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\begin{center}
\vskip 1cm{\LARGE\bf Greatest Common Divisors in \\
\vskip .1in
Shifted Fibonacci Sequences
}
\vskip 1cm
\large
Kwang-Wu Chen\\
Department of Mathematics and Computer Science Education\\
Taipei Municipal University of Education\\
No.\ $1$,  Ai-Kuo West Road \\
Taipei, Taiwan $100$, R.O.C.\\
\href{mailto:kwchen@tmue.edu.tw}{\tt kwchen@tmue.edu.tw} \\
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\vskip .2 in

\begin{abstract}
It is well known that successive members of the 
Fibonacci sequence are relatively prime.
Let 
$$
f_n(a)=\gcd(F_n+a,F_{n+1}+a).
$$
Therefore $(f_n(0))$ is the constant sequence $1,1,1,\ldots$, 
but Hoggatt in 1971 noted that $(f_n(\pm1))$ is unbounded.
In this note we prove that $(f_n(a))$ is bounded
if $a\neq\pm 1$.
\end{abstract}

\theoremstyle{plain}
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\newtheorem{corollary}{Corollary}
\newtheorem{lemma}{Lemma}

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\section{Introduction}
Let the generalized Fibonacci sequence be defined by 
$$
G_n=G_{n-1}+G_{n-2},\quad\mbox{for }n\geq 3,
$$
and $G_1=\alpha$, $G_2=\beta$. It is well known that \cite[p.\ 109]{K}
$$
G_n=\alpha F_{n-2}+\beta F_{n-1}.
$$
If $\alpha=\beta=1$, then the generalized Fibonacci sequence $G_n$ is 
the Fibonacci sequence $F_n$, \seqnum{A000045}, and
if $\alpha=1$ and $\beta=3$, $G_n$ is
the Lucas sequence $L_n$, \seqnum{A000032}.
It is well known that successive members of the 
Fibonacci sequence are relatively prime. Consider a slightly
different sequence, 
$$
(F_n+a),
$$
which we call a shifted Fibonacci sequence
by $a$, e.g., \seqnum{A000071}, \seqnum{A001611}, and \seqnum{A157725}. 
In 1971 Hoggatt \cite{DT} noted that
\begin{eqnarray*}
\gcd(F_{4n+1}+1,F_{4n+2}+1) &amp;=&amp; L_{2n},\\
\gcd(F_{4n+1}-1,F_{4n+2}-1) &amp;=&amp; F_{2n},\\
\gcd(F_{4n+3}-1,F_{4n+4}-1) &amp;=&amp; L_{2n+1}.
\end{eqnarray*}
That is to say, the successive members of 
the shifted Fibonacci sequence by $\pm 1$ are not always relatively
prime. Let 
$$
f_n(a)=\gcd(F_n+a,F_{n+1}+a).
$$
Therefore $(f_n(0))$ is the constant sequence $1,1,1,\ldots$, 
but $(f_n(\pm1))$ is unbounded above.

In 2003 Hern\'andez and Luca \cite{HL} proved that there exists
a constant $c$ such that 
$$
\gcd(F_m+a,F_n+a)&gt;\exp(cm),
$$
holds for infinitely many pairs of positive integers $m&gt;n$.

In this note we prove that $(f_n(a))$ is bounded above
if $a\neq\pm 1$.
In fact we prove the following two theorems in this note.
\begin{theorem} \label{thm.2}
For any integers $\alpha$, $\beta$, $n$ and $a$ with 
$\alpha^2+\alpha\beta-\beta^2-a^2\neq 0$, we have
\begin{equation} \label{eqn.1}
\gcd(G_{2n-1}+a,G_{2n}+a)\leq |\alpha^2+\alpha\beta-\beta^2-a^2|.
\end{equation}
\end{theorem}

\begin{theorem} \label{thm.3}
For any integers $\alpha$, $\beta$, $n$ and $a$ with
$\alpha^2+\alpha\beta-\beta^2+a^2\neq 0$, we have
\begin{equation} \label{eqn.2}
\gcd(G_{2n}+a,G_{2n+1}+a)\leq |\alpha^2+\alpha\beta-\beta^2+a^2|.
\end{equation}
\end{theorem}

Let $\alpha=\beta=1$ in Theorem \ref{thm.2} 
and Theorem \ref{thm.3}.
We can get the corollary.

