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\begin{center}
\vskip 1cm{\Large\bf 
The Greatest Common Divisor of \\
\vskip .1in
Shifted Fibonacci Numbers
}
\vskip 1cm
Le Duc Hieu\\
\'Ecole Polytechnique \\ Institut Polytechnique de Paris\\ Palaiseau \\
France \\
\href{mailto: hieu.le@polytechnique.edu }{\tt hieu.le@polytechnique.edu }\\
\ \\
J\"{u}rgen Spilker\\
Institute of Mathematics\\
University of Freiburg\\ 
79085 Freiburg im Breisgau \\
Germany\\
\href{mailto:juergen.spilker@t-online.de }{\tt juergen.spilker@t-online.de}\\
\ \\
Luu Ba Thang \\
Department of Mathematics and Informatics \\
Hanoi National University of Education \\
136 Xuan Thuy \\
Cau Giay \\
Hanoi\\
Vietnam \\
\href{mailto: thanglb@hnue.edu.vn}{\tt thanglb@hnue.edu.vn} 
\end{center}

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\def\modd#1 #2{#1\ \mbox{\rm (mod}\ #2\mbox{\rm )}}

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\begin{abstract}
Let $F_n$ and  $L_n$ denote the Fibonacci and Lucas
numbers, respectively. We generalize the well-known formula $\gcd
(F_{6n+3}+1,F_{6n+6}+1)=L_{3n+2}$ if $n$ is even. We find conditions
on $a$, $s$, and $t$ so that the functions $n\mapsto \gcd(F_n+s, F_{n+a}+s)$
and $n \mapsto \gcd(F_n+s, F_{n+1}+t)$  are unbounded. Finally, we generalize this
for three shifted Fibonacci numbers.
\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
$$ F_0=0, F_1=1;
F_n=F_{n-1}+F_{n-2}, n\geq 2;
$$ and $$L_0=2, L_1=1;\\
L_n=L_{n-1}+L_{n-2}, n\geq 2.
$$
Notice that $F_n$ and $L_n$ can be extended to integer indices by the recursion. 
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 $s^2+(-1)^n$.  If $s\ne \pm 1$ then the function $n \mapsto \gcd(F_n+s,F_{n+1}+s)$ is periodic. 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 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 $\gcd(W_n+s, W_{n+1}+t)$.

Let $\mathbb {N}$ denote the set of positive integers, $\mathbb N_0$ denote the set of non-negative  integers, and  $\mathbb Z$ denote the set of  integers.
In this article, we generalize these ideas to compute 
$$\gcd(F_{2an+2b+i}+(-1)^b, F_{2an+2b+i+k}+(-1)^bF_{k-1})$$
in Theorem \ref{thm1}. Then in Theorem \ref{thm6}, we investigate  necessary and sufficien conditions on $a\in \mathbb N, s \in \mathbb Z$, so that the function 
\begin{displaymath}
\begin{aligned}
D_{a,s}: \mathbb N_0 &amp;\to \mathbb N \\
n &amp;\mapsto \gcd(F_n+s, F_{n+a}+s) 
\end{aligned}
\end{displaymath}
is bounded. In Theorem \ref{thm10}, we treat the same problem for $C_{s,t}(n):=\gcd(F_n+s,F_{n+1}+t)$.  If $D_{a,s}(n)$ and $C_{s,t}(n)$ are bounded, they are periodic (Theorem \ref{thm7} and Theorem \ref{thm11}). Finally, we prove that the function 
\begin{displaymath}
\begin{aligned}
E_{a,s}: \mathbb N_0 &amp;\to \mathbb N\\
n &amp;\mapsto \gcd(F_n+s, F_{n+a}+s, F_{n+2a}+s) 
\end{aligned}
\end{displaymath}
is bounded without exceptions.
\section{Results}
Firstly, we have a theorem on the gcd of Fibonacci numbers with shift $1$ and with general shift $k\geq 1$.
\begin{theorem}\label{thm0} Let $a,b,i\in \mathbb Z$ with $ i = 2j+1$ and  set$$D:=\gcd(F_{2an+2b+i}+(-1)^b, F_{2an+2b+i+1}+(-1)^b).$$
Then 
\begin{displaymath}
D = 
\begin{cases}
F_{an+b+j}, &amp; \text{ if  $an+j$  is odd};\\ 
L_{an+b+j}, &amp; \text{ if  $an+j$  is even}
\end{cases}
\end{displaymath}
for all $n\in \mathbb N_0$.
\end{theorem}

