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\vskip 1cm{\LARGE\bf 
Generalization of Greatest Common Divisor\\ 
\vskip .1in
of Shifted Fibonacci Numbers
}
\vskip 1cm
\large
Tran Quang Do and Luu Ba Thang\footnote{Corresponding author.}\\  
Department of Mathematics and Informatics \\
Hanoi National University of Education \\
136 Xuan Thuy \\
Cau Giay \\
Hanoi\\
Vietnam \\
\href{mailto: quangdo1501@gmail.com }{\tt quangdo1501@gmail.com}\\
\href{mailto: thanglb@hnue.edu.vn}{\tt thanglb@hnue.edu.vn} 
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\begin{abstract}
    Let $F_n$ denote the $n$'th Fibonacci number and let $k$ be a  positive integer.  We find  necessary and sufficient conditions on $s$ and $t$ so that the function $n\mapsto \gcd(F_n+s, F_{n+k}+t)$ is unbounded. 
\end{abstract}

\section{Introduction}
The Fibonacci sequence $(F_n)_{n\geq 0}$ and Lucas sequence $(L_n)_{n\geq 0}$ are defined by the recursions
\begin{displaymath}
F_0=0, F_1=1;
F_n=F_{n-1}+F_{n-2}, n\geq 2;
\end{displaymath}
 and 
\begin{displaymath} L_0=2, L_1=1;
L_n=L_{n-1}+L_{n-2}, n\geq 2.
\end{displaymath}
In 1971, Dudley and Tucker \cite{D} showed that $\gcd (F_n+s, F_{n+1}+s)$ is unbounded for $s=\pm 1$. In 2011, Chen \cite{C} determined $\gcd(F_n+s,F_{n+1}+s)$ for $s\in \{\pm 1, \pm 2\}$ and proved that $\gcd(F_n+s,F_{n+1}+s)$ is bounded for $s\ne \pm 1$. In 2018, Rahn and Kreh \cite{R} determined $\gcd(F_n+s,F_{n+1}+s)$ for $s=\pm 3$. 
In 2016, Spilker \cite{S} proved that $\gcd(F_n+s,F_{n+1}+s)$ divides $s^2+(-1)^n$.  In 2020, Chen and Pan \cite{CP} considered general second-order linear homogeneous recurrence functions $W_n$. They proved a criterion of periodicity of $\gcd(W_n+s, W_{n+k}+t)$. They transformed also problems about the $\gcd(W_n+s, W_{n+k}+t)$ with arbitrarily $k$ into those with $k=1$ and  calculated on this way some $\gcd(W_n+s, W_{n+1}+t)$.
In 2022, Hieu, Spilker, and Thang \cite{HST} investigated the necessary and sufficient conditions on $k\in \mathbb N, s,t\in \mathbb Z$, so that the function 
$\gcd(F_n+s, F_{n+k}+s)$ and the function $\gcd(F_n+s,F_{n+1}+t)$ are unbounded. They proposed the open problem to characterize the function $\gcd(F_n+s, F_{n+k}+t)$ for $k\in \mathbb N, s,t \in \mathbb Z$.
Let $\mathbb {N}_0$ denote the set of non-negative  integers.  In this article, given a positive integer $k$, we 
investigate  necessary and sufficient conditions on $s, t \in \mathbb Z$ (depend on $k$),  so that the function 
\begin{displaymath}
\begin{aligned}
B^k_{s,t}: \mathbb N_0 &amp;\to \mathbb N \\
n &amp;\mapsto \gcd(F_n+s, F_{n+k}+t) 
\end{aligned}
\end{displaymath}
is bounded (Theorem \ref{thm1}). If $B^k_{s,t}$ is bounded, it is periodic (Theorem \ref{thm2}).  In more detail,
we obtain the following results:
\begin{theorem}\label{thm1}
Let $s,t \in \mathbb Z $ and $n\in \mathbb N_0$. The following assertions on $B^k_{s,t}(n)$ are equivalent:
\begin{enumerate}
\item \label{thm1a} $n\mapsto B^k_{s,t}(n)$ is unbounded on $\mathbb N_0$;
\item \label{thm1b} $(s,t)\in \mathcal{R}:=\{\pm (F_{j},F_{j+k}) : j\in \mathbb Z\}$;
\item \label{thm1c} $e^{k*}_{s,t}=0$,
\end{enumerate}
where $e^k_{s,t}(n):=t^2 - L_k st +(-1)^k s^2 - (-1)^nF_k^2$ and $e^{k*}_{s,t}:=e^k_{s,t}(0)e^k_{s,t}(1)$. 
\end{theorem}