\begin{corollary} \label{cor.14}
For integers $n$ and $a$,
\begin{eqnarray*}
\gcd(F_{2n-1}+a,F_{2n}+a) &amp;\leq &amp; |a^2-1|,\quad\mbox{if }a\neq \pm 1,\\
\gcd(F_{2n}+a,F_{2n+1}+a) &amp;\leq &amp; a^2+1.
\end{eqnarray*}
\end{corollary}

Hence we conclude that $(f_n(a))$ is bounded above 
if $a\neq\pm 1$.
Another easy corollary is that
$$
\ell_n(a)=\gcd(L_n+a,L_{n+1}+a)
$$
has only finitely many values.

\begin{corollary} \label{cor.15}
For integers $n$ and $a$,
\begin{eqnarray*}
\gcd(L_{2n-1}+a,L_{2n}+a) &amp;\leq &amp; a^2+5,\\
\gcd(L_{2n}+a,L_{2n+1}+a) &amp;\leq &amp; |a^2-5|.
\end{eqnarray*}
\end{corollary}

Similarly, let $\alpha=1$ and $\beta=3$ in 
Theorem \ref{thm.2} and Theorem \ref{thm.3}.
We conclude that $\ell_n(a)$ is bounded above 
for any integers $a$.

In the next section we will derive two basic lemmas.
From them, we determine $f_n(1)$, $f_n(2)$, $f_n(-1)$,
$f_n(-2)$, and $\ell_n(1)$, in Section 3, 4, and 5.
In the last section we prove Theorems 
\ref{thm.2} and \ref{thm.3}.







\section{Preliminaries}

\begin{lemma} \label{lma.5}
For integers $n$, $k$, and $a$, 
\begin{equation}\label{eqn.3}
\gcd(G_n+aF_k,G_{n-1}-aF_{k+1})=\gcd(G_{n-2}+aF_{k+2},G_{n-3}-aF_{k+3}).
\end{equation}
\end{lemma}
\begin{proof}
Since $\gcd(a,b)=\gcd(a+bc,b)$ for any integers $a$, $b$, and $c$, we have
\begin{eqnarray*}
\gcd(G_n+aF_k,G_{n-1}-aF_{k+1}) &amp;=&amp; \gcd(G_n+aF_k-(G_{n-1}-aF_{k+1}),G_{n-1}-aF_{k+1}) \\
&amp;=&amp; \gcd(G_{n-2}+aF_{k+2},G_{n-1}-aF_{k+1})\\
&amp;=&amp; \gcd(G_{n-2}+aF_{k+2},G_{n-1}-aF_{k+1}-(G_{n-2}+aF_{k+2})) \\
&amp;=&amp; \gcd(G_{n-2}+aF_{k+2},G_{n-3}-aF_{k+3}).
\end{eqnarray*}
\end{proof}

\begin{lemma} \label{lma.6}
For integers $m$, $k$, and $a$,
\begin{equation} \label{eqn.4}
\gcd(G_m+a,G_{m+1}+a)=\gcd(G_{m-(2k)}+aF_{2k-1},G_{m-(2k+1)}-aF_{2k}).
\end{equation}
\end{lemma}
\begin{proof}
We simplify $\gcd(G_m+a,G_{m+1}+a)$,
\begin{eqnarray*}
\gcd(G_m+a,G_{m+1}+a) &amp;=&amp; \gcd(G_m+a,G_{m+1}+a-(G_m+a))\\
&amp;=&amp; \gcd(G_m+a,G_{m-1}). 
\end{eqnarray*} 
Because $F_{-1}=1$ and $F_0=0$ we can write 
$$
\gcd(G_m+a,G_{m+1}+a)=\gcd(G_m+aF_{-1},G_{m-1}+aF_0),
$$
and applying (\ref{eqn.3}) $k$ times gives the result.
\end{proof}