\begin{theorem}\label{thm1} Let $a,b,i,k \in \mathbb Z$ with $ i = 2j - 1, k \geq 1$ and set $$D:=\gcd(F_{2an+2b+i}+(-1)^b, F_{2an+2b+i+k}+(-1)^bF_{k-1}).$$
Then 
\begin{displaymath}
D = 
\begin{cases}
F_{an+b+j}\gcd(L_{an+b+j-1}, F_k), &amp; \text{ if  $an+j$  is odd};\\ 
L_{an+b+j}\gcd(F_{an+b+j-1}, F_k), &amp; \text{ if  $an+j$  is even,}
\end{cases}
\end{displaymath}
for all $n\in \mathbb N_0$.
\end{theorem}


\begin{corollary}\label{thm2} Let $a,b,i \in \mathbb Z$ with $i = 2j - 1$.
Then
\begin{displaymath}
\gcd(F_{2an+2b+i}+(-1)^b, F_{2an+2b+i+1}) = 
\begin{cases}
F_{an+b+j}, &amp; \text{ if  $an+j$ is odd;}\\
L_{an+b+j}, &amp; \text{ if  $an+j $ is even,}
\end{cases}
\end{displaymath}
for all $n\in \mathbb N_0$.
\end{corollary}

\begin{corollary}\label{thm3} Let $a,b,i,j \in \mathbb Z$ with $i = 2j - 1$. 
Then 
\begin{displaymath}
\gcd(F_{2an+2b+i}+(-1)^b, F_{2an+2b+i+2}+(-1)^b) = 
\begin{cases}
F_{an+b+j}, &amp; \text{ if  $an+j$  is odd;}\\
L_{an+b+j}, &amp; \text{ if  $an+j $ is even,}
\end{cases}
\end{displaymath}
for all $n\in \mathbb N_0$.
\end{corollary}

\begin{corollary}\label{thm4} Let $a,b,i,j\in \mathbb Z$ with $i = 2j - 1$. 
Then 
\begin{displaymath}
\gcd(F_{2an+2b+i}+(-1)^b, F_{2an+2b+i+3}+(-1)^b) = 
\begin{cases}
\lambda F_{an+b+j},  &amp; \text{ if  $an+j$  is odd;}\\
\lambda L_{an+b+j}, &amp; \text{ if  $an+j$  is even,}
\end{cases}
\end{displaymath}
 for all $n\in \mathbb N_0$,
where 
\begin{displaymath}
\lambda := 
\begin{cases}
2,  &amp;\text{ if  $an+b+j \equiv \modd{1} {3}$;}\\
1, &amp; \text{ otherwise.}\\
\end{cases}
\end{displaymath}
\end{corollary}

\begin{corollary}\label{thm5}
Let $a,b,i \in \mathbb Z$ with $i = 2j - 1$. Then 
\begin{displaymath}
\gcd(F_{2an+2b+i}+(-1)^b, F_{2an+2b+i+4}+(-1)^b2) = 
\begin{cases}
\lambda F_{an+b+j}, &amp; \text{ if  $an+j$  is odd;}\\
\delta L_{an+b+j}, &amp; \text{ if $an+j$  is even,}
\end{cases}
\end{displaymath}
for $n\in \mathbb N_0$,
where
\begin{displaymath}
\lambda := 
\begin{cases}
3, &amp; \text{ if  $an+b+j \equiv \modd{3} {4}$;}\\
1, &amp; \text{ otherwise,}
\end{cases}
\end{displaymath}
and 
\begin{displaymath}
\delta := 
\begin{cases}
3, &amp; \text{ if  $an+b+j \equiv \modd{1} {4}$;}\\
1, &amp; \text{ otherwise}.
\end{cases}
\end{displaymath}
\end{corollary}
The next theorem characterizes the boundedness of $D_{a,s}(n):=\gcd(F_n+s, F_{n+a}+s)$.
\begin{theorem}\label{thm6}
Let $a \in \mathbb N, s \in \mathbb Z$, and set
$$
\mathcal{S}:=\{(1,1),(1,-1),(3,1),(3,-1)\} \cup \{(4k+2, F_{2k+1}):k \geq 0\} \cup \{(4k+2, -F_{2k+1}): k \geq 0\}.
$$
Then $D_{a,s}(n)$ is bounded on $\mathbb N_0$ if and only if $(a,s)\notin \mathcal{S}$.
\end{theorem}

