^c \equiv -1$.
	Hence, by the previous considerations, we have that $a = b = 2c$.
	This proves~\eqref{ite:orders:2}.

	Finally, we obtain~\eqref{ite:orders:3} as a direct consequence of~\eqref{ite:orders:1} and~\eqref{ite:orders:2}.
\end{proof}

\subsection{Second-order linear recurrences}\label{sec:second-order-recurrences}

In this section, we collect some results on linear recurrences in $\mathcal{L}(f)$.

\begin{lemma}\label{lem:binet}
	Let $\boldsymbol{x} \in \mathcal{L}(f)$.
	Then we have that
	\begin{equation*}
		x_n = \frac{(x_1 - \beta x_0) \alpha^n - (x_1 - \alpha x_0) \beta^n}{\alpha - \beta}
	\end{equation*}
	for every integer $n \geq 0$.
\end{lemma}
\begin{proof}
	This is a special case of the general expression of linear recurrences as generalized power sums~\cite[Section~1.1.6]{MR1990179} and can be easily proven by induction.
\end{proof}

\begin{lemma}\label{lem:period}
	Let $\boldsymbol{x} \in \mathcal{L}(f)$, let $p$ be a prime number not dividing $2(x_1^2 - a_1 x_1 x_0 - x_0^2) D$, let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ lying over $p$, and let $v \geq 1$ be an integer.
	Then we have that $\tau(\boldsymbol{x}; p^v) = 2\ord_{\mathfrak{p}^v}(\alpha^2)$.
\end{lemma}
\begin{proof}
	The claim can be derived from more general results on the period of linear recurrences modulo integers~\cite[Section~3.1]{MR1990179}, but we provide a short proof for completeness.

	Define the matrix
	\begin{equation*}
		M \coloneqq \begin{pmatrix}
				(x_1 - \beta x_0) & -(x_1 - \alpha x_0) \\
				(x_1 - \beta x_0)\alpha & -(x_1 - \alpha x_0)\beta \\
			\end{pmatrix}
	\end{equation*}
	and note that
	\begin{equation*}
		\det(M) = (x_1^2 - a_1 x_1 x_0 - x_0^2) (\alpha - \beta) .
	\end{equation*}
	Since $p \nmid (x_1^2 - a_1 x_1 x_0 - x_0^2) D$, we get that $\alpha - \beta$ and $M$ are invertible modulo $\mathfrak{p}^v$.
	Hence, by Lemma~\ref{lem:binet}, for every integer $n \geq 1$, we have that
	\begin{align*}
		\begin{cases}
			x_n &\!\!\!\!\!\!\!\!\equiv x_0 \\
			x_{n+1} &\!\!\!\!\!\!\!\!\equiv x_1 \\
		\end{cases} \pmod{p^v}
		\quad&\Longleftrightarrow\quad
		M \begin{pmatrix}
			\alpha^n \\
			\beta^n
		\end{pmatrix}
		\equiv
		(\alpha - \beta)\begin{pmatrix}
			x_0 \\
			x_1
		\end{pmatrix} \pmod{\mathfrak{p}^v} \\
		\quad&\Longleftrightarrow\quad
		\begin{pmatrix}
			\alpha^n \\
			\beta^n
		\end{pmatrix}
		\equiv
		M^{-1}(\alpha - \beta)\begin{pmatrix}
			x_0 \\
			x_1
		\end{pmatrix} \pmod{\mathfrak{p}^v}\\
		\quad&\Longleftrightarrow\quad
		\begin{pmatrix}
			\alpha^n \\
			\beta^n
		\end{pmatrix}
		\equiv
		\begin{pmatrix}
			1 \\
			1
		\end{pmatrix} \pmod{\mathfrak{p}^v} .
	\end{align*}
	Therefore, we get that $\tau(\boldsymbol{x}; p^v) = \lcm\!\big(\!\ord_{\mathfrak{p}^v}(\alpha), \ord_{\mathfrak{p}^v}(\beta)\big)$.
	Then, since $p$ is odd, the claim follows from Lemma~\ref{lem:orders}\eqref{ite:orders:3}.
\end{proof}