\begin{theorem}\label{thm2}
If $(s,t) \notin \mathcal{R} $, then the function $B^k_{s,t}(n)$
is simply periodic on $\mathbb {N}_0$, which means there exists a positive integer $p$ such that $B^k_{s,t}(n+p)=B^k_{s,t}(n)$ for all $n \geq 0$. A period $p\leq c^2$ can be chosen where $c:= \left|e^{k*}_{s,t}\right |$ such that
\begin{displaymath}
F_{p} \equiv \modd{F_0} {c} \; \text{and} \; F_{p+1} \equiv 
\modd{F_1} {c} .
\end{displaymath}
\end{theorem}
\section{Auxiliary results}
The following lemma contains some well-known facts on $(F_n)_{n\geq 0}$ and $(L_n)_{n\geq 0}$, which we have used. Notice that $F_n$ and $L_n$ can be extended to integer indices by the recursion.

\begin{lemma}[\cite{K}]\label{lemma1}
Let $k\in \mathbb N, n, m \in \mathbb Z$. Then we have
\begin{enumerate}
        \item [\rm (a)] $ F_n^2- F_{n-k}F_{n+k} = (-1)^{n-k} F_k^2$;  
        \item [\rm (b)] $F_{n+2k} = L_k F_{n+k}  - (-1)^k F_n$;
    \item [\rm (c)]    $F_{\frac{m-n}{2}}$ divides $F_m + F_n$ if $\dfrac{m-n}{2}$ is an odd integer;
    \item [\rm (d)] $F_{\frac{m-n}{2}}$ divides $F_m - F_n$ if $\dfrac{m-n}{2}$ is an even integer;
       \item [\rm (e)] $\dfrac{L_k^2}{4} - (-1)^k = \dfrac{5}{4} F_k^2$;
   \item [\rm (f)] $2F_{n+k} = L_k F_{n}+L_nF_k$.
\end{enumerate}
\end{lemma}
\section{Proof of Theorem~\ref{thm1} and \ref{thm2}}
Set $B:=B^k_{s,t}(n)$. We need the following lemma.
\begin{lemma}\label{lemma11} Let $s,t\in \mathbb Z, n\in \mathbb N_0$. Then
\begin{itemize}
\item [\rm (a)] $B$ divides $e^k_{s,t}(n)$;
\item [\rm (b)] $B$ divides $e^{k*}_{s,t}$.
\end{itemize}
\end{lemma}
\begin{proof} 
By using Lemma \ref{lemma1} (a) we have: $$ F_{n+k}^2- F_{n}F_{n+2k} = (-1)^{n} F_k^2 .$$
By using Lemma \ref{lemma1} (b) we have: $$F_{n+k}^2 - F_n (L_k F_{n+k} - (-1)^k F_n) = (-1)^n F_k^2. $$
Since $B^k_{s,t} (n) = \gcd(F_n + s , F_{n+k} +t) $, by applying 
\begin{displaymath}
 F_{n+k} \equiv \modd{-t} {B} \; \text{and}\;   
 F_{n} \equiv \modd{-s} {B},   
\end{displaymath}
we deduce that
\begin{displaymath}
 t^2 - L_k st + (-1)^k s^2 \equiv (-1)^n F_k^2 \pmod B. 
\end{displaymath}
Hence, $e^k_{s,t}(n) \equiv \modd{0} B$ and $e^{k*}_{s,t} \equiv \modd{0} B$.

\end{proof}
\begin{corollary}\label{cor12} If $s,t\in \mathbb Z$ and $e^{k*}_{s,t}\ne 0$, then the function $n\mapsto B^k_{s,t}(n)$ is bounded on $\mathbb {N}_0$.
\end{corollary}
\begin{proof}[Proof of Theorem \ref{thm1}]
Now, we give a proof of Theorem \ref{thm1} by the implications
$(\ref{thm1a}) \Rightarrow (\ref{thm1c}) \Rightarrow (\ref{thm1b}) \Rightarrow (\ref{thm1a})$.

\bigskip

\noindent $(\ref{thm1a}) \Rightarrow (\ref{thm1c})$: by Corollary \ref{cor12}.

\bigskip

\noindent $(\ref{thm1c}) \Rightarrow (\ref{thm1b})$: We have
\begin{displaymath}
 \begin{aligned}
t^2 - L_k s t +(-1)^k s^2 
&amp; = \left(t - \dfrac{L_k}{2} s \right)^2 + \left( (-1)^k - \dfrac{L_k^2}{4} \right) s^2 \\  &amp;=\left( t- \dfrac{L_k}{2}s \right)^2 - \dfrac{5}{4} F_k^2  s^2\;\;\text{(by Lemma \ref{lemma1} (e)).}
\end{aligned}
\end{displaymath}
Hence \begin{equation}\label{equ1} \left( t- \dfrac{L_k}{2} s \right)^2 - \dfrac{5}{4} F_k^2 s^2 = \pm F_k^2 . \end{equation} 

If $s$ is even, from \eqref{equ1} we have $F_k \mid t - \dfrac{L_k}{2}s$. Hence, 
$$ \dfrac{ t - \frac{L_k}{2}s}{F_k} - \dfrac{\sqrt{5}}{2} s= (\dfrac{t - \frac{L_k}{2}s}{F_k}+\dfrac{s}{2}) - s\dfrac{\sqrt{5}+1}{2}$$ is integer in the quadratic field $\mathbb{Q}(\sqrt{5})$.