Let $e_{s,t}(n):=s^2+st-t^2+(-1)^n $ and $e_{s,t}^*:=e_{s,t}(0)e_{s,t}(1)$. We characterize the boundedness of
$$C_{s,t}(n):=\gcd(F_n+s,F_{n+1}+t).$$

\begin{theorem}\label{thm10}Let $s,t \in \mathbb Z $ and $n\in \mathbb N_0$. The following assertions on $C_{s,t}(n)$ are equivalent:
\begin{itemize}
\item[\rm a)] $n\mapsto C_{s,t}(n)$ is unbounded on $\mathbb N_0$;
\item[\rm b)] $(s,t)\in \mathcal{R}:=\{\pm (F_{j-1},F_j), \pm (F_j,-F_{j-1}): j\in \mathbb N_0\};$
\item[\rm c)] $e^*(s,t)=0$.
\end{itemize}
\end{theorem}
The next two theorems show the boundedness of $D_{a,s}(n)$ and $C_{s,t}(n)$ respectively imply periodicity.
\begin{theorem}\label{thm7}
Let $a \in \mathbb N$ and $s \in \mathbb Z$. If $(a,s) \notin \mathcal{S} $, then the function $D_{a,s}(n)$
is simply periodic on $\mathbb N_0$, which means there exists a positive integer $p$ such that $D_{a,s}(n+p)=D_{a,s}(n)\text{ }$ for all $n \geq 0$. A period $p \leq c^2$ can be chosen such that
\begin{equation}\label{perio}
F_{p} \equiv \modd{0} {c}\; \text{and} \; F_{p+1} \equiv \modd{1} {c}
\end{equation}
where $c := |F_a^4 - s^4(L_a-1+(-1)^{a+1})^2|$.
\end{theorem}
\begin{theorem}\label{thm11} If $(s,t) \notin \mathcal{R} $, then the function $C_{s,t}(n)$
is simply periodic on $\mathbb N_0$. A period $p\leq c^2$ can be chosen by (\ref{perio}), where $c:=|(s^2 +st-t^2)^2-1|$.
\end{theorem}
A theorem on three shifted Fibonacci numbers is given in the following last main result.
\begin{theorem}\label{thm8}
Let $a \in \mathbb N$ and  $s \in \mathbb Z$ and set
$$E_{a,s}(n) := \gcd(F_n+s, F_{n+a}+s, F_{n+2a}+s), $$
for all $n\in \mathbb N_0$.
Then we have
\begin{enumerate}
    \item [\rm a)] $E_{a,s} (n)$ divides $d$ where \begin{displaymath} d:=  \begin{cases}s(L_a-1+(-1)^{a+1}), &amp; \text{ if  $s\neq 0$;}\\  F_a, &amp;  \text{ if  $s = 0$.} \end{cases}
\end{displaymath} 
    \item [\rm b)] The function $n \mapsto E_{a,s}(n)$ is simply periodic.
\end{enumerate}
\end{theorem}

\section{Auxiliary results}
In the following proofs we need some well-known facts on the sequences $(F_n)_{n\geq 0}$ and $(L_n)_{n\geq 0}$, which we have collected in the following lemma:
\begin{lemma}[\cite{K}]\label{lemma1}
Let $n, m \in \mathbb Z$. Then we have
\begin{enumerate}
    \item [\rm a)] $\gcd(F_n, F_{n+1}) = 1;$
    \item [\rm b)] $\gcd(L_n, L_{n+1}) = 1;$
    \item [\rm c)] $F_{-n} = (-1)^{n+1}F_n;$
    \item [\rm d)] $L_{-n} = (-1)^nL_n;$
    \item [\rm e)] $\gcd(F_m, F_n) = F_{\gcd(m, n)};$
    \item [\rm f)] $F_m \mid F_n$ if and only if $m \mid n;$
    \item [\rm g)] $F_{2n} = F_nL_n;$
    \item [\rm h)] $L_n = F_{n+1} + F_{n-1};$
    \item [\rm i)] $5F_n = L_{n+1}+L_{n-1};$
    \item [\rm j)] $F_{m+n} = F_mL_n + (-1)^{n+1}F_{m-n}$, $m \geq n;$
    \item [\rm k)] $F_{m+n} = F_nL_m + (-1)^nF_{m-n}$, $m \geq n;$
    \item [\rm l)] $5F_n = L_mL_{n-m+1}+L_{m-1}L_{n-m};$
    \item [\rm m)] $F_{m+n} = F_mF_{n+1} + F_{m-1}F_n;$
    \item [\rm n)] $F_{m+2n}F_m - F_{m+n}^2 = (-1)^{m+1}F_n^2;$
\item [\rm o)] $L_{4n+2}-2=L_{2n+1}^2;$
\item [\rm p)] $5F_{2n}^2=L_{4n}-2$.
\end{enumerate}
\end{lemma}
\section{Proofs  of Theorems $1$ and $2$ and Corollaries 3--6}
\begin{proof}[Proof of Theorem \ref{thm0}] Using the property $\gcd(u,v)=\gcd(u,v-u)$, observe that
$$ D = \gcd(F_{2an+2b+i}+(-1)^b, F_{2an+2b+2j}).$$
By Lemma \ref{lemma1}.g, we have
$$F_{2an+2b+2j} = F_{an+b+j}L_{an+b+j}.$$

