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\begin{center}
\vskip 1cm{\LARGE\bf Integer Solutions of Some Diophantine \\
\vskip .1in
Equations via Fibonacci and Lucas Numbers
}
\vskip 1cm
\large
Bahar Demirt\"{u}rk and Refik Keskin \\
Department of Mathematics \\
Sakarya University \\
54187 Sakarya \\
Turkey \\
\href{mailto:demirturk@sakarya.edu.tr}{\tt demirturk@sakarya.edu.tr} \\
\href{mailto:rkeskin@sakarya.edu.tr}{\tt rkeskin@sakarya.edu.tr} \\
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\vskip .2 in

\begin{abstract}
We study the problem of finding all integer solutions of
the Diophantine equations $x^{2}-5F_{n}xy-5\left( -1\right) ^{n}y^{2}=\pm
L_{n}^{2},$ $x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=\pm 5F_{n}^{2},$ and $%
x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=\pm F_{n}^{2}.$  Using these
equations, we also explore all integer solutions of some other Diophantine
equations.
\end{abstract}

\newtheorem{theorem}{Theorem}
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\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
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\newtheorem{defin}[theorem]{Definition}
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\section{Introduction}

In this paper we study some Diophantine equations involving the well-known
Fibonacci and Lucas sequences, which are defined as follows; $%
F_{0}=0,F_{1}=F_{2}=1$ and $F_{n}=F_{n-1}+F_{n-2}$ for $n\geq 2$, $L_{0}=2$,
$L_{1}=1$ and $L_{n}=L_{n-1}+L_{n-2}$ for $n\geq 2$.

Both the Fibonacci and Lucas sequences may be extended backwards, i.e., $%
F_{-1}=F_{1}-F_{0}$, $F_{-2}=F_{0}-F_{-1}$,\ldots , $L_{-1}=L_{1}-L_{0}$, $%
L_{-2}=L_{0}-L_{-1}$ and so on. In general for $n&gt;0$, we set $F_{-n}=\left(
-1\right) ^{n+1}F_{n}$ and $L_{-n}=\left( -1\right) ^{n}L_{n}$ \cite{HGG, VRB}.

Fibonacci and Lucas numbers possess many interesting and important
properties. To begin with, we shall give some of them. Perhaps the most
important one is Binet's formula, which allows one to compute
$F_{n}$ directly
without computing the previous Fibonacci numbers. Binet's formula is
obtained by solving the following quadratic equation for $x$:
\begin{equation}
x^{2}-x-1=0.  \label{eq:1.3}
\end{equation}%
The two solutions of (\ref{eq:1.3}) are $\alpha =\left( 1+\sqrt{5}\right) /2$
and $\beta =\left( 1-\sqrt{5}\right) /2$. Clearly $\alpha +\beta =1$, $%
\alpha -\beta =\sqrt{5}$ and $\alpha \beta =-1$. So, for Fibonacci numbers
Binet's formula is given by $F_{n}=\left( \alpha ^{n}-\beta ^{n}\right)
/\left( \alpha -\beta \right) $ with $n\in \mathbb{Z}$. Also the
corresponding formula for Lucas numbers is given by $L_{n}=\alpha ^{n}+\beta
^{n}$ with $n\in \mathbb{Z}$. In addition to the above formulas, $\alpha
^{n}=\alpha F_{n}+F_{n-1}$ and $\beta ^{n}=\beta F_{n}+F_{n-1}$ for all $%
n\in \mathbb{Z}$.

Now we compile some identities and basic theorems involving Fibonacci and
Lucas numbers from various sources to use in the following theorems \cite%
{HGG, KSH, VJD}.

The first identity is
\begin{equation}
F_{n}^{2}-F_{n}F_{n-1}-F_{n-1}^{2}=\left( -1\right) ^{n+1}\text{ for all }%
n\in \mathbb{Z}  \label{eq:1.4}
\end{equation}%
and is known as Cassini's identity. A similar identity is given for Lucas
numbers as%
\begin{equation}
L_{n}^{2}-L_{n}L_{n-1}-L_{n-1}^{2}=\left( -1\right) ^{n}5\text{ for all }%
n\in \mathbb{Z}.  \label{eq:1.5}
\end{equation}%
The other identities can be listed as follows:%
\begin{equation}
L_{m}F_{n}-F_{m}L_{n}=2\left( -1\right) ^{m}F_{n-m}  \label{eq:1.6}
\end{equation}%
\begin{equation}
L_{m}L_{n}-5F_{m}F_{n}=2\left( -1\right) ^{m}L_{n-m}  \label{eq:1.7}
\end{equation}%
\begin{equation}
F_{m+1}L_{n}+L_{n-1}F_{m}=L_{n+m}  \label{eq:1.8}
\end{equation}%
\begin{equation}
L_{m}L_{n}+5F_{m}F_{n}=2L_{n+m}  \label{eq:1.9}
\end{equation}%
\begin{equation}
L_{m}F_{n}+F_{m}L_{n}=2F_{n+m}  \label{eq:1.10}
\end{equation}%
\begin{equation}
F_{n-1}+F_{n+1}=L_{n}  \label{eq:1.11}
\end{equation}%
\begin{equation}
L_{n-1}+L_{n+1}=5F_{n}  \label{eq:1.12}
\end{equation}%
\begin{equation}
L_{n}^{2}-5F_{n}^{2}=\left( -1\right) ^{n}4  \label{eq:1.13}
\end{equation}%
for all $m,$ $n\in \mathbb{Z}.$

We will give the following theorem without proof since it can be found in
\cite{HRD}.

\begin{theorem}
\label{t:1.1} The set of units of the ring $\mathbb{Z[\alpha ]=}\left\{
a\alpha +b:a,b\in \mathbb{Z}\right\} $ is
\begin{equation*}
\left\{ \pm \alpha ^{n}:n\in \mathbb{Z}\right\} .
\end{equation*}
\end{theorem}

The proof of the following theorem can be found in \cite{VJD}, but for the
sake of completeness we will give its proof.

\begin{theorem}
\label{t:1.2} All integer solutions of the equation $x^{2}-xy-y^{2}=\pm 1$
are given by $\left( x,y\right) =\pm \left( F_{n},F_{n-1}\right) $ with $%
n\in \mathbb{Z}$.
\end{theorem}

\begin{proof}
If $\left( x,y\right) =\pm \left( F_{n},F_{n-1}\right) $, it is clear from (%
\ref{eq:1.4}) that $x^{2}-xy-y^{2}=\pm 1$. Assume that $x^{2}-xy-y^{2}=\pm 1$
for some integer $x$ and $y$. Then it is seen that $\pm 1=\alpha \beta
x^{2}+\left( \alpha +\beta \right) xy+y^{2}=\left( \alpha x+y\right) \left(
\beta x+y\right) $. Hence, it follows that $\alpha x+y$ is a unit in $%
\mathbb{Z[\alpha ]}$. Thus from {Theorem }\ref{t:1.1}, we have $\alpha
x+y=\pm \alpha ^{n}$ for some $n\in \mathbb{Z}$. Since $\alpha ^{n}=\alpha
F_{n}+F_{n-1}$, we get $\alpha x+y=\pm \left( \alpha F_{n}+F_{n-1}\right) $.
Therefore it is seen that $\left( x,y\right) =\pm \left(
F_{n},F_{n-1}\right) $.%
\end{proof}

\begin{corollary}
\label{c:1} All integer solutions of the equations $x^{2}-xy-y^{2}=-1$ and $%
x^{2}-xy-y^{2}=1$ are given by $\left( x,y\right) =\pm \left(
F_{2n},F_{2n-1}\right) $ and $\left( x,y\right) =\pm \left(
F_{2n+1},F_{2n}\right) $ with $n\in \mathbb{Z}$, respectively.
\end{corollary}