If $s$ is odd,  from \eqref{equ1} we have $\dfrac{t-\frac{L_k}{2}s}{F_k} = \dfrac{c}{2}$ where $c$ is odd. Hence, 
$$ \dfrac{ t - \frac{L_k}{2}}{F_k} - \dfrac{\sqrt{5}}{2} s = \dfrac{c-s} {2}+s\dfrac{\sqrt{5}+1}{2}$$ is integer in the quadratic field $\mathbb{Q}(\sqrt{5})$. By dividing $F_k^2$ of two sides in \eqref{equ1}, we have 
\begin{displaymath}
\begin{aligned}
   \pm 1 &amp;=  \left( \dfrac{ t- \frac{L_k}{2} s}{F_k} \right)^2 - \dfrac{5}{4}s^2 \\
&amp;=\left( \dfrac{ t- \frac{L_k}{2} s}{F_k} - \dfrac{\sqrt{5}}{2}s  \right)\left( \dfrac{ t- \frac{L_k}{2} s}{F_k} + \dfrac{\sqrt{5}}{2}s  \right) .
\end{aligned}
\end{displaymath}
 These fractions are units in the quadratic field $\mathbb Q(\sqrt 5)$, hence by \cite[Theorem 257] {H}
\begin{displaymath}
     \dfrac{ t- \frac{L_k}{2} s}{F_k} + \dfrac{\sqrt{5}}{2}s = \pm \alpha^j,\; \text{ where }\; \alpha: = \dfrac{1+\sqrt{5}}{2} , j \in \mathbb{Z} .
 \end{displaymath}
Since $ \alpha^j = \dfrac{L_j + F_j \sqrt{5}}{2}$, we get $s = \pm F_j$ and  
\begin{displaymath}t = \pm \dfrac{F_j L_k + F_k L_j}{2} = \pm  F_{j+k} \;\;\text{(by Lemma \ref{lemma1}.f).}
\end{displaymath}
This implies $(s,t) \in \mathcal{R}$.

\bigskip

\noindent $(\ref{thm1b}) \Rightarrow (\ref{thm1a})$:  If $(s,t) = (F_j, F_{j+k})$, we choose $n$ such that $\dfrac{n-j}{2}$ is an odd integer. Then 
\begin{displaymath}
F_{\frac{n-j}{2} } \mid F_n + F_j \mbox{ and }   F_{\frac{n-j}{2} } \mid F_{n+k} + F_{j+k} \;\; \text{(by Lemma \ref{lemma1} (c)).}
\end{displaymath}
If $(s,t) = -(F_j, F_{j+k})$, we
choose $n$ such that $\dfrac{n-j}{2}$ is an even integer, then 
\begin{displaymath}
F_{\frac{n-j}{2} } \mid F_{n+k} - F_{j+k} \mbox{ and }   F_{\frac{n-j}{2} } \mid F_n - F_j \;\; \text{(by Lemma \ref{lemma1} (d) ).} \end{displaymath}
Hence, $B^k_{s,t}(n)$ is unbounded. The proof of Theorem \ref{thm1} is completed.
\end{proof} 

\begin{proof}[Proof of Theorem \ref{thm2}] The proof of Theorem \ref{thm2} is completely analogous to the one in \cite{S} with $k=1$ and $s=t$. Set 
$e^k_{s,t}(n):=t^2 - L_k st +(-1)^k s^2 - (-1)^nF_k^2$ and $c:=\left | e^k_{s,t}(0)e^k_{s,t}(1)\right |$. 
Then $c\ne 0$ and $B^k_{s,t}(n)$ divides $c$ for $(s,t)\not\in \mathcal R$. Then there is a positive integer $p \leq c^2$ such that 
\begin{displaymath}
F_n \equiv \modd{F_{n+p}} {c} \; \text{for all}\; n \in \mathbb{N}.
\end{displaymath}
We have $B^k_{s,t}(n)= \gcd(F_n +s, F_{n+k} +t) \mid c$ and $F_{n+p} - F_n , F_{n+k+p} - F_{n+k}$ are divisible by $c$, then 
\begin{displaymath}
B^k_{s,t}(n) \mid ((F_{n+p} - F_n)+(F_n+s))\; \text{and}\;  B^k_{s,t}(n) \mid ((F_{n+k+p} - F_{n+k})+(F_{n+k}+t)).
\end{displaymath}
Thus,
\begin{displaymath}
B^k_{s,t}(n)\mid (F_{n+p}+s)  \; \text{and}\;  B^k_{s,t}(n)\mid (F_{n+k+p}+t).
\end{displaymath}
This implies $B^k_{s,t}(n)\mid  \gcd(F_{n+k} +s, F_{n+p+k} +t)$. Hence,  $B^k_{s,t}(n)\mid B^k_{s,t}(n+p)$. Similarly, we have also 
$B^k_{s,t}(n+p)\mid B^k_{s,t}(n)$. Therefore,  $B^k_{s,t}(n+p)=B^k_{s,t}(n)$. 