If $an+j$ is odd, then by Lemma \ref{lemma1}.k, it follows that
$$F_{2an+2b+i}+(-1)^b= F_{an+b+j}L_{an+b+j+1}.$$
This implies
$$D=F_{an+b+j}\gcd(L_{an+b+j+1},L_{an+b+j}),$$
and hence by Lemma \ref{lemma1}.b, we have $D=F_{an+b+j}$.

If $an+j$ is even, then
$$F_{2an+2b+i}+(-1)^b= F_{an+b+j+1}L_{an+b+j}.$$
and 
$$D=L_{an+b+j}\gcd(F_{an+b+j+1},F_{an+b+j}) =L_{an+b+j}.$$
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm1}]
Using Lemma \ref{lemma1}.m, we deduce that
$$F_{2an+2b+i+k} = F_kF_{2an+2b+i+1}+F_{k-1}F_{2an+2b+i}.$$
Thus, by $\gcd(u,v)=\gcd(u,v-F_{k-1}u)$, we have
$$D = \gcd(F_{2an+2b+i}+(-1)^b, F_kF_{2an+2b+i+1}).$$
Using Lemma \ref{lemma1}.g, we deduce that
$$F_{2an+2b+i+1} = F_{an+b+j}L_{an+b+j}.$$
Now we consider two cases: $an+j$ is odd or $an+j$ is even.

 If $an+j$ is odd, then by Lemma \ref{lemma1}.j, we have
$$F_{2an+2b+i}+(-1)^b = F_{an+b+j}L_{an+b+j-1}+(-1)^{an+b+j}+(-1)^b=F_{an+b+j}L_{an+b+j-1}.$$
Thus, combining with Lemma \ref{lemma1}.b, we have 
$$D = F_{an+b+j}\gcd(L_{an+b+j-1}, F_kL_{an+b+j}) = F_{an+b+j}\gcd(L_{an+b+j-1}, F_k).$$

If $an+j$ is even, then by Lemma \ref{lemma1}.k,  we deduce that
$$F_{2an+2b+i}+(-1)^b = L_{an+b+j}F_{an+b+j-1}+(-1)^{an+b+j-1}+(-1)^b=L_{an+b+j}F_{an+b+j-1}.$$
Thus, combining with Lemma \ref{lemma1}.a, we have that
$$D = L_{an+b+j}\gcd(F_{an+b+j-1}, F_kF_{an+b+j}) = L_{an+b+j}\gcd(F_{an+b+j-1}, F_k).$$
\end{proof} 

\begin{proof}[Proof of Corollary \ref{thm2} and Corollary \ref{thm3}] 
Notice that we have $$\gcd(L_{an+b+j-1}, F_1) = \gcd(F_{an+b+j-1}, F_1) = 1,$$
and 
$$\gcd(L_{an+b+j-1}, F_2) = \gcd(F_{an+b+j-1}, F_2) = 1.$$
So we obtain Corollary \ref{thm2} and Corollary \ref{thm3} from Theorem \ref{thm1} by setting $k=1$ or $k=2$.
\end{proof}

\begin{proof}[Proof of Corollary \ref{thm4}]
We have 
\begin{displaymath}
\begin{cases}
F_m \equiv \modd{0} {2}, &amp;\text{ if and only if } m \equiv \modd{0} {3};\\
L_m \equiv \modd{0} {2}, &amp;\text{ if and only if } m\equiv \modd{0} {3};
\end{cases}
\end{displaymath}
and 
\begin{displaymath}
\gcd(F_{an+b+j-1}, F_3) =
\begin{cases}
2, &amp;\text{ if $ an + b + j \equiv \modd{1} {3}$;}\\
1, &amp;\text{ otherwise;}
\end{cases}
\end{displaymath}
and
\begin{displaymath}
\gcd(L_{an+b+j-1}, F_3) =
\begin{cases}
2, &amp; \text{ if  $an + b + j \equiv \modd{1} {3}$;}\\
1, &amp;\text{ otherwise.}
\end{cases}
\end{displaymath}
Therefore, we conclude that Corollary  \ref{thm4} is the case $k = 3$ in Theorem \ref{thm1}.
\end{proof}
\begin{example} If $a=3,b=0, i=3$ and $n$ is even, we get
$$\gcd(F_{6n+3}+1, F_{6n+6}+1)=L_{3n+2}.$$
\end{example}