\begin{lemma}\label{lem:case-n-odd-D-p-equal-1}
	Let $n \geq 3$ be integer.
	Suppose that $p$ is an odd primitive divisor of $\ell_n$ such that $\big(\frac{D}{p}\big) = 1$.
	Then there exists $\boldsymbol{x} \in \mathcal{L}(f)$ such that $\tau(\boldsymbol{x}; p) = \rho(\boldsymbol{x}; p) = 2n$ and all the residues of $\boldsymbol{x}$ modulo $p$ are nonzero.
	Moreover, if $n$ is odd, then there exists $\boldsymbol{y} \in \mathcal{L}(f)$ such that $\tau(\boldsymbol{y}; p) = \rho(\boldsymbol{y}; p) = n$ and all the residues of $\boldsymbol{y}$ modulo $p$ are nonzero.
\end{lemma}
\begin{proof}
	Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ lying over $p$.
	Since $p$ is a primitive divisor of $\ell_n$, by Lemma~\ref{lem:prim-div-order-alpha2} we have that $\ord_{\mathfrak{p}}(\alpha^2) = n$.
	Since $p$ is odd and $\big(\frac{D}{p}\big) = 1$, we have that $f$ splits modulo $p$, that is, there exist integers $a,b$ such that $a \equiv \alpha \pmod{\mathfrak{p}}$ and $b \equiv \beta \pmod{\mathfrak{p}}$.
	In particular, we have that $\ord_p(a) = \ord_{\mathfrak{p}}(\alpha)$ and $\ord_p(b) = \ord_{\mathfrak{p}}(\beta)$, where $\ord_p$ denotes the multiplicative order modulo $p$.
	By Lemma~\ref{lem:orders}\eqref{ite:orders:1}--\eqref{ite:orders:2}, we get that $\ord_p(a) = 2n$ or $\ord_p(b) = 2n$.
	By swapping $a$ and $b$, we can assume that $\ord_p(a) = 2n$.
	Let $\boldsymbol{x} \in \mathcal{L}(f)$ such that $x_0 = 1$ and $x_1 = a$.
	Since $f(a) \equiv 0 \pmod p$, it follows easily by induction that $x_n \equiv a^n \pmod p$ for every integer $n \geq 0$.
	Hence, we get that $\tau(\boldsymbol{x}; p) = \rho(\boldsymbol{x}; p) = 2n$ and all the residues of $\boldsymbol{x}$ modulo $p$ are nonzero, as desired.

	If $n$ is odd, then by Lemma~\ref{lem:orders}\eqref{ite:orders:1}, we have that $\ord_p(b) = n$.
	Let $\boldsymbol{y} \in \mathcal{L}(f)$ such that $y_0 = 1$ and $y_1 = b$.
	Then, reasoning as for $\boldsymbol{x}$, we get that $\tau(\boldsymbol{y}; p) = \rho(\boldsymbol{y}; p) = n$ and all the residues of $\boldsymbol{y}$ modulo $p$ are nonzero, as desired.
\end{proof}

\begin{lemma}\label{lem:zero-term}
 Let $\boldsymbol{x} \in \mathcal{L}(f)$ with $x_0 = 1$, let $p$ be a prime number not dividing $D$, let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ lying over $p$, let $v \geq 1$ be an integer, let $n \coloneqq \ord_{\mathfrak{p}}(\alpha^2)$, and let $s \geq 0$ be an integer.
 Then we have that $x_s \equiv 0 \pmod{p^v}$ if and only if
 \begin{equation}\label{equ:zero-term:1}
 x_1 \equiv \frac{\beta - \alpha y_s}{1 - y_s} \pmod {\mathfrak{p}^v}
 \end{equation}
 and $1 - y_s$ is invertible modulo $\mathfrak{p}^v$, where $y_s \coloneqq (-\alpha^2)^{-s}$.
\end{lemma}
\begin{proof}
 Since $p \nmid D$, we get that $\alpha - \beta$ is invertible modulo $\mathfrak{p}^v$.
 Hence, employing Lemma~\ref{lem:binet} and the fact that $\beta = - \alpha^{-1}$, it follows easily that $x_s \equiv 0 \pmod{p^v}$ is equivalent to
 \begin{equation}\label{equ:zero-term:2}
 x_1 (1 - y_s) \equiv \beta - \alpha y_s \pmod {\mathfrak{p}^v} .
 \end{equation}
	We claim that if~\eqref{equ:zero-term:2} holds then $1 - y_s$ is invertible modulo $\mathfrak{p}^v$.
 Indeed, if $1 - y_s$ is not invertible modulo $\mathfrak{p}^v$, then $y_s \equiv 1 \pmod{\mathfrak{p}}$, and so~\eqref{equ:zero-term:2} implies that $\alpha - \beta \equiv 0 \pmod{\mathfrak{p}}$, which is a contradiction, since $p \nmid D$.
	Hence, we have that $1 - y_s$ is invertible modulo $\mathfrak{p}^v$.
 Consequently, we get that~\eqref{equ:zero-term:2} is equivalent to~\eqref{equ:zero-term:1} and $1 - y_s$ being invertible modulo $\mathfrak{p}^v$, as desired.
\end{proof}