Since the proof of the following theorem can be seen easily we omit its
proof.

\begin{theorem}
\label{t:1.3} All nonnegative integer solutions of the equation $%
x^{2}-xy-y^{2}=\pm 1$ are given by $\left( x,y\right) =\left(
F_{n},F_{n-1}\right) $ with $n\geq 0$.
\end{theorem}

\begin{corollary}
\label{c:2} All nonnegative integer solutions of the equations $%
x^{2}-xy-y^{2}=-1$ and $x^{2}-xy-y^{2}=1$ are given by $\left( x,y\right)
=\left( F_{2n},F_{2n-1}\right) $ and $\left( x,y\right) =\left(
F_{2n+1},F_{2n}\right) $ with $n\geq 0$, respectively.
\end{corollary}

\begin{theorem}
\label{t:1.4} All nonnegative integer solutions of the equation $%
u^{2}-5v^{2}=\pm 4$ are given by $\left( u,v\right) =\left(
L_{n},F_{n}\right) $ with $n\geq 0$.
\end{theorem}

\begin{proof}
It is clear that if $\left( u,v\right) =\left( L_{n},F_{n}\right) ,$ then by
(\ref{eq:1.13}) we get $u^{2}-5v^{2}=\pm 4.$ Assume that $u^{2}-5v^{2}=\pm 4$%
. Then $u$ and $v$ have the same parity. Let $x=\left( u+v\right) /2$ and $%
y=v$. Then it follows that
\begin{equation*}
x^{2}-xy-y^{2}=\left( \left( u+v\right) /2\right) ^{2}-\left( \left(
u+v\right) /2\right) v-v^{2}=\left( u^{2}-5v^{2}\right) /4=\pm 1.
\end{equation*}%
From {Theorem }\ref{t:1.3}, it is seen that $\left( x,y\right) =\left(
F_{n+1},F_{n}\right) $ for some $n\geq 0$. Thus, $\left( u+v\right)
/2=F_{n+1}$ and $y=F_{n}$. Therefore we get $\left( u,v\right) =\left(
L_{n},F_{n}\right) $.%
\end{proof}

\begin{corollary}
\label{c:3} All nonnegative integer solutions of the equations $%
u^{2}-5v^{2}=-4$ and $u^{2}-5v^{2}=4$ are given by $\left( u,v\right)
=\left( L_{2n+1},F_{2n+1}\right) $ and $\left( u,v\right) =\left(
L_{2n},F_{2n}\right) $ with $n\geq 0$, respectively.
\end{corollary}

\section{Identities And Solutions of Some Diophantine Equations}

There are various methods for deriving identities for Fibonacci and Lucas
numbers, such as the use of Binet's formula, induction, matrices, etc. The
usage of matrices enables us to obtain easily a large number of new
identities \cite{KSH}.

In this section we introduce three kinds of matrices including Fibonacci and
Lucas numbers. Also using the identities given in section 1, we derive some
new identities for Fibonacci and Lucas numbers. Let us give them in the
following theorems.

\begin{theorem}
\label{t:2.1} Let $k,m,n\in \mathbb{Z}$. Then
\begin{equation*}
L_{n+m}^{2}-5(-1)^{n+k+1}F_{k-n}L_{n+m}F_{m+k}-5\left( -1\right)
^{n+k}F_{m+k}^{2}=\left( -1\right) ^{m+k}L_{k-n}^{2}.
\end{equation*}
\end{theorem}

\begin{proof}
For the proof of the theorem, by using (\ref{eq:1.9}) and (\ref{eq:1.10}) we
can consider the matrix multiplication given below. That is,%
\begin{equation*}
\left[
\begin{array}{cc}
L_{n}/2 &amp; 5F_{n}/2 \\
F_{k}/2 &amp; L_{k}/2%
\end{array}%
\right] \left[
\begin{array}{c}
L_{m} \\
F_{m}%
\end{array}%
\right] =\left[
\begin{array}{c}
L_{n+m} \\
F_{m+k}%
\end{array}%
\right] .
\end{equation*}%
By (\ref{eq:1.7}),%
\begin{equation*}
\left\vert
\begin{array}{cc}
L_{n}/2 &amp; 5F_{n}/2 \\
F_{k}/2 &amp; L_{k}/2%
\end{array}%
\right\vert =\dfrac{L_{n}L_{k}-5F_{n}F_{k}}{4}=\dfrac{\left( -1\right)
^{n}L_{k-n}}{2}\neq 0,
\end{equation*}%
and therefore we can write,%
\begin{equation*}
\left[
\begin{array}{c}
L_{m} \\
F_{m}%
\end{array}%
\right] =\left[
\begin{array}{cc}
L_{n}/2 &amp; 5F_{n}/2 \\
F_{k}/2 &amp; L_{k}/2%
\end{array}%
\right] ^{-1}\left[
\begin{array}{c}
L_{n+m} \\
F_{m+k}%
\end{array}%
\right] .
\end{equation*}%
From here we get,%
\begin{eqnarray*}
L_{m} &amp;=&amp;\frac{\left( -1\right) ^{n}\left( L_{k}L_{n+m}-5F_{n}F_{m+k}\right)
}{L_{k-n}} \\
F_{m} &amp;=&amp;\frac{\left( -1\right) ^{n}\left( L_{n}F_{m+k}-F_{k}L_{n+m}\right)
}{L_{k-n}}.
\end{eqnarray*}%
Since $L_{m}^{2}-5F_{m}^{2}=\left( -1\right) ^{m}4$, we get
\begin{equation*}
\left( L_{k}L_{n+m}-5F_{n}F_{m+k}\right) ^{2}-5\left(
L_{n}F_{m+k}-F_{k}L_{n+m}\right) ^{2}=\left( -1\right) ^{m}4L_{k-n}^{2}.
\end{equation*}%
By using (\ref{eq:1.6}) and (\ref{eq:1.13}), we obtain%
\begin{equation}
L_{n+m}^{2}-5(-1)^{n+k+1}F_{k-n}L_{n+m}F_{m+k}-5\left( -1\right)
^{n+k}F_{m+k}^{2}=\left( -1\right) ^{m+k}L_{k-n}^{2}.  \label{eq:2.1}
\end{equation}%
\end{proof}

\begin{theorem}
\label{t:2.2} Let $k,m,n\in \mathbb{Z}$ and $k\neq n$. Then%
\begin{equation*}
L_{n+m}^{2}-\left( -1\right) ^{k+n}L_{k-n}L_{n+m}L_{m+k}+\left( -1\right)
^{n+k}L_{m+k}^{2}=\left( -1\right) ^{m+k+1}5F_{k-n}^{2}.
\end{equation*}
\end{theorem}