\section{The sequence $(f_n(1))$}
\begin{theorem} \label{thm.7}
For any integer $n$, we have
\begin{eqnarray}
\gcd(F_{4n-1}+1,F_{4n}+1) &amp;=&amp; F_{2n-1},\label{eqn.5}\\
\gcd(F_{4n}+1,F_{4n+1}+1) &amp;=&amp; \begin{cases}
2, &amp; \text{if }n\equiv 1\pmod 3,\\
1, &amp; \text{otherwise,}\end{cases}\label{eqn.6}\\
\gcd(F_{4n+1}+1,F_{4n+2}+1) &amp;=&amp; L_{2n},\label{eqn.7}\\
\gcd(F_{4n+2}+1,F_{4n+3}+1) &amp;=&amp; \begin{cases}
2, &amp; \text{if }n\equiv 2\pmod 3,\\
1, &amp; \text{otherwise.}\end{cases}\label{eqn.8}
\end{eqnarray}
\end{theorem}
\begin{proof}
Let $m=4n-1$, $k=n$, and $a=1$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(F_{4n-1}+1,F_{4n}+1) &amp;=&amp; 
\gcd(F_{2n-1}+F_{2n-1},F_{2n-2}-F_{2n})\\
&amp;=&amp; \gcd(2F_{2n-1},-F_{2n-1})\\
&amp;=&amp; F_{2n-1},
\end{eqnarray*}
giving (\ref{eqn.5}). 
Let $m=4n+1$, $k=n$, and $a=1$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(F_{4n+1}+1,F_{4n+2}+1) &amp;=&amp; 
\gcd(F_{2n+1}+F_{2n-1},F_{2n}-F_{2n})\\
&amp;=&amp; F_{2n+1}+F_{2n-1}\\
&amp;=&amp; L_{2n},
\end{eqnarray*}
giving (\ref{eqn.7}). 
Let $m=4n$, $k=n$, and $a=1$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(F_{4n}+1,F_{4n+1}+1) &amp;=&amp; 
\gcd(F_{2n}+F_{2n-1},F_{2n-1}-F_{2n})\\
&amp;=&amp; \gcd(F_{2n+1},-F_{2n-2}).
\end{eqnarray*}
Since $\gcd(F_{qn+r},F_n)=\gcd(F_n,F_r)$ for integers $q$, $r$, and $n$.
This gives
$$
\gcd(F_{4n}+1,F_{4n+1}+1) = \gcd(F_{2n-2},F_3).
$$
Because $\gcd(F_k,F_r)=F_{\gcd(k,r)}$ for integers $k$ and $r$,
\begin{eqnarray*}
\gcd(F_{4n}+1,F_{4n+1}+1) &amp;=&amp; \gcd(F_{2n-2},F_3)\\
&amp;=&amp; F_{\gcd(2n-2,3)}\\
&amp;=&amp;\begin{cases}
F_3=2, &amp;n\equiv 1\pmod 3,\\
F_1=1, &amp;\text{otherwise,}\end{cases}
\end{eqnarray*}
which is (\ref{eqn.6}). 
Let $m=4n+2$, $k=n+1$, and $a=1$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(F_{4n+2}+1,F_{4n+3}+1) &amp;=&amp; 
\gcd(F_{2n}+F_{2n+1},F_{2n-1}-F_{2n+2})\\
&amp;=&amp; \gcd(F_{2n+2},F_{2n-1}-F_{2n+2})\\
&amp;=&amp; \gcd(F_{2n+2},F_{2n-1})\\
&amp;=&amp; \gcd(F_3,F_{2n-1})\\
&amp;=&amp; F_{\gcd(3,2n-1)}\\
&amp;=&amp;\begin{cases}
F_3=2, &amp;n\equiv 2\pmod 3,\\
F_1=1, &amp;\text{otherwise,}\end{cases}
\end{eqnarray*}
which is (\ref{eqn.8}). 
\end{proof}