\begin{lemma}\label{lem:coincidence}
	Let $\boldsymbol{x} \in \mathcal{L}(f)$ with $x_0 = 1$, let $p$ be a prime number not dividing $D$, let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ lying over $p$, let $v \geq 1$ be an integer, let $n \coloneqq \ord_{\mathfrak{p}}(\alpha^2)$, and let $s, t$ be integers such that $0 \leq s < t < 2n$ and $s \equiv t \pmod 2$.
	Then we have that $x_s \equiv x_t \pmod{p^v}$ if and only if
	\begin{equation}\label{equ:coincidence:5}
		x_1 \equiv \frac{\beta - \alpha z_{s,t}}{1 - z_{s,t}} \pmod {\mathfrak{p}^v}
	\end{equation}
	and $1 - z_{s,t}$ is invertible modulo $\mathfrak{p}^v$, where $z_{s,t} \coloneqq (-1)^{t + 1} \alpha^{-(s + t)}$.
\end{lemma}
\begin{proof}
	Since $p \nmid D$, we get that $\alpha - \beta$ is invertible modulo $\mathfrak{p}^v$.
	Hence, employing Lemma~\ref{lem:binet}, we have that $x_s \equiv x_t \pmod{p^v}$ is equivalent to
	\begin{equation}\label{equ:coincidence:2}
		(x_1 - \beta)\alpha^s - (x_1 - \alpha) \beta^s \equiv (x_1 - \beta)\alpha^t - (x_1 - \alpha) \beta^t \pmod {\mathfrak{p}^v} .
	\end{equation}
	Since $0 \leq s < t < 2n$, $s \equiv t \pmod 2$, and $n \coloneqq \ord_{\mathfrak{p}}(\alpha^2)$, we have that $1 - \alpha^{t-s}$ is invertible modulo $\mathfrak{p}^v$.
	Hence, it follows that~\eqref{equ:coincidence:2} is equivalent to
	\begin{equation}\label{equ:coincidence:3}
		x_1 - \beta \equiv \frac{\beta^t(\beta^{s-t} - 1)}{\alpha^s (1 - \alpha^{t-s})} \, (x_1 - \alpha) \pmod {\mathfrak{p}^v} .
	\end{equation}
	Since $\beta = -\alpha^{-1}$ and $s \equiv t \pmod 2$, we get that
	\begin{equation*}
		\frac{\beta^t(\beta^{s-t} - 1)}{\alpha^s (1 - \alpha^{t-s})}
		= (-1)^{t + 1} \alpha^{-(s + t)} = z_{s,t} .
	\end{equation*}
	Therefore, we have that~\eqref{equ:coincidence:3} is equivalent to
	\begin{equation}\label{equ:coincidence:4}
		x_1 (1 - z_{s,t}) \equiv \beta - \alpha z_{s,t} \pmod {\mathfrak{p}^v} .
	\end{equation}
	We claim that if~\eqref{equ:coincidence:4} holds then $1 - z_{s,t}$ is invertible modulo $\mathfrak{p}^v$.
 Indeed, if $1 - z_{s,t}$ is not invertible modulo $\mathfrak{p}^v$, then $z_{s,t} \equiv 1 \pmod{\mathfrak{p}}$, and so~\eqref{equ:coincidence:4} implies that $\alpha - \beta \equiv 0 \pmod{\mathfrak{p}}$, which is a contradiction, since $p \nmid D$.
	Hence, we have that $1 - z_{s,t}$ is invertible modulo $\mathfrak{p}^v$.
	Consequently, we get that~\eqref{equ:coincidence:4} is equivalent to~\eqref{equ:coincidence:5} and $1 - z_{s,t}$ being invertible modulo $\mathfrak{p}^v$, as desired.
\end{proof}