\begin{proof}[Proof of Corollary \ref{thm5}]
The proof is similar to the proof of Corollary \ref{thm4} by using the fact that $F_4 = 3$ and 
\begin{displaymath}
\begin{cases}
F_m \equiv \modd{0} {3} &amp; \text{ if and only if } m \equiv \modd{0} {4};\\
L_m \equiv \modd{0} {3} &amp;\text{ if and only if } m \equiv \modd{2} {4}.
\end{cases}
\end{displaymath}
\end{proof}

\section{Proof of Theorem \ref{thm6}}
Let $D= D_{a,s}(n)$. We need two lemmas.

\begin{lemma}\label{lemma3} For all $a\in \mathbb N, s\in \mathbb Z$, and $n\in \mathbb N_0$, we have
$D_{a,s}(n)$ divides the value \\ $F_a^2+(-1)^n s^2(L_a-1+(-1)^{a+1})$.
\end{lemma}

\begin{proof} 
We apply Lemma \ref{lemma1}.n and get on the one hand
$$F_{n+2a} F_n-F_{n+a}^2 = (-1)^{n+1}F_a^2,$$
hence 
$$-sF_{n+2a}-s^2\equiv (-1)^{n+1}F_a^2\pmod D.$$
On the other hand, we have 
$$F_{n+2a}=F_{n+a}L_a+(-1)^{a+1}F_n$$ by Lemma \ref{lemma1}.j and 
$$F_{n+2a}\equiv -s L_a+(-1)^{a}s \pmod D.$$
Together, this gives us
$$s^2(L_a+(-1)^{a+1}-1)\equiv (-1)^{n+1}F_a^2\pmod D.$$

\end{proof}

\begin{lemma}\label{lemma4}
 Let $(a,s)\in \mathbb Z^2 $ and $a&gt;0$. Then 
\begin{itemize}
\item[\rm 1)]  $F_a^2 + s^2(L_a-1+(-1)^{a+1})=0$ is impossible.
\item [\rm 2)] $ F_a^2 - s^2(L_a-1+(-1)^{a+1})=0$ implies $(a,s)\in \mathcal S$.
\end{itemize}
 \end{lemma}

 \begin{proof}  Let $(a,s)$ be a solution of the  equation  $F_a^2+s^2(L_a-1+(-1)^{a+1})=0$. If $a=1$, then $1+s^2=0$, a contradiction.
If $a&gt;1$, then $L_a-1+(-1)^{a+1}\geq L_a-2&gt;0$; hence $F_a=0$ and $s=0$, also a contradiction.

Let $(a,s)$ be a solution of the equation  $F_a^2-s^2(L_a-1+(-1)^{a+1})=0$. Then we have 

$$ F_a^2= s^2(L_a-1+(-1)^{a+1}). $$
 Now we consider two cases for $a$.

\bigskip

\noindent{\it First case:} $a$ is odd, that is $F_a^2=s^2L_a$.  If $a=1$, then $1=s^2$ and $(a,s)=(1,\pm 1)\in \mathcal{S}$. We can assume $a\geq 3$. Let $p$ a prime divisor of $L_a$, hence of $F_a$. Since 
$$\gcd(F_a, L_a)=\gcd(F_a, L_a-F_a)=\gcd (F_a, 2F_{a-1})\in \{1,2\},$$
we get $p=2, L_a=2^{2\alpha}, F_a=\left | s \right | 2^\alpha, \alpha \geq 1$. Since $a$ is odd and $F_a$ is even, we see from Table~\ref{table11} that
$a\equiv \modd{3} {6}$ and $F_a\not\equiv \modd{0} {4}$; hence $\alpha =1, L_a=4$ and $(a,s)=(3, \pm 1)\in \mathcal{S}$.
\begin{table}[H]
\centering 
\begin{tabular}{c | c c c c c}
\hline\hline 
$a$ &amp; 3  &amp; 5 &amp; 7 &amp; 9 &amp; 11\\ [0.5ex] 
\hline 
$F_a$ &amp;  2 &amp; 5 &amp; 13 &amp; 34 &amp; 89 \\ 
$F_a \mod 2$ &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 1 \\
$F_a \mod 4$ &amp; 2 &amp; 1 &amp; 3 &amp; 2 &amp; 1 \\
 [1ex] 
\hline
\end{tabular}
\caption{Residue} 
\label{table11} 
\end{table}