\begin{proof}
By using (\ref{eq:1.9}) we can consider the following matrix multiplication
for the proof of the theorem. That is,%
\begin{equation*}
\left[
\begin{array}{cc}
L_{n}/2 &amp; 5F_{n}/2 \\
L_{k}/2 &amp; 5F_{k}/2%
\end{array}%
\right] \left[
\begin{array}{c}
L_{m} \\
F_{m}%
\end{array}%
\right] =\left[
\begin{array}{c}
L_{n+m} \\
L_{m+k}%
\end{array}%
\right] .
\end{equation*}%
By (\ref{eq:1.6}),
\begin{equation*}
\left\vert
\begin{array}{cc}
L_{n}/2 &amp; 5F_{n}/2 \\
L_{k}/2 &amp; 5F_{k}/2%
\end{array}%
\right\vert =\dfrac{5\left( L_{n}F_{k}-L_{k}F_{n}\right) }{4}=\dfrac{5\left(
-1\right) ^{n}F_{k-n}}{2}
\end{equation*}%
and therefore for $k\neq n,$ we get%
\begin{equation*}
\left[
\begin{array}{c}
L_{m} \\
F_{m}%
\end{array}%
\right] =\left[
\begin{array}{cc}
L_{n}/2 &amp; 5F_{n}/2 \\
L_{k}/2 &amp; 5F_{k}/2%
\end{array}%
\right] ^{-1}\left[
\begin{array}{c}
L_{n+m} \\
L_{m+k}%
\end{array}%
\right] .
\end{equation*}%
Hence we have%
\begin{eqnarray*}
L_{m} &amp;=&amp;\frac{\left( -1\right) ^{n}\left( F_{k}L_{n+m}-F_{n}L_{m+k}\right)
}{F_{k-n}} \\
F_{m} &amp;=&amp;\frac{\left( -1\right) ^{n}\left( L_{n}L_{m+k}-L_{k}L_{n+m}\right)
}{5F_{k-n}}.
\end{eqnarray*}%
Since $L_{m}^{2}-5F_{m}^{2}=\left( -1\right) ^{m}4$, we get
\begin{equation*}
5\left( F_{k}L_{n+m}-F_{n}L_{m+k}\right) ^{2}-\left(
L_{n}L_{m+k}-L_{k}L_{n+m}\right) ^{2}=\left( -1\right) ^{m}20F_{k-n}^{2}.
\end{equation*}%
Using (\ref{eq:1.7}) and (\ref{eq:1.13}), we obtain
\begin{equation}
L_{n+m}^{2}-\left( -1\right) ^{k+n}L_{k-n}L_{n+m}L_{m+k}+\left( -1\right)
^{n+k}L_{m+k}^{2}=\left( -1\right) ^{m+k+1}5F_{k-n}^{2}.  \label{eq:2.2}
\end{equation}%
\end{proof}

\bigskip

Using
\begin{equation*}
\left[
\begin{array}{cc}
F_{n}/2 &amp; L_{n}/2 \\
F_{k}/2 &amp; L_{k}/2%
\end{array}%
\right] \left[
\begin{array}{c}
L_{m} \\
F_{m}%
\end{array}%
\right] =\left[
\begin{array}{c}
F_{n+m} \\
F_{m+k}%
\end{array}%
\right]
\end{equation*}%
and (\ref{eq:1.13}) we can give the following theorem.

\begin{theorem}
\label{t:2.3} Let $k,m,n\in \mathbb{Z}$ and $k\neq n$. Then%
\begin{equation}
F_{n+m}^{2}-L_{n-k}F_{n+m}F_{m+k}+\left( -1\right) ^{n+k}F_{m+k}^{2}=\left(
-1\right) ^{m+k}F_{n-k}^{2}.  \label{eq:2.3}
\end{equation}
\end{theorem}

\bigskip

The equations given in {Theorem }\ref{t:2.1}, {Theorem }\ref{t:2.2}, and {%
Theorem }\ref{t:2.3} induced us to explore the solutions of Diophantine
equations;%
\begin{equation*}
x^{2}-5F_{n}xy-5\left( -1\right) ^{n}y^{2}=\pm L_{n}^{2}
\end{equation*}%
\begin{equation*}
x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=\pm 5F_{n}^{2}
\end{equation*}%
\begin{equation*}
x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=\pm F_{n}^{2},
\end{equation*}%
where $n\geq 1$ is an integer. Our aim is to show that, the solutions of
these equations are pairs of Fibonacci or Lucas numbers. Let us give the
solutions of these equations in the following theorems. From now on we will
assume that $n$ is an integer greater than zero.

\begin{theorem}
\label{t:2.4} All integer solutions of the equations $x^{2}-5F_{n}xy-5\left(
-1\right) ^{n}y^{2}=-L_{n}^{2}$ and $x^{2}-5F_{n}xy-5\left( -1\right)
^{n}y^{2}=L_{n}^{2}$ are given by $\left( x,y\right) =\pm \left(
L_{n+m},F_{m}\right) $ for some odd integer $m$ and for some even integer $m$%
, respectively.
\end{theorem}

\begin{proof}
Assume that $x^{2}-5F_{n}xy-5\left( -1\right) ^{n}y^{2}=-L_{n}^{2}$. Then
\begin{equation*}
\left( 2x-5F_{n}y\right) ^{2}-\left( 25F_{n}^{2}+20\left( -1\right)
^{n}\right) y^{2}=-4L_{n}^{2}.
\end{equation*}
Using (\ref{eq:1.13}) we get $\left( 2x-5F_{n}y\right)
^{2}-5L_{n}^{2}y^{2}=-4L_{n}^{2}$. From here it follows that $%
L_{n}|2x-5F_{n}y$. Therefore taking
\begin{equation*}
u=\left( \left( \left( 2x-5F_{n}y\right) /L_{n}\right) +y\right) /2=\left(
x-L_{n-1}y\right) /L_{n}\text{ and }v=y
\end{equation*}%
we get,%
\begin{eqnarray*}
u^{2}-uv-v^{2} &amp;=&amp;\left( \left( x-L_{n-1}y\right) /L_{n}\right) ^{2}-\left(
\left( x-L_{n-1}y\right) /L_{n}\right) y-y^{2} \\
&amp;=&amp;\left( x^{2}-(L_{n-1}+L_{n+1})xy-y^{2}\left(
L_{n}^{2}-L_{n}L_{n-1}-L_{n-1}^{2}\right) \right) /L_{n}^{2}
\end{eqnarray*}%
and using the identities (\ref{eq:1.5}) and (\ref{eq:1.13}), we obtain%
\begin{equation*}
u^{2}-uv-v^{2}=\left( x^{2}-5F_{n}xy-5\left( -1\right) ^{n}y^{2}\right)
/L_{n}^{2}=-L_{n}^{2}/L_{n}^{2}=-1.
\end{equation*}%
Then from {Corollary }\ref{c:1}, it follows that $\left( u,v\right) =\pm
\left( F_{m+1},F_{m}\right) $ for some odd integer $m$. Hence we get%
\begin{equation*}
\left( x-L_{n-1}y\right) /L_{n}=\pm F_{m+1}\text{ and }y=\pm F_{m}.
\end{equation*}%
\bigskip Then we have $x=\pm \left( F_{m+1}L_{n}+L_{n-1}F_{m}\right) $ and $%
y=\pm F_{m}.$ Using (\ref{eq:1.8}) we get
\begin{equation*}
\left( x,y\right) =\pm \left( L_{n+m},F_{m}\right)
\end{equation*}%
for some odd integer $m.$

Now assume that $x^{2}-5F_{n}xy-5\left( -1\right) ^{n}y^{2}=L_{n}^{2}.$ Then
similarly, taking
\begin{equation*}
u=\left( x-L_{n-1}y\right) /L_{n}\text{ and }v=y,
\end{equation*}%
we get $u^{2}-uv-v^{2}=1$. From {Corollary }\ref{c:1}, it follows that $%
\left( u,v\right) =\pm \left( F_{m+1},F_{m}\right) $ for some even integer $%
m $. Hence we get $\left( x-L_{n-1}y\right) /L_{n}=\pm F_{m+1}$ and $y=\pm
F_{m}$. Using (\ref{eq:1.8}) we get $\left( x,y\right) =\pm \left(
L_{n+m},F_{m}\right) $ for some even integer $m.$