\begin{lemma}\label{lem:parity}
	Let $n \geq 4$ be an even integer.
	Suppose that $p$ is a high primitive divisor of $\ell_n$ such that $\big(\frac{D}{p}\big) = -1$, and put $v \coloneqq \nu_p(\ell_n)$.
	Then there exists $\boldsymbol{x} \in \mathcal{L}(f)$ such that $\tau(\boldsymbol{x}; p^v) = 2n$, $x_s \not\equiv x_t \pmod {p^v}$ for all integers $s, t$ with $0 \leq s < t < 2n$ and $s \equiv t \pmod 2$, and all the residues of $\boldsymbol{x}$ modulo $p^v$ are nonzero.
\end{lemma}
\begin{proof}
	Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ lying over $p$.
	Since $p$ is a primitive divisor of $\ell_n$ and $p^v \mid \ell_n$, from Lemma~\ref{lem:prim-div-order-alpha2} we have that $n = \ord_{\mathfrak{p}}(\alpha^2) = \ord_{\mathfrak{p}^v}(\alpha^2)$.

	We claim that there exists an integer $c$ that satisfies
	\begin{equation}\label{equ:excluded:1}
		c \not\equiv \frac{\beta - \alpha^{2k+1}}{1 - \alpha^{2k}} \pmod{\mathfrak{p}^v}
	\end{equation}
	for every integer $k \geq 0$ such that $1 - \alpha^{2k}$ is invertible modulo ${\mathfrak{p}^v}$.
	Indeed, we have $p^v$ choices for $c$ modulo $\mathfrak{p}^v$, while the right-hand side of~\eqref{equ:excluded:1} takes at most $n = \ord_{\mathfrak{p}^v}(\alpha^2)$ values modulo $\mathfrak{p}^v$.
 Since $p$ is a high primitive divisor of $\ell_n$, we have that $p^v > n$ and so it is possible to choose a value of $c$ with the desired property.

	Let $\boldsymbol{x} \in \mathcal{L}(f)$ with $x_0 = 1$ and $x_1 = c$.
	Since $\big(\frac{D}{p}\big) = -1$, we have that $p \nmid D$ and that $f$ has no roots modulo $p$.
	Hence, also since $p$ is odd, we get that $p \nmid 2(x_1^2 - a_1 x_1 - 1)D$.
	Consequently, by Lemma~\ref{lem:period}, we get that $\tau(\boldsymbol{x}; p^v) = 2n$, as desired.

	Since $n = \ord_{\mathfrak{p}^v}(\alpha^2)$, we have that $\alpha^{2n} \equiv 1 \pmod{\mathfrak{p}^v}$.
 Consequently, by Lemma~\ref{lem:sqrt-1}, we get that $\alpha^n \equiv \pm 1 \pmod{\mathfrak{p}^v}$.
 In fact, since $n$ is even and $n = \ord_{\mathfrak{p}^v}(\alpha^2)$, it follows that $\alpha^n \equiv -1 \pmod{\mathfrak{p}^v}$.
 Hence, we have that $-1$ is equal to a power of $\alpha^2$ modulo $\mathfrak{p}^v$.

	Suppose that there exist integers $s, t$ such that $0 \leq s < t < 2n$, $s \equiv t \pmod 2$, and $x_s \equiv x_t \pmod {p^v}$.
	By Lemma~\ref{lem:coincidence}, we get that
	\begin{equation}\label{equ:excluded:2}
		c \equiv x_1 \equiv \frac{\beta - \alpha z_{s,t}}{1 - z_{s,t}} \pmod{\mathfrak{p}^v} ,
	\end{equation}
	where $z_{s,t} \coloneqq (-1)^{t + 1} \alpha^{-(s + t)}$ is equal to a power of $\alpha^2$ modulo $\mathfrak{p}^v$.
	But then~\eqref{equ:excluded:1} and~\eqref{equ:excluded:2} are in contradiction.
	Therefore, it follows that $x_s \not\equiv x_t \pmod {p^v}$ for all integers $s, t$ with $0 \leq s < t < 2n$ and $s \equiv t \pmod 2$.