\section{The sequence $(f_n(2))$}
\begin{theorem} \label{thm.8}
For any integer $n$, we have
\begin{eqnarray}
\gcd(F_{4n-1}+2,F_{4n}+2) &amp;=&amp; 1,\label{eqn.9}\\
\gcd(F_{4n}+2,F_{4n+1}+2) &amp;=&amp; 1,\label{eqn.10}\\
\gcd(F_{4n+1}+2,F_{4n+2}+2) &amp;=&amp;
\begin{cases}
3, &amp; \text{if }n\equiv 0\pmod 2,\\
1, &amp; \text{if }n\equiv 1\pmod 2,\end{cases}\label{eqn.11}\\
\gcd(F_{4n+2}+2,F_{4n+3}+2) &amp;=&amp; 
\begin{cases}
5, &amp;\text{if }n\equiv 1\pmod 5,\\
1, &amp;\text{otherwise.}\end{cases}\label{eqn.12}
\end{eqnarray}
\end{theorem}
\begin{proof}
Let $m=4n-1$, $k=n$, and $a=2$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(F_{4n-1}+2,F_{4n}+2) &amp;=&amp; 
\gcd(F_{2n-1}+2F_{2n-1},F_{2n-2}-2F_{2n})\\
&amp;=&amp; \gcd(3F_{2n-1},F_{2n-1}+F_{2n})\\
&amp;=&amp; \gcd(3F_{2n-1},F_{2n+1}).
\end{eqnarray*}
Since $\gcd(a,bc)=\gcd(a,\gcd(a,b)c)$ and 
$\gcd(F_{2n-1},F_{2n+1})=\gcd(F_{2n-1},F_2)=1$,
we have
\begin{eqnarray*}
\gcd(F_{4n-1}+2,F_{4n}+2) &amp;=&amp;
\gcd(3\gcd(F_{2n-1},F_{2n+1}),F_{2n+1})\\
&amp;=&amp; \gcd(3,F_{2n+1})=\gcd(F_4,F_{2n+1})\\
&amp;=&amp; F_{\gcd(4,2n+1)}=F_1=1,
\end{eqnarray*}
which is (\ref{eqn.9}). 
Let $m=4n$, $k=n$, and $a=2$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(F_{4n}+2,F_{4n+1}+2) &amp;=&amp; 
\gcd(F_{2n}+2F_{2n-1},F_{2n-1}-2F_{2n})\\
&amp;=&amp; \gcd(F_{2n-1}+F_{2n+1},-F_{2n}-F_{2n-2})\\
&amp;=&amp; \gcd(L_{2n},L_{2n-1})\\
&amp;=&amp; 1,
\end{eqnarray*}
which is (\ref{eqn.10}). 
Let $m=4n+1$, $k=n$, and $a=2$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(F_{4n+1}+2,F_{4n+2}+2) &amp;=&amp; 
\gcd(F_{2n+1}+2F_{2n-1},F_{2n}-2F_{2n})\\
&amp;=&amp; \gcd(F_{2n+1}+2F_{2n-1}+2F_{2n},F_{2n})\\
&amp;=&amp; \gcd(3F_{2n+1},F_{2n})\\
&amp;=&amp; \gcd(3,F_{2n})= \gcd(F_4,F_{2n})\\
&amp;=&amp; F_{\gcd(4,2n)}\\
&amp;=&amp;\begin{cases}
F_4=3, &amp;\text{if }n\equiv 0\pmod 2,\\
F_1=1, &amp;\text{if }n\equiv 1\pmod 2,\end{cases}
\end{eqnarray*}
which is (\ref{eqn.11}).
Let $m=4n+2$, $k=n$, and $a=2$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(F_{4n+2}+2,F_{4n+3}+2) &amp;=&amp; 
\gcd(F_{2n+2}+2F_{2n-1},F_{2n+1}-2F_{2n})\\
&amp;=&amp; \gcd(F_{2n+2}+2F_{2n-1},-F_{2n}+F_{2n-1})\\
&amp;=&amp; \gcd(F_{2n+2}+2F_{2n-1},-F_{2n-2})\\
&amp;=&amp; \gcd(F_{2n-2},F_{2n+2}+2F_{2n}).
\end{eqnarray*}
Since $F_{2n+2}+2F_{2n}=F_{2n+1}+3F_{2n}=4F_{2n}+F_{2n-1}$,
we have
\begin{eqnarray*}
\gcd(F_{4n+2}+2,F_{4n+3}+2) &amp;=&amp; \gcd(F_{2n-2},4F_{2n}+F_{2n-1})\\
&amp;=&amp; \gcd(F_{2n-2},5F_{2n})\\
&amp;=&amp; \gcd(F_{2n-2},5\gcd(F_{2n-2},F_{2n}))\\
&amp;=&amp; \gcd(F_{2n-2},5) = \gcd(F_{2n-2},F_5)\\
&amp;=&amp; F_{\gcd(2n-2,5)}\\
&amp;=&amp;\begin{cases}
F_5=5, &amp;\text{if }n\equiv 1\pmod 5\\
F_1=1, &amp;\text{otherwise,}\end{cases}
\end{eqnarray*}
which is (\ref{eqn.12}). 
\end{proof}