Conversely, if $\left( x,y\right) =\pm \left( L_{n+m},F_{m}\right) $ for
some odd integer $m,$ then by (\ref{eq:2.1}) it follows that $%
x^{2}-5F_{n}xy-5(-1)^{n}y^{2}=-L_{n}^{2}$ and if $\left( x,y\right) =\pm
\left( L_{n+m},F_{m}\right) $ for some even integer $m,$ then by (\ref%
{eq:2.1}) it follows that $x^{2}-5F_{n}xy-5(-1)^{n}y^{2}=L_{n}^{2}.$%
\end{proof}

\begin{theorem}
\label{t:2.5} All integer solutions of both equations $x^{2}-L_{n}xy+\left(
-1\right) ^{n}y^{2}=-5F_{n}^{2}$ and $x^{2}-L_{n}xy+\left( -1\right)
^{n}y^{2}=5F_{n}^{2}$ are given by $\left( x,y\right) =\pm \left(
L_{n+m},L_{m}\right) $ for some even integer $m$ and for some odd integer $m$%
, respectively.
\end{theorem}

\begin{proof}
Assume that $x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=-5F_{n}^{2}$. Then
multiplying both sides of the equation by 4 and using the identity (\ref%
{eq:1.13}) we get $\left( 2x-L_{n}y\right) ^{2}-5F_{n}^{2}y^{2}=-20F_{n}^{2}$%
. From here it follows that $5F_{n}|2x-L_{n}y$. Similarly it is seen that $%
5F_{n}|x+L_{n-1}y$. Then taking $u=\left( x+L_{n-1}y\right) /5F_{n}$ and $%
v=\left( 2x-L_{n}y\right) /5F_{n}$, we see that%
\begin{equation*}
u^{2}-uv-v^{2}=-5\left( x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}\right)
/25F_{n}^{2}=1.
\end{equation*}%
Then it follows from {Corollary }\ref{c:1} that $\left( u,v\right) =\pm
\left( F_{m+1},F_{m}\right) $ for some even integer $m.$ Thus, $\left(
x+L_{n-1}y\right) /5F_{n}=\pm F_{m+1}$ and $\left( 2x-L_{n}y\right)
/5F_{n}=\pm F_{m}$. Using (\ref{eq:1.8}), (\ref{eq:1.11}), and (\ref{eq:1.12}%
), we get $\left( x,y\right) =\pm \left( L_{n+m},L_{m}\right) $ for some
even integer $m$.

Assume that $x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=5F_{n}^{2}.$ Then, in
a similar way it is seen that $\left( x,y\right) =\pm \left(
L_{n+m},L_{m}\right) $ for some odd integer $m.$

Conversely, if $\left( x,y\right) =\pm \left( L_{n+m},L_{m}\right) $ for
some even integer $m,$ then by (\ref{eq:2.2}) it follows that $%
x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=-5F_{n}^{2}$ and if $\left(
x,y\right) =\pm \left( L_{n+m},L_{m}\right) $ for some odd integer $m,$ then
by (\ref{eq:2.2}) it follows that $x^{2}-L_{n}xy+\left( -1\right)
^{n}y^{2}=5F_{n}^{2}.$%
\end{proof}

\begin{theorem}
\label{t:2.6} All integer solutions of the equations $x^{2}-L_{n}xy+\left(
-1\right) ^{n}y^{2}=-F_{n}^{2}$ and $x^{2}-L_{n}xy+\left( -1\right)
^{n}y^{2}=F_{n}^{2} $ are given by $\left( x,y\right) =\pm \left(
F_{n+m},F_{m}\right) $ for some odd integer $\ m$ and for some even integer $%
m$, respectively.
\end{theorem}

\begin{proof}
Assume that $x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=-F_{n}^{2}$. Then we
get $\left( 2x-L_{n}y\right) ^{2}-5F_{n}^{2}y^{2}=-4F_{n}^{2}$. It is seen
that $F_{n}|2x-L_{n}y$. It follows that $\left( \left( 2x-L_{n}y\right)
/F_{n}\right) ^{2}-5y^{2}=-4.$ Thus taking $u=\left( \left( \left(
2x-L_{n}y\right) /F_{n}\right) +y\right) /2=\left( x-F_{n-1}\right) y/F_{n}$
and $\ v=y$ we get $u^{2}-uv-v^{2}=-1$. Therefore from {Corollary }\ref{c:1}
we have $\left( u,v\right) =\pm \left( F_{m+1},F_{m}\right) $ for some odd
integer $m$. Then it follows that $\left( x,y\right) =\pm \left(
F_{n+m},F_{m}\right) $ for some odd integer $m.$

Assume that $x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=F_{n}^{2}.$ Following
the same process with the solution of the above equation it can be seen that
$\left( x,y\right) =\pm \left( F_{n+m},F_{m}\right) $ for some even integer $%
m$.

Conversely, if $\left( x,y\right) =\pm \left( F_{n+m},F_{m}\right) $ for
some odd integer $m,$ then by (\ref{eq:2.3}) it follows that $%
x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=-F_{n}^{2}$ and if $\left(
x,y\right) =\pm \left( F_{n+m},F_{m}\right) $ for some even integer $m,$
then by (\ref{eq:2.3}) it follows that $x^{2}-L_{n}xy+\left( -1\right)
^{n}y^{2}=F_{n}^{2}.$%
\end{proof}

\bigskip Now we recall some divisibility properties of Fibonacci and Lucas
numbers in our interjection of solutions of Diophantine equations. These
divisibility properties are given in several sources such as \cite{CRL, KSH,
VJD}. Also we gave different proofs of the following theorems in \cite{KSK}.
So, now we give them without proof.

\begin{theorem}
\label{t:2.7} Let $m,n\in \mathbb{Z}$ and $n\geq 3$. Then $F_{n}|F_{m}$ if
and only if $n|m $.
\end{theorem}

\begin{theorem}
\label{t:2.8} Let $m,n\in \mathbb{Z}$ and $n\geq 2$. Then $L_{n}|F_{m}$ if
and only if $n|m $ and $m/n$ is an even integer.
\end{theorem}

\begin{theorem}
\label{t:2.9} Let $m,n\in \mathbb{Z}$ and $n\geq 2$. Then $L_{n}|L_{m}$ if
and only if $n|m $ and $m/n$ is an odd integer.
\end{theorem}

By these theorems the following identities can be obtained easily;%
\begin{equation}
2|L_{n}\text{ if and only if }3|n  \label{eq:2.4}
\end{equation}%
\begin{equation}
3|L_{n}\text{ if and only if }n=4k+2\text{ for some }k\in \mathbb{Z}\text{.}
\label{eq:2.5}
\end{equation}

\bigskip

Now we turn to our problem of finding the solutions of different Diophantine
equations benefiting from {Theorem }\ref{t:2.4}, {Theorem }\ref{t:2.5}, and {%
Theorem }\ref{t:2.6}.