 Suppose that $x_s \equiv 0 \pmod{p^v}$ for some integer $s \geq 0$.
 Then, by Lemma~\ref{lem:zero-term}, we have that
	\begin{equation}\label{equ:excluded:3}
 c \equiv x_1 \equiv \frac{\beta - \alpha y_s}{1 - y_s} \pmod{\mathfrak{p}^v} ,
 \end{equation}
 where $y_s \coloneqq (-\alpha^2)^{-s}$ is equal to a power of $\alpha^2$ modulo $\mathfrak{p}^v$.
	But then~\eqref{equ:excluded:1} and~\eqref{equ:excluded:3} are in contradiction.
 Therefore, it follows that $x_s \not\equiv 0 \pmod {p^v}$ for all integers $s \geq 0$.
\end{proof}

For every $\boldsymbol{x} \in \mathcal{L}(f)$ and for all integers $m, d \geq 1$ and $r$, let us define
\begin{equation*}
	\rho(\boldsymbol{x}; m, r, d) \coloneqq \#\big\{x_n \bmod m : n\geq 0,\; n \equiv r \!\!\!\pmod d \big\} .
\end{equation*}
We need the following lemma.

\begin{lemma}\label{lem:combining-sequences}
	Let $m_1, m_2 \geq 1$ be coprime integers, let $\boldsymbol{x}^{(1)}, \boldsymbol{x}^{(2)} \in \mathcal{L}(f)$, and put $\tau_i \coloneqq \tau(\boldsymbol{x}^{(i)}; m_i)$ for each $i \in \{1,2\}$.
	If $\rho(\boldsymbol{x}^{(2)}; m_2) = \tau_2$ then there exists $\boldsymbol{x} \in \mathcal{L}(f)$ such that $\boldsymbol{x} \equiv \boldsymbol{x}^{(i)} \pmod {m_i}$ for each $i \in \{1,2\}$, and
	\begin{equation*}
		\rho(\boldsymbol{x}; m_1 m_2) = \frac{\tau_2}{d} \sum_{r = 0}^{d-1} \rho(\boldsymbol{x}^{(1)}; m_1, r, d) ,
	\end{equation*}
	where $d \coloneqq \gcd(\tau_1, \tau_2)$.
\end{lemma}
\begin{proof}
	Since $m_1, m_2$ are coprime, we can pick $\boldsymbol{x} \in \mathcal{L}(f)$ such that $x_j \equiv x_j^{(i)} \pmod {m_i}$ for each $i \in \{1, 2\}$ and $j \in \{0,1\}$.
	Then, it follows easily by induction that $x_j \equiv x_j^{(i)} \pmod {m_i}$ for each $i \in \{1, 2\}$ and for every integer $j \geq 0$.
	At this point, the rest of the claim is a consequence of~\cite[Lemma~6]{MR4125906}.
\end{proof}

\section{Proof of Theorem~\ref{thm:main}}\label{sec:proof-main}

We collected in Table~\ref{tab:special-cases} some values of $\tau(\boldsymbol{x}; m)$ and $\rho(\boldsymbol{x}; m)$, for $\boldsymbol{x} \in \mathcal{L}(f)$ and $m \in \mathbb{Z}^+$, that are needed for the proof.
We begin with the following lemma.

\begin{lemma}\label{lem:case-n-4}
 There exist $\boldsymbol{x} \in \mathcal{L}(f)$ and $m \in \mathbb{Z}^+$ such that $\rho(\boldsymbol{x}; m) = 4$ and all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero.
 In particular, we have that $4 \in \mathcal{R}(f)$.
\end{lemma}
\begin{proof}
 If $a_1 = 1$ or $a_1 = 2$, then the claim follows from rows 2 or 13 of Table~\ref{tab:special-cases}, respectively.

 If $a_1$ is odd and $a_1 \geq 3$, then pick $\boldsymbol{x} \in \mathcal{L}(f)$ with $x_0 = 1$, $x_1 = 2$, and let $m \coloneqq 2 a_1$.
 Since $2 a_1 \equiv 0 \pmod m$ and $a_1^2 \equiv a_1 \pmod m$, we get that the first terms of $\boldsymbol{x}$ are congruent to
 \begin{equation*}
 1,\, 2,\, 1,\, a_1 + 2,\, a_1 + 1,\, a_1 + 2,\, 1,\, 2, \, \dots \pmod m .
 \end{equation*}
 Noting that $m \geq 6$ and $a_1 \not\equiv -2, -1, 0, 1 \pmod m$, it follows easily that $\tau(\boldsymbol{x}; m) = 6$, $\rho(\boldsymbol{x}; m) = 4$ and that all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero.