This $p$ is a period of $B^k_{s,t}(n)$. However,  a period $p$ that we get  by this method is not necessarily the smallest positive one. For example, if we choose $s=t=2$ and $k=1$ then  we have $c=15$ and $F_{n+80}\equiv \modd{F_n} {15}$. Hence $p=80$ but $40$ is also a period of $B^1_{1,2}(n)$  and it is the smallest one.
  \end{proof}
\begin{remark}If $s=t$, we obtain Theorem 7 in \cite{HST}.  If $k=1$, we obtain Theorem 8 in \cite{HST}.
\end{remark}
\begin{remark} The same arguments as in the proof of Theorem \ref{thm1} give the following result on Lucas numbers.
\end{remark}
\begin{theorem}
Let $k \in \mathbb{N}, s,t \in \mathbb{Z}$. The following assertions on $B^{k*}_{s,t}(n)$ are equivalent:
\begin{enumerate}
\item  The function $B^{k*}_{s,t}(n) := \gcd(L_n +s , L_{n+k} +t  ) $ is unbounded on $\mathbb {N}_0$;
\item   $(s,t)\in \mathcal{S}:=\{\pm (L_{j},L_{j+k}) : j\in \mathbb {Z}\}$;
\item  $t^2 - L_k st + (-1)^k s^2 = \pm 5F_k^2$. 
\end{enumerate}
\end{theorem}

\section{Acknowledgment}
The authors would like to thank the referee for useful comments and suggestions.

\begin{thebibliography}{9}
\bibitem {C}  K.-W. Chen, Greatest common divisors in shifted Fibonacci sequences, {\it J. Integer Sequences} \textbf{14} (2011), \href{https://cs.uwaterloo.ca/journals/JIS/VOL14/Chen/chen70.html}{Article 11.4.7}.

\bibitem{CP} K.-W. Chen and Y.-R. Pan,  Greatest common divisors of shifted Horadam sequences, {\it J. Integer Sequences} \textbf{23} (2020), \href{https://cs.uwaterloo.ca/journals/JIS/VOL23/Pan/pan32.html}{Article 20.5.8}.

\bibitem {D} U. Dudley and B. Tucker, Greatest common divisors in altered Fibonacci sequences, {\it Fibonacci Quart.} \textbf{9} (1971), 89--91.

\bibitem{H} G. H. Hardy and E. M. Wright, {\it An Introduction to the Theory of Numbers}, Oxford University Press, 1954.

\bibitem {HST}  L. D. Hieu, J. Spilker, and L. B. Thang, The greatest common divisor of shifted Fibonacci numbers, {\it J. Integer Sequences} \textbf{25} (2022), \href{https://cs.uwaterloo.ca/journals/JIS/VOL25/Thang/thang42.html}{Article 22.1.7}.

\bibitem{K} T. Koshy, {\it Fibonacci and Lucas Numbers With Applications},
Wiley, 2001.

\bibitem {R} A. Rahn and M. Kreh,  Greatest common divisors of shifted Fibonacci sequence revisited, {\it J. Integer Sequences} \textbf{21} (2018), \href{https://cs.uwaterloo.ca/journals/JIS/VOL21/Kreh/kreh6.html}{Article 18.6.7}.

\bibitem {S} J. Spilker, The GCD of the shifted Fibonacci sequence, in J\"{u}rgen Sander, J\"{o}rn Steuding, and Rasa Steuding, eds., {\it From Arithmetic to Zeta-Functions}, Springer, 2016,   pp.~473--483.

\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11B37.

\noindent \emph{Keywords: } Fibonacci number, Lucas number, recurrence,
greatest common divisor.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000045} and
\seqnum{A000204}.)

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\vspace*{+.1in}
\noindent
Received October 18 2022;
revised versions received November 8 2022; November 11 2022.
Published in {\it Journal of Integer Sequences}, November 13 2022.

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