\begin{theorem}
\label{t:2.10} If $n\geq 3$ is an odd integer, then all integer solutions of
the equations $x^{2}-L_{n}xy-y^{2}=-1$ and $x^{2}-L_{n}xy-y^{2}=1$ are given
by $\left( x,y\right) =\pm \left( F_{\left( 2k+2\right) n}/F_{n},F_{\left(
2k+1\right) n}/F_{n}\right) $ and $\left( x,y\right) =\pm \left( F_{\left(
2k+1\right) n}/F_{n},F_{2kn}/F_{n}\right) $ with $k\in \mathbb{Z},$
respectively. If $n$ is an even integer$,$ then all integer solutions of the
equation $x^{2}-L_{n}xy+y^{2}=1$ are given by $\left( x,y\right) =\pm \left(
F_{\left( k+1\right) n}/F_{n},F_{kn}/F_{n}\right) $ with $k\in \mathbb{Z}.$
\end{theorem}

\begin{proof}
Assume that $x^{2}-L_{n}xy-y^{2}=-1$. Multiplying both sides of the equation
by $F_{n}^{2}$, we get%
\begin{equation*}
\left( F_{n}x\right) ^{2}-L_{n}\left( F_{n}x\right) \left( F_{n}y\right)
-\left( F_{n}y\right) ^{2}=-F_{n}^{2}.
\end{equation*}%
From {Theorem }\ref{t:2.6}, we obtain $F_{n}x=\pm F_{n+m}$ and $F_{n}y=\pm
F_{m}$ for some odd integer $m.$ Then it follows that $x=\pm F_{n+m}/F_{n}$
and $y=\pm F_{m}/F_{n}$ for some odd integer $m.$ \ By {Theorem }\ref{t:2.7}%
, it is known that for $n\geq 3,$ $F_{n}|F_{m}$ if and only if $n|m.$ Since $%
n|m$ and $m$ is an odd integer we get $m=\left( 2k+1\right) n,$ for some $%
k\in \mathbb{Z}.$ Thus we obtain%
\begin{equation*}
\left( x,y\right) =\pm \left( F_{\left( 2k+2\right) n}/F_{n},F_{\left(
2k+1\right) n}/F_{n}\right) .
\end{equation*}

If $n$ is an odd integer, then similarly it can be seen that all integer
solutions of the equation $x^{2}-L_{n}xy-y^{2}=1$ are given by $\left(
x,y\right) =\pm \left( F_{\left( 2k+1\right) n}/F_{n},F_{2kn}/F_{n}\right) $
with $k\in \mathbb{Z}$.

Assume that $x^{2}-L_{n}xy+y^{2}=1$ and $n$ is an even integer$.$ From {%
Theorem }\ref{t:2.6}, we obtain $\left( x,y\right) =\pm \left(
F_{n+m}/F_{n},F_{m}/F_{n}\right) $ for some even integer $m.$ If $n=2,$ then
$F_{2}=1$ and by {Theorem }\ref{t:2.6} we get $\left( x,y\right) =\pm \left(
F_{m+2},F_{m}\right) $ for some even integer $m.$ If $n&gt;2$, then by {Theorem
}\ref{t:2.7}, it follows that $n|m.$ Since $n$ and $m$ are even, we get $%
m=kn $ for some $k\in \mathbb{Z}.$ Therefore we obtain
\begin{equation*}
\left( x,y\right) =\pm \left( F_{\left( k+1\right)
n}/F_{n},F_{kn}/F_{n}\right) .
\end{equation*}

Conversely, if $n\geq 3$ is an odd integer and $\left( x,y\right) =\pm
\left( F_{\left( 2k+2\right) n}/F_{n},F_{\left( 2k+1\right) n}/F_{n}\right) $
for some $k\in \mathbb{Z},$ then by (\ref{eq:2.3}) it follows that $%
x^{2}-L_{n}xy-y^{2}=-1$ and if $n$ is an odd integer and $\left( x,y\right)
=\pm \left( F_{\left( 2k+1\right) n}/F_{n},F_{2kn}/F_{n}\right) $ for some $%
k\in \mathbb{Z}$, then by (\ref{eq:2.3}) it follows that $%
x^{2}-L_{n}xy-y^{2}=1.$ Furthermore, if $n&gt;2$ is an even integer and $\left(
x,y\right) =\pm \left( F_{\left( k+1\right) n}/F_{n},F_{kn}/F_{n}\right) $
for some $k\in \mathbb{Z},$ then by (\ref{eq:2.3}) it follows that $%
x^{2}-L_{n}xy+y^{2}=1.$%
\end{proof}

\begin{corollary}
\label{c:4} If $n$ is an even integer greater than $2$, then the equation $%
x^{2}-L_{n}xy+y^{2}=-1$ has no integer solutions.
\end{corollary}

\begin{proof}
Let $x^{2}-L_{n}xy+y^{2}=-1.$ Then we get $\left( x,y\right) =\pm \left(
F_{n+m}/F_{n},F_{m}/F_{n}\right) $ for some odd integer $m.$ Using {Theorem }%
\ref{t:2.7}, it can be seen that $(x,y)$ is a pair of integer if and only if
$n|m$. But the fact that $n$ is even and $m$ is odd gives a contradiction.
Therefore $x^{2}-L_{n}xy+y^{2}=-1$ has no integer solutions.%
\end{proof}

\begin{theorem}
\label{t:2.11} All integer solutions of the equation $x^{2}-5F_{n}xy-5\left(
-1\right) ^{n}y^{2}=1$ are given by $\left( x,y\right) =\pm \left( L_{\left(
2k+1\right) n}/L_{n},F_{2kn}/L_{n}\right) $ with $k\in \mathbb{Z}$.
\end{theorem}

\begin{proof}
Assume that $x^{2}-5F_{n}xy-5\left( -1\right) ^{n}y^{2}=1.$ Multiplying both
sides of the equation by $L_{n}^{2},$ we get%
\begin{equation*}
\left( L_{n}x\right) ^{2}-5F_{n}\left( L_{n}x\right) \left( L_{n}y\right)
-5\left( -1\right) ^{n}\left( L_{n}y\right) ^{2}=L_{n}^{2}.
\end{equation*}%
From {Theorem }\ref{t:2.4}, it is seen that $L_{n}x$ $=\pm L_{n+m}$ and $%
L_{n}y=\pm F_{m}$ for some even integer $m.$ That is, $\left( x,y\right)
=\pm \left( L_{n+m}/L_{n},F_{m}/L_{n}\right) .$ By {Theorem }\ref{t:2.8} and
{Theorem }\ref{t:2.9}, it follows that $m/n$ is an even integer. So that, $%
m=2kn$, for some $k\in \mathbb{Z}.$ Therefore we obtain $\left( x,y\right)
=\pm \left( L_{\left( 2k+1\right) n}/L_{n},F_{2kn}/L_{n}\right) .$

Conversely, if $\left( x,y\right) =\pm \left( L_{\left( 2k+1\right)
n}/L_{n},F_{2kn}/L_{n}\right) $ for some $k\in \mathbb{Z},$ then by (\ref%
{eq:2.1}) it follows that $x^{2}-5F_{n}xy-5\left( -1\right) ^{n}y^{2}=1.$%
\end{proof}

\begin{corollary}
\label{c:5} If $n&gt;1$, then the equation $x^{2}-5F_{n}xy-5\left( -1\right)
^{n}y^{2}=-1$ has no integer solutions.
\end{corollary}

\begin{theorem}
\label{t:2.12}All integer solutions of the equation $x^{2}-L_{n}xy+\left(
-1\right) ^{n}y^{2}=-5$ and $x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=5$ are
given by $\left( x,y\right) =\pm \left( L_{n+m}/F_{n},L_{m}/F_{n}\right) $
with even integer $m$ and $\left( x,y\right) =\pm \left(
L_{n+m}/F_{n},L_{m}/F_{n}\right) $ with odd integer $m,$ respectively, where
$F_{n}\mid L_{m}$. In particular, $F_{n}\mid L_{m}$ if and only if $n=1$, $%
F_{n}=1$, $m=k$; $n=2$, $F_{n}=1$, $m=k$; $n=3$, $F_{n}=2$, $m=3k$; $n=4$, $%
F_{n}=3$, $m=4k+2$, where $k$ is an integer.
\end{theorem}

\begin{proof}
The results concerning when $F_{n}\mid L_{m}$ are given in Theorem 1 of \cite%
{HLT}. Assume that $x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=-5$ for some
integer $x$ and $y.$ Multiplying this equation by $F_{n}^{2},$ we get
\begin{equation*}
\left( F_{n}x\right) ^{2}-L_{n}\left( F_{n}x\right) \left( F_{n}y\right)
+\left( -1\right) ^{n}\left( F_{n}y\right) ^{2}=-5F_{n}^{2}.
\end{equation*}%
From {Theorem }\ref{t:2.5}, it follows that $\left( x,y\right) =\pm \left(
L_{n+m}/F_{n},L_{m}/F_{n}\right) $ for some even integer $m$. Similarly it
can be seen that all integer solutions of the equation $x^{2}-L_{n}xy+\left(
-1\right) ^{n}y^{2}=5$ are given by $\left( x,y\right) =\pm \left(
L_{n+m}/F_{n},L_{m}/F_{n}\right) $ for some odd integer $m.$