 If $a_1$ is even and $a_1 \geq 4$, then pick $\boldsymbol{x} \in \mathcal{L}(f)$ with $x_0 = 1$, $x_1 = 3$, and let $m \coloneqq 2 a_1$.
 Since $a_1^2 \equiv 2 a_1 \equiv 0 \pmod m$, we get that the first terms of $\boldsymbol{x}$ are congruent to
 \begin{equation*}
 1,\, 3,\, a_1 + 1,\, a_1 + 3,\, 1,\, 3,\, \dots \pmod m .
 \end{equation*}
 Noting that $m \geq 8$ and $a_1 \not\equiv -3,-2,-1,0,2 \pmod m$, it follows easily that $\tau(\boldsymbol{x}; m) = \rho(\boldsymbol{x}; m) = 4$ and that all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero.
\end{proof}

\begin{table}[h]\centering

 \caption{Values of $\tau(\boldsymbol{x}; m)$ and $\rho(\boldsymbol{x}; m)$ for $\boldsymbol{x} \in \mathcal{L}(X^2 - a_1 X - 1)$ with initial values $x_0, x_1$.
 The symbol $*$ means that all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero.}\label{tab:special-cases}
 
		\begin{tabular}{cccccccc}
			\toprule
			row n. & $a_1$ & $x_0$ & $x_1$ & $m$ & $\tau(\boldsymbol{x}; m)$ & $\rho(\boldsymbol{x}; m)$ \\
			\midrule
			$1$ & $1$ & $0$ & $1$ & $3$ & $8$ & $3$ & \\
			$2$ & $1$ & $1$ & $3$ & $5$ & $4$ & $4$ & $*$ \\
			$3$ & $1$ & $1$ & $3$ & $8$ & $12$ & $6$ & $*$ \\
			$4$ & $1$ & $1$ & $3$ & $10$ & $12$ & $8$ & $*$ \\
			$5$ & $1$ & $1$ & $3$ & $13$ & $28$ & $12$ & $*$ \\
			$6$ & $1$ & $1$ & $3$ & $17$ & $36$ & $16$ & $*$ \\
			$7$ & $1$ & $1$ & $3$ & $28$ & $48$ & $20$ & $*$ \\
			$8$ & $1$ & $1$ & $3$ & $26$ & $84$ & $24$ & $*$ \\
			$9$ & $1$ & $1$ & $3$ & $56$ & $48$ & $28$ & $*$ \\
			$10$ & $1$ & $1$ & $3$ & $52$ & $84$ & $36$ & $*$ \\
			$11$ & $1$ & $1$ & $3$ & $78$ & $168$ & $48$ & $*$ \\
			\midrule
			$12$ & $2$ & $1$ & $1$ & $4$ & $4$ & $2$ & $*$ \\
			$13$ & $2$ & $1$ & $1$ & $5$ & $12$ & $4$ & $*$ \\
			$14$ & $2$ & $1$ & $1$ & $28$ & $12$ & $8$ & $*$ \\
			$15$ & $2$ & $1$ & $1$ & $13$ & $28$ & $12$ & $*$ \\
			$16$ & $2$ & $1$ & $1$ & $39$ & $56$ & $24$ & $*$ \\
			\midrule
			$17$ & $3$ & $1$ & $1$ & $9$ & $6$ & $3$ & $*$ \\
			$18$ & $3$ & $1$ & $1$ & $8$ & $12$ & $6$ & $*$ \\
			$19$ & $3$ & $1$ & $1$ & $17$ & $16$ & $12$ & $*$ \\
			\bottomrule
		\end{tabular}
\end{table}

We have to prove that $\mathcal{R}(f) = \mathbb{Z}^+$.
First, we prove that $\{n, 2n\} \subseteq \mathcal{R}(f)$ for every odd integer $n \geq 1$.
We have that $1 \in \mathcal{R}(f)$, since $\rho(\boldsymbol{x}; 1) = 1$ for every $\boldsymbol{x} \in \mathcal{L}(f)$; and $2 \in \mathcal{R}(f)$, since $\rho(\boldsymbol{x}; 2) = 2$ for every $\boldsymbol{x} \in \mathcal{L}(f)$ with $x_0 \not\equiv x_1 \pmod 2$.
If $a_1 = 1$, or $a_1 = 3$, then $\{3, 6\} \subseteq \mathcal{R}(f)$ by rows 1 and 3, or 17 and 18, of Table~\ref{tab:special-cases}, respectively.
Hence, assume that $n \geq 3$ is an odd integer, and $a_1 \notin \{1, 3\}$ or $n \neq 3$.
By Lemma~\ref{lem:existence-of-primitive-divisor}, we have that $\ell_n$ has an odd primitive divisor $p$.
Moreover, since $n$ is odd, from Lemma~\ref{lem:legendre-D-p-equal-1} it follows that $\big(\frac{D}{p}\big) = 1$.
Therefore, from Lemma~\ref{lem:case-n-odd-D-p-equal-1} we obtain that $\{n, 2n\} \subseteq \mathcal{R}(f)$.