\section{The sequences $(f_n(-1))$, $(f_n(-2))$, and $(\ell_n(1))$}
Applying the same methods we get
\begin{theorem} \label{thm.9}
For any integer $n$, we have
\begin{eqnarray*}
\gcd(F_{4n-1}-1,F_{4n}-1) &amp;=&amp; L_{2n-1},\\
\gcd(F_{4n}-1,F_{4n+1}-1) &amp;=&amp; 
\begin{cases}
2, &amp; \text{if }n\equiv 1\pmod 3,\\
1, &amp; \text{otherwise,}\end{cases}\\
\gcd(F_{4n+1}-1,F_{4n+2}-1) &amp;=&amp; F_{2n},\\
\gcd(F_{4n+2}-1,F_{4n+3}-1) &amp;=&amp; 
\begin{cases}
2, &amp; \text{if }n\equiv 2\pmod 3,\\
1, &amp; \text{otherwise.}\end{cases}
\end{eqnarray*}
\end{theorem}

\begin{theorem} \label{thm.10}
For any integer $n$, we have
\begin{eqnarray*}
\gcd(F_{4n-1}-2,F_{4n}-2) &amp;=&amp; 1,\\
\gcd(F_{4n}-2,F_{4n+1}-2) &amp;=&amp; 
\begin{cases}
5, &amp;\text{if }n\equiv 4\pmod 5,\\
1, &amp;\text{otherwise,.}\end{cases}\\
\gcd(F_{4n+1}-2,F_{4n+2}-2) &amp;=&amp;
\begin{cases}
1, &amp; \text{if }n\equiv 0\pmod 2,\\
3, &amp; \text{if }n\equiv 1\pmod 2,\end{cases}\\
\gcd(F_{4n+2}-2,F_{4n+3}-2) &amp;=&amp; 1.
\end{eqnarray*}
\end{theorem}

\begin{theorem} \label{thm.11}
For any integer $n$, we have
\begin{eqnarray*}
\gcd(L_{4n-1}+1,L_{4n}+1) &amp;=&amp; 
\begin{cases}
3, &amp; \text{if }n\equiv 0\pmod 6,\\
1, &amp; \text{if }n\equiv 1\pmod 6,\\
6, &amp; \text{if }n\equiv 2\pmod 6,\\
1, &amp; \text{if }n\equiv 3\pmod 6,\\
3, &amp; \text{if }n\equiv 4\pmod 6,\\
2, &amp; \text{if }n\equiv 5\pmod 6.
\end{cases}\\
\gcd(L_{4n}+1,L_{4n+1}+1) &amp;=&amp;
\begin{cases}
4, &amp; \text{if }n\equiv 1\pmod 3,\\
1, &amp; \text{otherwise,}\end{cases}\\
\gcd(L_{4n+1}+1,L_{4n+2}+1) &amp;=&amp;
\begin{cases}
2, &amp; \text{if }n\equiv 0\pmod 3,\\
1, &amp; \text{otherwise,}\end{cases}\\
\gcd(L_{4n+2}+1,L_{4n+3}+1) &amp;=&amp; 
\begin{cases}
4, &amp;\text{if }n\equiv 2\pmod 3,\\
1, &amp;\text{otherwise.}\end{cases}
\end{eqnarray*}
\end{theorem}






\section{The proofs of Theorems \ref{thm.2} and \ref{thm.3}}
First we give the proof of Theorem \ref{thm.2}.
Let $m=4n-1$ and $k=n$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(G_{4n-1}+a,G_{4n}+a) &amp;=&amp; \gcd(G_{2n-1}+aF_{2n-1},G_{2n-2}-aF_{2n})\\
&amp;=&amp; \gcd(\alpha F_{2n-3}+\beta F_{2n-2}+aF_{2n-1},
\alpha F_{2n-4}+\beta F_{2n-3}-aF_{2n}).
\end{eqnarray*}
Using the recursion relation for $F_n$, let
$$
a_n=\alpha F_{2n-3}+\beta F_{2n-2}+aF_{2n-1}=(\alpha+a)F_{2n-3}+(\beta+a)F_{2n-2}
$$
and
$$
b_n=\alpha F_{2n-4}+\beta F_{2n-3}-aF_{2n} = (-\alpha+\beta-a)F_{2n-3}+(\alpha-2a)F_{2n-2}.
$$
Since $\gcd(a_n,b_n)$ divides $\gamma a_n+\theta b_n$ for 
any integers $\gamma$ and $\theta$, and
\begin{eqnarray*}
(\alpha+a)b_n-(-\alpha+\beta-a)a_n &amp;=&amp; (\alpha^2+\alpha\beta-\beta^2-a^2)F_{2n-2}\\
(\alpha-2a)a_n-(\beta+a)b_n &amp;=&amp; (\alpha^2+\alpha\beta-\beta^2-a^2)F_{2n-3},
\end{eqnarray*}
we see that if $\alpha^2+\alpha\beta-\beta^2-a^2\neq 0$, then 
the greatest common divisor of the two numbers is 
$|\alpha^2+\alpha\beta-\beta^2-a^2|$.
Therefore $\gcd(a_n,b_n)$ divides $\alpha^2+\alpha\beta-\beta^2-a^2$.
That is to say, 
$$
\gcd(G_{4n-1}+a,G_{4n}+a)\leq |\alpha^2+\alpha\beta-\beta^2-a^2|.
$$
%===
If we let $m=4n+1$ and $k=n$ in (\ref{eqn.4}) we have,
in exactly the same way, that 
$$
\gcd(G_{4n+1}+a,G_{4n+2}+a)\leq |\alpha^2+\alpha\beta-\beta^2-a^2|.
$$
\qed