Conversely, if $\left( x,y\right) =\pm \left(
L_{n+m}/F_{n},L_{m}/F_{n}\right) $ for some even integer $m,$ then by (\ref%
{eq:2.2}) it follows that $x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=-5$ and
if $\left( x,y\right) =\pm \left( L_{n+m}/F_{n},L_{m}/F_{n}\right) $ for
some odd integer $m,$ then by (\ref{eq:2.2}) it follows that $%
x^{2}-L_{n}xy+\left( -1\right) ^{n}y^{2}=5.$%
\end{proof}

\begin{theorem}
\label{t:2.13} If $n$ is an even integer, then all integer solutions of the
equation $x^{2}-L_{2n}xy+y^{2}=-5F_{n}^{2}$ are given by $\left( x,y\right)
=\pm \left( L_{\left( 2k+3\right) n}/L_{n},L_{(2k+1)n}/L_{n}\right) $ with $%
k\in \mathbb{Z}$, if $n$ is an odd integer, then all integer solutions of
the equation $x^{2}-L_{2n}xy+y^{2}=5F_{n}^{2}$ are given by $\left(
x,y\right) =\pm \left( L_{\left( 2k+3\right)
n}/L_{n},L_{(2k+1)n}/L_{n}\right) $ with $k\in \mathbb{Z}$.
\end{theorem}

\begin{proof}
Assume that $n$ is an even integer and $x^{2}-L_{2n}xy+y^{2}=-5F_{n}^{2}$
for some integer $x$ and $\ y.$ Then multiplying this equation by $L_{n}^{2}$
and using {Theorem }\ref{t:2.5} and noting that $F_{2n}=F_{n}L_{n}$, we get $%
L_{n}x=\pm L_{2n+m}$ and $L_{n}y=\pm L_{m},$ for some even integer $m$.
Therefore it follows that $\left( x,y\right) =\pm \left(
L_{2n+m}/L_{n},L_{m}/L_{n}\right) $ for some even integer $m.$ Furthermore
by {Theorem }\ref{t:2.9}, it is seen that $L_{n}|L_{m}$ if and only if $m/n$
is an odd integer. Therefore $m=\left( 2k+1\right) n$, for some $k\in
\mathbb{Z}$. Thus $\left( x,y\right) =\pm \left( L_{\left( 2k+3\right)
n}/L_{n},L_{\left( 2k+1\right) n}/L_{n}\right) .$

Assume that $n$ is an odd integer and $x^{2}-L_{2n}xy+y^{2}=5F_{n}^{2}$ for
some integer $x$ and $y.$ Then by {Theorem }\ref{t:2.5}, we get $\left(
x,y\right) =\pm \left( L_{2n+m}/L_{n},L_{m}/L_{n}\right) $ for some odd
integer $m.$ Also by {Theorem }\ref{t:2.9} it follows that $m=\left(
2k+1\right) n$ for some $k\in \mathbb{Z}$. So that, $\left( x,y\right) =\pm
\left( L_{\left( 2k+3\right) n}/L_{n},L_{(2k+1)n}/L_{n}\right) $ with $k\in
\mathbb{Z}$. If $n=1,$ then all integer solutions of the equation $%
x^{2}-3xy+y^{2}=5$ are given by $\left( x,y\right) =\pm \left(
L_{m+2},L_{m}\right) $ with odd integer $m.$

Conversely, if $n$ is an even integer and $\left( x,y\right) =\pm \left(
L_{\left( 2k+3\right) n}/L_{n},L_{(2k+1)n}/L_{n}\right) $ for some $k\in
\mathbb{Z}$, then by (\ref{eq:2.2}) it follows that $%
x^{2}-L_{2n}xy+y^{2}=-5F_{n}^{2}$ and if $n&gt;1$ is an odd integer and $\left(
x,y\right) =\pm \left( L_{\left( 2k+3\right)
n}/L_{n},L_{(2k+1)n}/L_{n}\right) $ for some $k\in \mathbb{Z},$ then by (\ref%
{eq:2.2}) it follows that $x^{2}-L_{2n}xy+y^{2}=5F_{n}^{2}.$%
\end{proof}

\begin{corollary}
\label{c:12} If $n&gt;1$ is an odd integer, then the equation $%
x^{2}-L_{2n}xy+y^{2}=-5F_{n}^{2}$ has no integer solutions and if $n$ is an
even integer, then the equation $x^{2}-L_{2n}xy+y^{2}=5F_{n}^{2}$ has no
integer solutions.
\end{corollary}

\begin{theorem}
\label{t:2.14} All integer solutions of the equation $%
x^{2}-L_{2n}xy+y^{2}=F_{n}^{2}$ are given by $\left( x,y\right) =\pm \left(
F_{\left( 2k+2\right) n}/L_{n},F_{2kn}/L_{n}\right) $ with $k\in \mathbb{Z}$.
\end{theorem}

\begin{proof}
Assume that $n\geq 2$ and $x^{2}-L_{2n}xy+y^{2}=F_{n}^{2}$ for some integer $%
x$ and $y$. Multiplying this equation by $L_{n}^{2}$ we get%
\begin{equation*}
\left( L_{n}x\right) ^{2}-L_{2n}\left( L_{n}x\right) \left( L_{n}y\right)
+\left( L_{n}y\right) ^{2}=F_{2n}^{2}.
\end{equation*}%
Then by {Theorem }\ref{t:2.6} it follows that $\left( x,y\right) =\pm \left(
F_{2n+m}/L_{n},F_{m}/L_{n}\right) $ for some even integer $m.$ Hence using {%
Theorem }\ref{t:2.8} it is seen that $L_{n}|F_{m}$ if and only if $m/n$ is
an even integer. Then we have $m=2kn$ for some $k\in \mathbb{Z}.$ Therefore $%
\left( x,y\right) =\pm \left( F_{\left( 2k+2\right)
n}/L_{n},F_{2kn}/L_{n}\right) .$ If $n=1,$ then it can be seen that all
integer solutions of the equation $x^{2}-3xy+y^{2}=1$ are given by $\left(
x,y\right) =\pm \left( F_{m+2},F_{m}\right) $ with even integer $m.$