\noindent{\it Second case:}  $a$ is even, that is $F_a^2=s^2(L_a-2)$. If $a\equiv \modd{0} {4}$, say $a=4k, k&gt;0$, then $F_a^2=s^2 5 F_{2k}^2$ by Lemma \ref{lemma1}.p and $5$ is square, a contradiction. Now we can assume $a\equiv \modd{2} {4}$, say $a=4k+2, k\geq 0$. Then $L_a-2=L_{2k+1}^2$ by Lemma \ref{lemma1}.o and 
$$F_{2k+1}^2(L_{4k+2}-2)=F_{2k+1}^2L_{2k+1}^2=F_{4k+2}^2$$
by Lemma \ref{lemma1}.g. This implies $(a,s)=(4k+2, \pm F_{2k+1})\in \mathcal {S}$. The proof is completed.

\end{proof}

 \begin{proof}[Proof of Theorem \ref{thm6}]
First, let $(a,s) \in \mathbb N \times \mathbb Z, (a,s) \notin S$. By Lemma \ref{lemma4}, we have 
$$F_a^2+(-1)^ns^2(L_a-1+(-1)^{a+1}) \neq 0 \; \text{for} \; n\in \{0,1\}$$
and by Lemma \ref{lemma3}, the function $D_{a,s}(n)$ is bounded.

On the other hand, if $(a,s) \in S$, then we see that $D_{a,s}$ is unbounded, since for $n\in \mathbb N$
by Theorem \ref{thm0}, we have
$$
D_{1,1}(4n+1) =L_{2n}\; \text{and} \; D_{1,-1}(4n-1) =F_{2n},
$$
and by Corollary \ref{thm4}, we have
$$
D_{3,1}(4n+1) =\lambda F_{2n+1}\geq F_{2n+1}\; \text{and}\; 
D_{3,-1}(4n+1) =\lambda L_{2n+1}\geq L_{2n+1}.
$$


If $n=4m+2k+3$ then
\begin{displaymath}
\begin{aligned}
\gcd(F_n+F_{2k+1}, F_{n+4k+2}+F_{2k+1})
&amp;=\gcd(F_n+F_{2k+1}, F_{n+4k+2}-F_n)\\
&amp;=\gcd(F_{4m+2k+3}+F_{2k+1}, F_{4m+6k+5}-F_{4m+2k+3})\\
&amp;=\gcd(F_{2m+1}L_{2m+2k+2}, F_{4m+4k+4}L_{2k+1}) \\ &amp;\;\;\;\; \;\text{( by Lemma \ref{lemma1}.j and \ref{lemma1}.k)}\\
&amp;=L_{2m+2k+2}\gcd(F_{2m+1},F_{2m+2k+2}L_{2k+1}) \text{ (by Lemma \ref{lemma1}.g) .}
\end{aligned}
\end{displaymath}

Then $L_{2m+2k+2}$ divides  $D_{4k+2, F_{2k+1}}(n)$. Hence, $D_{4k+2, F_{2k+1}}(n)$ is unbounded.

Same argumentation, we obtain the unboundedness of  $D_{4k+2, -F_{2k+1}}(n)$. So all $(a,s) \in \mathcal {S}$ generate  unbounded functions $D_{a,s}(n)$.
\end{proof}

\begin{remark} The same arguments as in the proof of Theorem \ref{thm6} give the following result on Lucas numbers.
\end{remark}

\begin{theorem}
Let $a\in \mathbb N, s\in \mathbb Z$. Then the conditions a) and b) are equivalent:
\begin{itemize}
\item [\rm a)] The function $D^*_{a,s}(n)=\gcd (L_n+s, L_{n+a}+s)$ is unbounded on $\mathbb N_0$;
\item [\rm b)] $a=4b, s=\pm L_{2b}$ for $b\in \mathbb N$.
\end{itemize}
\end{theorem}