In the following we give the proof of Theorem \ref{thm.3}.
Let $m=4n$ and $k=n$ in (\ref{eqn.4}):
\begin{eqnarray*}
\gcd(G_{4n}+a,G_{4n+1}+a) &amp;=&amp; \gcd(G_{2n}+aF_{2n-1},G_{2n-1}-aF_{2n})\\
&amp;=&amp; \gcd(\alpha F_{2n-2}+\beta F_{2n-1}+aF_{2n-1},
\alpha F_{2n-3}+\beta F_{2n-2}-aF_{2n}).
\end{eqnarray*}
Using the recursion relation for $F_n$, let 
$$
a_n=\alpha F_{2n-2}+\beta F_{2n-1}+aF_{2n-1}=\alpha F_{2n-2}+(\beta+a)F_{2n-1}
$$
and
$$
b_n=\alpha F_{2n-3}+\beta F_{2n-2}-aF_{2n} = (-\alpha+\beta-a)F_{2n-2}+(\alpha-a)F_{2n-1}.
$$
Since $\gcd(a_n,b_n)$ divides $\gamma a_n+\theta b_n$ for 
any integers $\gamma$ and $\theta$, and
\begin{eqnarray*}
(\alpha-a)a_n-(a+\beta)b_n &amp;=&amp; (\alpha^2+\alpha\beta-\beta^2+a^2)F_{2n-2}\\
\alpha b_n-(\beta-\alpha-a)a_n &amp;=&amp; (\alpha^2+\alpha\beta-\beta^2+a^2)F_{2n-1},
\end{eqnarray*}
we see that if $\alpha^2+\alpha\beta-\beta^2+a^2\neq 0$, then 
the greatest common divisor of the two numbers is 
$|\alpha^2+\alpha\beta-\beta^2+a^2|$.
Therefore $\gcd(a_n,b_n)$ divides $\alpha^2+\alpha\beta-\beta^2+a^2$.
That is to say, 
$$
\gcd(G_{4n}+a,G_{4n+1}+a)\leq |\alpha^2+\alpha\beta-\beta^2+a^2|.
$$
%===
If we let $m=4n+2$ and $k=n$ in (\ref{eqn.4}) we have,
in exactly the same way, that 
$$
\gcd(G_{4n+2}+a,G_{4n+3}+a)\leq |\alpha^2+\alpha\beta-\beta^2+a^2|.
$$
\qed

\section{Acknowledgement}
The author would like to thank the referee for some useful comments and suggestions.

\begin{thebibliography}{99}

\bibitem[1]{DT}
U. Dudley and B. Tucker, Greatest common divisors in altered Fibonacci sequences, 
\textit{Fibonacci Quart.} \textbf{9} (1971), 89--91.

\bibitem[2]{HL}
S. Hern\'andez and F. Luca, Common factors of shifted Fibonacci numbers, 
\textit{Period. Math. Hungar.} \textbf{47} (2003), 95--110.

\bibitem[3]{K}
T. Koshy, \textit{Fibonacci and Lucas Numbers with Applications},
Wiley-Interscience, 2001.

\end{thebibliography}


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11A05.

\noindent \emph{Keywords: } Fibonacci numbers, Lucas numbers,
generalized Fibonacci numbers.

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\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000071},
\seqnum{A001611}, and
\seqnum{A157725}.)

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\vspace*{+.1in}
\noindent
Received January 31 2011;
revised version received March 26 2011. 
Published in {\it Journal of Integer Sequences}, March 26 2011.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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