Conversely, if $\left( x,y\right) =\pm \left( F_{\left( 2k+2\right)
n}/L_{n},F_{2kn}/L_{n}\right) $ for some $k\in \mathbb{Z}$, then by (\ref%
{eq:2.3}) it follows that $x^{2}-L_{2n}xy+y^{2}=F_{n}^{2}.$%
\end{proof}

\begin{theorem}
\label{t:2.15} For all $n\geq 2$, the equation $%
x^{2}-L_{2n}xy+y^{2}=-F_{n}^{2}$ has no integer solutions.
\end{theorem}

\begin{proof}
Assume that $x^{2}-L_{2n}xy+y^{2}=-F_{n}^{2}$ for some integer $x$ and $y$.
Then by {Theorem }\ref{t:2.6}, we get $\left( x,y\right) =\pm \left(
F_{2n+m}/L_{n},F_{m}/L_{n}\right) $ for some odd integer $m.$ Using {Theorem
}\ref{t:2.8} it is seen that $L_{n}|F_{m}$ if and only if $m=2kn$ for some $%
k\in \mathbb{Z}.$ Since $m$ is odd, this is impossible. Therefore $%
x^{2}-L_{2n}xy+y^{2}=-F_{n}^{2}$ has no integer solutions.%
\end{proof}

\begin{theorem}
\label{t:2.16} Let $n\geq 3$ be an odd integer. Then all integer solutions
of the equation $x^{2}-L_{2n}xy+y^{2}=-L_{n}^{2}$ are given by $\left(
x,y\right) =\pm \left( F_{\left( 2k+3\right) n}/F_{n},F_{\left( 2k+1\right)
n}/F_{n}\right) $ with $k\in \mathbb{Z}$.
\end{theorem}

\begin{proof}
Assume that $x^{2}-L_{2n}xy+y^{2}=-L_{n}^{2}$ for some integer $x$ and $y.$
Then by {Theorem }\ref{t:2.6}, it is seen that $\left( x,y\right) =\pm
\left( F_{2n+m}/F_{n},F_{m}/F_{n}\right) $ for some odd integer $m.$
Furthermore by {Theorem }\ref{t:2.7} it follows that $F_{n}|F_{m}$ if and
only if $n|m.$ Since both $m$ and $n$ are odd integers it is seen that $%
m=\left( 2k+1\right) n$ for some $k\in \mathbb{Z}.$ Then this shows that $%
\left( x,y\right) =\pm \left( F_{\left( 2k+3\right) n}/F_{n},F_{\left(
2k+1\right) n}/F_{n}\right) $.

Conversely, if $n\geq 3$ is an odd integer and $\left( x,y\right) =\pm
\left( F_{\left( 2k+3\right) n}/F_{n},F_{\left( 2k+1\right) n}/F_{n}\right) $
for some $k\in \mathbb{Z}$, then by (\ref{eq:2.3}) it follows that $%
x^{2}-L_{2n}xy+y^{2}=-L_{n}^{2}.$%
\end{proof}

\bigskip

We can give the following corollary easily.

\begin{corollary}
\label{c:13} If $n$ is an even integer, then the equation $%
x^{2}-L_{2n}xy+y^{2}=-L_{n}^{2} $ has no integer solutions.
\end{corollary}

\begin{corollary}
\label{c:14} Let $n$ be an even integer. Then all integer solutions of the
equation $x^{2}-L_{2n}xy+y^{2}=L_{n}^{2}$ are given by $\left( x,y\right)
=\pm \left( F_{\left( k+2\right) n}/F_{n},F_{kn}/F_{n}\right) $ with $k\in
\mathbb{Z}$. Let $\ n$ be an odd integer. Then all integer solutions of the
equation $x^{2}-L_{2n}xy+y^{2}=L_{n}^{2}$ are given by $\left( x,y\right)
=\pm \left( F_{\left( 2k+2\right) n}/F_{n},F_{2kn}/F_{n}\right) $ with $k\in
\mathbb{Z}$.
\end{corollary}

\begin{proof}
Assume that $n$ is an even integer and $x^{2}-L_{2n}xy+y^{2}=L_{n}^{2}$ for
some integer $x$ and $y$. Then by {Theorem }\ref{t:2.6}, it is seen that $%
\left( x,y\right) =\pm \left( F_{2n+m}/F_{n},F_{m}/F_{n}\right) $ for some
even integer $m.$ Also by {Theorem }\ref{t:2.7} it follows that $n|m.$ Since
both $m$ and $n$ are even integers we have $m=kn$ for some $k\in \mathbb{Z}$%
. Then it is seen that $\left( x,y\right) =\pm \left( F_{\left( k+2\right)
n}/F_{n},F_{kn}/F_{n}\right) .$

Now assume that $n$ is an odd integer and $x^{2}-L_{2n}xy+y^{2}=L_{n}^{2}$
for some integer $x$ and $y.$ Then $\left( x,y\right) =\pm \left(
F_{2n+m}/F_{n},F_{m}/F_{n}\right) $ for some even integer $m$ and by {%
Theorem }\ref{t:2.7}, $n|m.$ Thus we get $m=2kn$ for some $k\in \mathbb{Z}.$
Hence it follows that $\left( x,y\right) =\pm \left( F_{\left( 2k+2\right)
n}/F_{n},F_{2kn}/F_{n}\right) .$%
\end{proof}

\begin{theorem}
\label{t:2.17} All integer solutions of the equation $%
x^{2}-L_{2n}xy+y^{2}=-5L_{n}^{2}$ and $x^{2}-L_{2n}xy+y^{2}=5L_{n}^{2}$ are
given by $\left( x,y\right) =\pm \left( L_{2n+m}/F_{n},L_{m}/F_{n}\right) $
with even integer $m$ and $\left( x,y\right) =\pm \left(
L_{2n+m}/F_{n},L_{m}/F_{n}\right) $ with odd integer $m$, respectively,
where $F_{n}\mid L_{m}$. In particular, $F_{n}\mid L_{m}$ if and only if $%
n=1 $, $F_{n}=1$, $m=k$; $n=2$, $F_{n}=1$, $m=k$; $n=3$, $F_{n}=2$, $m=3k$; $%
n=4$, $F_{n}=3$, $m=4k+2$, where $k$ is an integer.
\end{theorem}

\begin{proof}
The results concerning when $F_{n}\mid L_{m}$ are given in Theorem 1 of \cite%
{HLT}. Assume that $x^{2}-L_{2n}xy+y^{2}=-5L_{n}^{2}.$ Then by {Theorem }\ref%
{t:2.5}, it is seen that $\left( x,y\right) =\pm \left(
L_{2n+m}/F_{n},L_{m}/F_{n}\right) $ for some even integer $m$.

Assume that $x^{2}-L_{2n}xy+y^{2}=5L_{n}^{2}.$ It follows that $\left(
x,y\right) =\pm \left( L_{2n+m}/F_{n},L_{m}/F_{n}\right) $ for some odd
integer $m$, by {Theorem }\ref{t:2.5}.

Conversely, if $\left( x,y\right) =\pm \left(
L_{2n+m}/F_{n},L_{m}/F_{n}\right) $ for some even integer $m$, then by (\ref%
{eq:2.2}) it follows that $x^{2}-L_{2n}xy+y^{2}=-5L_{n}^{2}$ and if $\left(
x,y\right) =\pm \left( L_{2n+m}/F_{n},L_{m}/F_{n}\right) $ for some odd
integer $m,$ then by (\ref{eq:2.2}) it follows that $%
x^{2}-L_{2n}xy+y^{2}=5L_{n}^{2}.$%
\end{proof}

\bigskip

Furthermore we explore some Diophantine equations different from the
previous ones. Here are some of them.

\begin{theorem}
\label{t:2.18} Let $k\geq 0.$ Then all nonnegative integer solutions of the
equation $u^{2}-5v^{2}=4^{k}$ are given by $\left( u,v\right) =\left(
2^{k-1}L_{2m},2^{k-1}F_{2m}\right) $ and all nonnegative integer solutions
of the equation $u^{2}-5v^{2}=-4^{k}$ are given by $\left( u,v\right)
=\left( 2^{k-1}L_{2m+1},2^{k-1}F_{2m+1}\right) $ with $m\geq 0.$
\end{theorem}

\begin{proof}
Since the proof is obvious from mathematical induction, we omit it.%
\end{proof}

\begin{theorem}
\label{t:2.19} All nonnegative integer solutions of the equation $%
u^{2}-5v^{2}=1$ are given by $\left( u,v\right) =\left(
L_{6m}/2,F_{6m}/2\right) $ with $m\geq 0$ and all nonnegative integer
solutions of the equation $u^{2}-5v^{2}=-1$ are given by $\left( u,v\right)
=\left( L_{6m+3}/2,F_{6m+3}/2\right) $ with $m\geq 0.$
\end{theorem}

\begin{proof}
Assume that $u^{2}-5v^{2}=1.$ Then $\left( 2^{k}u\right) ^{2}-5\left(
2^{k}v\right) ^{2}=4^{k}.$ From {Theorem }\ref{t:2.18}, it is seen that $%
2^{k}u=2^{k-1}L_{2n}$ and $2^{k}v=2^{k-1}F_{2n}.$ Therefore we get $%
u=L_{2n}/2$ and $v=F_{2n}/2.$ Using {Theorem }\ref{t:2.7}, it is seen that $%
2|F_{2n}$ if and only if $3|2n$. So that, $3|n$ and that is $n=3m$ for some $%
m\in \mathbb{N}$. Hence we get $\left( u,v\right) =\left(
L_{6m}/2,F_{6m}/2\right) .$