It remains to prove that $2n \in \mathcal{R}(f)$ for every even integer $n \geq 2$.
By Lemma~\ref{lem:case-n-4}, we have that $4 \in \mathcal{R}(f)$.
Hence, we can assume that $n \geq 4$.

If $a_1 = 1$ and $n \in \{4, 6, 8, 10, 12, 14, 18, 24\}$, or $a_1 = 2$ and $n \in \{4, 6, 12\}$, or $a_1 = 3$ and $n = 6$, then $2n \in \mathcal{R}(f)$ by rows 4--11, 14--16, or 19 of Table~\ref{tab:special-cases}, respectively.

Hence, assume that $n \geq 4$ is an even integer, and that $a_1 = 1$ and $n \notin \{4, 6, 8, 10,$ $12, 14, 18, 24\}$, or $a_1 = 2$ and $n \notin \{4, 6, 12\}$, or $a_1 = 3$ and $n \neq 6$, or $a_1 \geq 4$.
Then, by Lemma~\ref{lem:existence-of-primitive-divisor-high-power}, we have that $\ell_n$ has a high primitive divisor $p$.

If $\big(\frac{D}{p}\big) = 1$ then Lemma~\ref{lem:case-n-odd-D-p-equal-1} yields that $2n \in \mathcal{R}(f)$, as desired.
Hence, suppose that $\big(\frac{D}{p}\big) = -1$.
Thanks to Lemma~\ref{lem:parity}, there exists $\boldsymbol{x}^{(1)} \in \mathcal{L}(f)$ such that $\tau(\boldsymbol{x}^{(1)}; m_1) = 2n$ and $\rho(\boldsymbol{x}^{(1)}; m_1, r, 2) = n$ for each $r \in \{0, 1\}$, where $m_1 \coloneqq p^{\nu_p(\ell_n)}$.
Note that, since $n$ is even, we also get that $\rho(\boldsymbol{x}^{(1)}; m_1, r, 4) = n / 2$ for each $r \in \{0, 1, 2, 3\}$.
We define $\boldsymbol{x}^{(2)} \in \mathcal{L}(f)$ and $m_2 \in \mathbb{Z}^+$ as follows.
If $a_1 = 1$ then we pick $\boldsymbol{x}^{(2)}$ and $m_2$ from rows 2 of Table~\ref{tab:special-cases}.
If $a_1 \geq 2$ then we let $x_0^{(2)} \coloneqq 0$, $x_1^{(2)} \coloneqq 1$, and put $m_2 \coloneqq a_1$.
It follows easily that $\tau(\boldsymbol{x}^{(2)}; m_2) = \rho(\boldsymbol{x}^{(2)}; m_2) \in \{2, 4\}$.
Since $p$ is a primitive divisor of $\ell_n$, we have that $p \nmid a_1 D$.
Hence, we get that $m_1, m_2$ are coprime integers.
(Note that $m_2 = D = 5$ if $a_1 = 1$.)
Put $\tau_i \coloneqq \tau(\boldsymbol{x}^{(i)}; m_i)$ for each $i \in \{1, 2\}$, and let $d \coloneqq \gcd(\tau_1, \tau_2)$.
Note that, since $n$ is even, we have that $d = \tau_2 \in \{2, 4\}$.
Therefore, applying Lemma~\ref{lem:combining-sequences} we get that there exists $\boldsymbol{x} \in \mathcal{L}(f)$ such that
\begin{equation*}
	\rho(\boldsymbol{x}; m_1 m_2) = \frac{\tau_2}{d} \sum_{r = 0}^{d-1} \rho(\boldsymbol{x}^{(1)}; m_1, r, d) = 1 \cdot \sum_{r = 0}^{d-1} \frac{2n}{d} = 2n ,
\end{equation*}
so that $2n \in \mathcal{R}(f)$, as desired.
The proof that $\mathcal{R}(f) = \mathbb{Z}^+$ is complete.