\section{Proof of Theorem \ref{thm10}}
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 1)] $C=C_{s,t}(n)$ divides $e_{s,t}(n)$;
\item [\rm 2)] $C$ divides $e^*_{s,t}$.
\end{itemize}
\end{lemma}
\begin{proof} Let $n\in \mathbb N_0$ and define $M(n)$ and $Q$ as follows:
\begin{displaymath}
M(n):= \begin{pmatrix}F_{n+2} &amp; F_{n+1} \\ F_{n+1} &amp; F_n \end{pmatrix}\; \text{and}\; Q:= \begin{pmatrix}1 &amp; 1 \\ 1 &amp; 0 \end{pmatrix}.
\end{displaymath}
Then $M(n+1)=QM(n)$ and $M(n)=Q^n M(0)$. Taking determinants  gives 
\begin{displaymath}
\begin{aligned}
(-1)^{n+1}=\det M(n)&amp; =F_{n+2}F_n-F_{n+1}^2 \\
&amp; = (F_{n+1}+F_n)F_n - F_{n+1}^2\\
&amp; \equiv  (t+s)s-t^2 \pmod C .
\end{aligned}
\end{displaymath}
Hence  $e_{s,t}(n)\equiv \modd{0} {C}$ and $e^*_{s,t}\equiv \modd{0} {C}$.
\end{proof}
\begin{corollary}\label{cor12} If $s,t\in \mathbb Z$ and $e^*_{s,t}\ne 0$, then the function $n\mapsto C_{s,t}(n)$ is bounded on $\mathbb N$.
\end{corollary}
\begin{proof}
Now, we give a proof of Theorem \ref{thm10} by the circle $(a) \Rightarrow (c) \Rightarrow (b) \Rightarrow (a)$.

 $(a) \Rightarrow (c)$ by Corollary \ref{cor12}.

 $(c) \Rightarrow (b)$ We have 
\begin{displaymath}
\begin{aligned}
\pm 1 &amp;= s^2+st-t^2 \\
           &amp; = (s+\dfrac t 2)^2 - 5(\dfrac t 2)^2\\
 &amp; =\dfrac{2s+t+t\sqrt 5} 2  \cdot \dfrac{2s+t-t\sqrt 5} 2.
\end{aligned}
\end{displaymath}
These fractions are units in the quadratic field $\mathbb Q(\sqrt 5)$; hence by \cite[Theorem 257] {H}
$$\dfrac {2s+t+t\sqrt 5} 2=\pm \alpha^j, \ \text{where} \ \alpha:= \dfrac {1+\sqrt 5} 2,\ j\in \mathbb Z.$$
Since $\alpha^j = \dfrac{ L_j+F_j\sqrt 5} 2 $, we get $t=\pm F_j$ and 
$$s=\pm \dfrac{L_j-t} 2=\pm \dfrac{F_{j+1}+F_{j-1}-F_j} 2 =\pm F_{j-1}, j\in \mathbb Z.$$
Since $F_{-j}=(-1)^{j+1}F_j$, we get $(s,t)\in \mathcal R$.

$(b) \Rightarrow (a)$ Let $i\geq 2$. By Jacobson \cite{J} we have
$$ F_{2^i3}\equiv \modd{0} {2^{i+1}}; \quad F_{2^i3-1}\equiv \modd{1} {2^{i+1}}.$$
Hence, by Lemma \ref{lemma1}.m
\begin{displaymath}
\begin{aligned}
F_{2^i3+j} &amp;= F_{2^i3}F_{j+1}+F_{2^i3-1}F_j\\
            &amp;\equiv F_j \pmod  {2^{i+1}}
\end{aligned}
\end{displaymath}
and $F_{2^i3-j}\equiv F_{-j}\equiv \modd{(-1)^{j+1} F_j} {2^{i+1}}$.  We see that the function $n\mapsto C_{s,t}(n)$ is unbounded on $\mathbb N$ if $(s,t)=-(F_j, F_{j+1})$ and $(s,t)=(-1)^j(F_j, -F_{j-1}), j\in \mathbb N_0$. If $(s,t)=(F_j, F_{j-1}), j\in \mathbb N_0$, then 
\begin{displaymath}
\begin{aligned}
C_{s,t}(n) &amp;= \gcd(F_n+F_j, F_{n+1}+F_{j+1})\\
           &amp; =\gcd (F_{n-1}+F_{j-1}, F_n+F_j )\\
&amp;=\gcd (F_{n-2}+F_{j-2}, F_{n-1}+F_{j-1})\\
 &amp; =\gcd (F_{n-j}, F_{n-j+1}+1).
\end{aligned}
\end{displaymath}
Since $F_{2m}=F_mL_m$ and $F_{2m+1}+1=F_mL_{m+1}$ if $m$ is odd, we have for $m$ is odd
$$ C_{s,t}(2m+j) =\gcd(F_{2m}, F_{2m+1}+1)=F_m\gcd(L_m,L_{m+1})=F_m,$$
so $C_{s,t}(n)$ is unbounded. The last case $(s,t)=(-1)^{j+1}(F_j, -F_{j-1})=(F_{-j}, F_{-(j-1)}), j \in \mathbb N_0$ is proved in a similar way. 
Thus, the proof of Theorem \ref{thm10} is completed.
\begin{remark} It is an open problem to characterize the general function $$n\mapsto \gcd (F_n+s, F_{n+a}+t), a\in \mathbb N, s,t \in \mathbb Z.$$
\end{remark}
\section{Proof of periodicity (Theorem \ref{thm7} and \ref{thm11})}
The proof of Theorem \ref{thm7}  and Theorem \ref{thm11} are completely analogous to the one in \cite{S} with $a=1$. Set $c_n:= F_a^2+(-1)^ns^2(L_a-1+(-1)^{a+1})$ and $c:=\left | c_0c_1\right |, $ then $c\ne 0$ and $D_{a,s}(n)$ divides $c$ for $(a,s)\not\in \mathcal S$. Then there is an integer $1\leq p\leq c^2$ such that $F_{n+p}\equiv \modd{F_n} {c}$ for all $n\in \mathbb N$. This $p$ is a period of $D_{a,s}(n)$.
Same argumentation, we obtain the proof of Theorem \ref{thm11}.
\end{proof}