Now assume that $u^{2}-5v^{2}=-1.$ Then we get $\left( 2^{k}u\right)
^{2}-5\left( 2^{k}v\right) ^{2}=-4^{k}.$ From {Theorem }\ref{t:2.18}, it
follows that $2^{k}u=2^{k-1}L_{2n+1}$ and $2^{k}v=2^{k-1}F_{2n+1}.$
Therefore we get $u=L_{2n+1}/2$ and $v=F_{2n+1}/2.$ By {Theorem }\ref{t:2.7}%
, it follows that $2|F_{2n+1}$ if and only if $3|2n+1$. That is $n=3m+1$ for
some $m\in \mathbb{N}$. Hence we get $\left( u,v\right) =\left(
L_{6m+3}/2,F_{6m+3}/2\right) .$%
\end{proof}

\begin{theorem}
\label{t:2.20} Let $k\geq 0.$ Then all nonnegative integer solutions of the
equation $x^{2}-xy-y^{2}=-4^{k}$ are given by $\left( x,y\right) =\left(
2^{k}F_{2m+2},2^{k}F_{2m+1}\right) $ and all integer solutions of the
equation $x^{2}-xy-y^{2}=4^{k}$ are given by $\left( x,y\right) =\left(
2^{k}F_{2m+1},2^{k}F_{2m}\right) $ with $m\geq 0$.
\end{theorem}

\begin{proof}
Proof follows from mathematical induction.\bigskip
\end{proof}

\begin{theorem}
\label{t:2.21} Let $k\geq 0.$ Then all nonnegative integer solutions of the
equation $u^{2}-5v^{2}=4\cdot 5^{k}$ are given by
\begin{equation*}
\left( u,v\right) =\left\{
\begin{array}{c}
\left( 5^{\left( k+1\right) /2}F_{2m+1},5^{\left( k-1\right)
/2}L_{2m+1}\right) ,\text{ }k\text{ is an odd integer;} \\
\left( 5^{k/2}L_{2m},5^{k/2}F_{2m}\right) ,\text{ }k\text{ is an even integer%
}%
\end{array}%
\right.
\end{equation*}%
and all nonnegative integer solutions of the equation $u^{2}-5v^{2}=-4\cdot
5^{k}$ are given by
\begin{equation*}
\left( u,v\right) =\left\{
\begin{array}{c}
\left( 5^{\left( k+1\right) /2}F_{2m},5^{\left( k-1\right) /2}L_{2m}\right) ,%
\text{ }k\text{ is an odd integer;} \\
\left( 5^{k/2}L_{2m+1},5^{k/2}F_{2m+1}\right) ,\text{ }k\text{ is an even
integer}%
\end{array}%
\right.
\end{equation*}%
where $m\geq 0$.
\end{theorem}

\begin{proof}
Proof is obvious from mathematical induction.\bigskip \bigskip
\end{proof}

\begin{theorem}
\label{t:2.22} Let $k\geq 0.$ Then all nonnegative integer solutions of the
equation $x^{2}-xy-y^{2}=5^{k}$ are given by
\begin{equation*}
\left( x,y\right) =\left\{
\begin{array}{c}
\left( 5^{\left( k-1\right) /2}L_{2m+2},5^{\left( k-1\right)
/2}L_{2m+1}\right) ,\text{ }k\text{ is an odd integer;} \\
\left( 5^{k/2}F_{2m+1},5^{k/2}F_{2m}\right) ,\text{ }k\text{ is an even
integer}%
\end{array}%
\right.
\end{equation*}%
and all nonnegative integer solutions of the equation $x^{2}-xy-y^{2}=-5^{k}$
are given by
\begin{equation*}
\left( x,y\right) =\left\{
\begin{array}{c}
\left( 5^{\left( k-1\right) /2}L_{2m+1},5^{\left( k-1\right)
/2}L_{2m}\right) ,\text{ }k\text{ is an odd integer;} \\
\left( 5^{k/2}F_{2m+2},5^{k/2}F_{2m+1}\right) ,\text{ }k\text{ is an even
integer}%
\end{array}%
\right.
\end{equation*}%
where $m\geq 0.$
\end{theorem}

\begin{proof}
Assume that $x^{2}-xy-y^{2}=5^{k}.$ Then $x&gt;y$ and we get $\left(
2x-y\right) ^{2}-5y^{2}=4\cdot 5^{k}.$ By {Theorem }\ref{t:2.21}, we obtain
\begin{equation*}
\left( 2x-y,y\right) =\left\{
\begin{array}{c}
\left( 5^{\left( k+1\right) /2}F_{2m+1},5^{\left( k-1\right)
/2}L_{2m+1}\right) ,\text{ }k\text{ is an odd integer;} \\
\left( 5^{k/2}L_{2m},5^{k/2}F_{2m}\right) ,\text{ }k\text{ is an even
integer.}%
\end{array}%
\right.
\end{equation*}%
Thus it follows that
\begin{equation*}
\left( x,y\right) =\left\{
\begin{array}{c}
\left( 5^{\left( k-1\right) /2}L_{2m+2},5^{\left( k-1\right)
/2}L_{2m+1}\right) ,\text{ }k\text{ is an odd integer;} \\
\left( 5^{k/2}F_{2m+1},5^{k/2}F_{2m}\right) ,\text{ }k\text{ is an even
integer.}%
\end{array}%
\right.
\end{equation*}

In a similar way, it can be shown that all nonnegative integer solutions of
the equation $x^{2}-xy-y^{2}=-5^{k}$ are given by
\begin{equation*}
\left( x,y\right) =\left\{
\begin{array}{c}
\left( 5^{\left( k-1\right) /2}L_{2m+1},5^{\left( k-1\right)
/2}L_{2m}\right) ,\text{ }k\text{ is an odd integer;} \\
\left( 5^{k/2}F_{2m+2},5^{k/2}F_{2m+1}\right) ,\text{ }k\text{ is an even
integer.}%
\end{array}%
\right.
\end{equation*}
\end{proof}

\section{Acknowledgements}

The authors would like to thank the anonymous referee for careful reading
and suggestions that improved the clarity of the manuscript.

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\bibitem{KSH} T. Koshy, \textit{Fibonacci and Lucas Numbers With Applications%
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\bibitem{VJD} S. Vajda, \textit{Fibonacci and Lucas Numbers and The Golden
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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification:} Primary 11B37;
Secondary 11B39,11B50.

\noindent {\it Keywords:}  Fibonacci number; Lucas number; Diophantine
equation.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences \seqnum{A000032} and \seqnum{A000045}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 17 2009;
revised version received  December 8 2009.
Published in {\it Journal of Integer Sequences}, December 8 2009.
Minor correction to equation (13) and the line before it, and
to the statements of Theorems 15 and 16, December
16 2009.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


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