At this point, note that for each integer $n \geq 4$, and for each integer $n \geq 3$ if $a_1 \geq 2$, the sequence $\boldsymbol{x} \in \mathcal{L}(f)$ and the modulo $m$ that we constructed so that $\rho(\boldsymbol{x}; m) = n$ have the additional property that all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero.
This follows from Table~\ref{tab:special-cases}, Lemma~\ref{lem:case-n-odd-D-p-equal-1}, Lemma~\ref{lem:case-n-4}, and Lemma~\ref{lem:parity}--Lemma~\ref{lem:combining-sequences} (note that $\boldsymbol{x}$ has no zero modulo $m_1 m_2$ since $\boldsymbol{x}^{(1)}$ has no zero modulo $m_1$).
Moreover, if $a_1 \geq 2$, then picking $\boldsymbol{x} \in \mathcal{L}(f)$ with $x_0 = 1$, $x_1 = 1$, and $m \coloneqq a_1$, we get that $\rho(\boldsymbol{x}; m) = 1$ and all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero.
If $a_1 = 2$ or $a_1 \geq 3$, then picking $\boldsymbol{x}$ and $m$ as in row 12 of Table~\ref{tab:special-cases}, or taking $x_0 = 1$, $x_1 = 2$, and $m = a_1$, we get that $\rho(\boldsymbol{x}; m) = 2$ and all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero.

It remains to prove that if $a_1 = 1$ and $n \in \{1, 2, 3\}$ then there exist no $\boldsymbol{x} \in \mathcal{L}(f)$ and $m \in \mathbb{Z}^+$ such that $\rho(\boldsymbol{x}; m) = n$ and all the residues of $\boldsymbol{x}$ modulo $m$ are nonzero.
This is done in the last paragraph of~\cite[p.~122]{MR4125906}.

The proof is complete.

\section{Proof of Corollary~\ref{cor:fractional-parts}}
\label{sec:proof-fractional-parts}

If $a_1 = 1$ then the result is in fact~\cite[Corollary~2]{MR4125906}.
Hence, assume that $a_1 \geq 2$.

We order $\alpha$ and $\beta$ so that $\alpha = \big(a_1 + \sqrt{D}\,\big) / 2$ and $\beta = \big(a_1 - \sqrt{D}\,\big) / 2$.
In particular, note that $\alpha > 1$ and $-1 < \beta < 0$.

Let $k \geq 1$ be an integer.
By Theorem~\ref{thm:main}, there exist $\boldsymbol{x} \in \mathcal{L}(f)$ and $m \in \mathbb{Z}^+$ such that $\rho(\boldsymbol{x}; m) = k$ and $x_n \not\equiv 0 \pmod m$ for every integer $n \geq 0$.
Moreover, by Lemma~\ref{lem:binet}, we have that $x_n = c_1 \alpha^n - c_2 \beta^n$ for all integers $n \geq 0$, where
\begin{equation*}
 c_1 \coloneqq \frac{x_1 - \beta x_0}{\alpha - \beta} \quad \text{ and } \quad c_2 \coloneqq \frac{x_1 - \alpha x_0}{\alpha - \beta} .
\end{equation*}
By eventually replacing $\boldsymbol{x}$ with $-\boldsymbol{x} \coloneqq (-x_n)_{n \geq 0}$, we can assume that $c_1 > 0$.
Furthermore, note that $c_2 \neq 0$.
Let $\xi \coloneqq c_1 / m$, so that $\xi \in \mathbb{Q}(\alpha)$.
Then, we have that
\begin{equation}\label{equ:xi-alpha-n}
 \xi \alpha^n - \frac{x_n}{m} = \frac{c_2 \beta^n}{m} \to 0 ,
\end{equation}
as $n \to +\infty$, since $|\beta| < 1$.
Let $r_1, \dots, r_k \in \mathbb{Z}$, with $0 < r_1 < \dots < r_k < m$, be the residues of $\boldsymbol{x}$ modulo $m$.
From~\eqref{equ:xi-alpha-n} it follows easily that the set of limit points of $\big(\!\fracpart(\xi \alpha^n)\big)_{n \geq 0}$ is equal to $\{r_1 / m, \dots, r_k / m\}$.
Hence, we get that $\big(\!\fracpart(\xi \alpha^n)\big)_{n \geq 0}$ has exactly $k$ limit points.

The proof is complete.

\section*{Acknowledgments}

The authors would like to thank the anonymous referee for carefully reading the paper and for providing valuable suggestions.


\section*{Declaration of interests}
The authors do not work for, advise, own shares in, or receive funds from
any organization that could benefit from this article, and have declared no affiliations other than their research organizations.


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