 A period $p$ that we get  by this method is not necessarily the smallest positive one. For example, if $a=1,s=2$ in Theorem \ref{thm7} or $t=s=2$ in Theorem \ref{thm11}, then  we have $c=15$ and $F_{n+80}\equiv \modd{F_n} {15}$. Hence $p=80$, but $40$ is also a period of $D_{1,2}(n)$  and it is the smallest one \cite{C}.

 One cannot always get the smallest period by this method. For example, if $a=4,s=-2$ in Theorem \ref{thm7}; hence $c=319$.   Then $10$ is the smallest period, but $F_{10}\not \equiv \modd{F_0} {319}$. If $s=7, t=2$ in Theorem \ref{thm11}, then we have $c=3480$ and $p=840$, but $120$ is the smallest period and  $F_{120}\not \equiv \modd{F_0} {3480}$.
 

\section{Gcd of three shifted Fibonacci numbers}
\begin{proof}[Proof of Theorem \ref{thm8}]
Fix $a&gt;0, n \geq 0$ and $s\in \mathbb Z$, and set $E(n)=E_{a,s}(n)$. Then by Lemma \ref{lemma1}.j, we have 
$$F_{n+2a} = F_{n+a}L_a + (-1)^{a+1}F_n,$$
which gives us that
$$-s \equiv -sL_a - (-1)^{a+1}s \pmod {E(n)}.$$
So $E(n)$ divides $s(L_a-1+(-1)^{a+1})$.   If $s = 0$, then by Lemma \ref{lemma1}.e we have 
$$E(n) = \gcd(F_n,F_{n+a}, F_{n+2a}) = F_{\gcd(n,n+a,n+2a)} = F_{\gcd(n,a)}=\gcd(F_n, F_a).$$
Hence, $E(n)$ divides $F_a$. Let 
\begin{displaymath} d:= \begin{cases}s(L_a-1+(-1)^{a+1}), &amp; \text{ if  $s\neq 0$;}\\  F_a, &amp;  \text{ if  $s = 0$.} \end{cases}
\end{displaymath}
Then $d\neq 0$, and we deduce that
$E(n)$ divides $d$.
In particular, the function $E(n)$ is bounded. As in the proof of Theorem \ref{thm7}, we conclude that this function is simply periodic.
\end{proof}
\begin{remark}
If $s=0$, then $E(a) = \gcd(F_a, F_{2a}, F_{3a}) = F_a$ so arbitrarily large values are possible.
\end{remark}

\section{Acknowledgments}
The authors would like to thank the referee and the editor-in-chief for many useful comments and suggestions. 
The third author would like to thank Hanoi National University of Education for the support during his work, No.~SPHN 21-03 TT.

\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{J} E. Jacobson, Distribution of the Fibonacci numbers mod $2^k$, {\it Fibonacci Quart.} \textbf{30} (1992), 211--215.

\bibitem{K} T. Koshy, {\it Fibonacci and Lucas Numbers With Applications}, A Wiley---Interscience Series of Text, 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, recurrence.

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\noindent (Concerned with sequence
\seqnum{A000045} and
\seqnum{A000204}.)


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\vspace*{+.1in}
\noindent
Received June 15 2020;
revised versions received  October 30 2020; January 3 2021; July 4 2021; 
July 22 2021; November 25 2021; January 21 2022.
Published in {\it Journal of Integer Sequences}, January 26 2022.

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