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\begin{center}
\vskip 1cm{\LARGE\bf On Fibonacci and Lucas Numbers \\
\vskip .1in
of the Form $cx^{2}$
}
\vskip 1cm
\large
Refik Keskin and Zafer Yosma \\
Department of Mathematics \\
Faculty of Science and Arts\\
Sakarya University \\
Sakarya\\
Turkey \\
\href{mailto:rkeskin@sakarya.edu.tr}{\tt rkeskin@sakarya.edu.tr} \\
\href{mailto:d085016004@sakarya.edu.tr}{\tt d085016004@sakarya.edu.tr} \\
\end{center}

\def\func#1{{\rm #1\ }}

\begin{abstract}
In this paper, by using some congruences concerning with Fibonacci and Lucas
numbers, we completely solve the Diophantine equations
$L_{n}=2L_{m}x^{2}$, $F_{n}=2F_{m}x^{2}$,
$L_{n}=6L_{m}x^{2}$, $F_{n}=3F_{m}x^{2}$, and $F_{n}=6F_{m}x^{2}$.
\end{abstract}


\section{Introduction}

Fibonacci and Lucas sequences are defined as follows; $F_{0}=0$, $F_{1}=1$, $%
F_{n}=F_{n-1}+F_{n-2}$ for $n\geq 2$ and $L_{0}=2,L_{1}=1,$ $%
L_{n}=L_{n-1}+L_{n-2}$ for $n\geq 2,$ respectively. $F_{n}$ is called the $n$%
-th Fibonacci number and $L_{n}$ is called the $n$-th Lucas number.
Fibonacci and Lucas numbers for negative subscripts are given by $%
F_{-n}=\left( -1\right) ^{n+1}F_{n}$ \ for $n\geq 1$ and $L_{-n}=\left(
-1\right) ^{n}L_{n}$ \ for $n\geq 1.$ It can be seen that $%
L_{n}=F_{n-1}+F_{n+1}$ and $L_{n-1}+L_{n+1}=5F_{n}$ for every $n\in \mathbb{Z%
}$. For more information about Fibonacci and Lucas sequences, one can
consult \cite{CH}, \cite{VAJDA}.

Let $\alpha $ and $\beta $ denote the roots of the equation $x^{2}-x-1=0.$
Then $\alpha =(1+\sqrt{5})/2$ \ and $\beta =(1-\sqrt{5})/2.$ It can be seen
that $\alpha \beta =-1$ and $\alpha +\beta =1$. Moreover it is well known
and easy to show that

\begin{equation*}
F_{n}=\left( \alpha ^{n}-\beta ^{n}\right) /\sqrt{5}
\end{equation*}%
and%
\begin{equation*}
L_{n}=\alpha ^{n}+\beta ^{n}
\end{equation*}%
for every $n\in \mathbb{Z}.$

In the following section, we will give some congruences concerning with
Fibonacci and Lucas numbers. By using these congruences, we may prove many
properties known before.

\section{Preliminaries}

The problem of characterizing
the square Fibonacci numbers was first introduced in the book
by Ogilvy \cite[p.\ 100]{Ogi}.
In 1963, both Moser and Carlitz \cite{Mos}, and Rollet
\cite{Rol} proposed this problem. In 1964, the square conjecture
was proved by Cohn \cite{Chn1} and independently by Wyler \cite{Wy}.
Later the problem of characterizing
the square Lucas numbers was solved by Cohn \cite{COHN}
and by Alfred \cite{Alf}. Moreover in 1965, Cohn solved the Diophantine
equations $F_{n}=2x^{2}$ and $L_{n}=2x^{2}$ in \cite{COHN}.

We give the following theorem from \cite{CHNN}.

\begin{theorem}
\label{t:2.1}If $F_{n}=x^{2}$, then $n=1,2,12$. If $F_{n}=2x^{2}$, then $n=3$%
, $6$. If $L_{n}=x^{2}$, then $n=1,3$ and if $L_{n}=2x^{2}$, then $n=6$.
\end{theorem}

The proofs of the following two theorems are given in \cite{RK}.

\begin{theorem}
\label{t:2.3}Let $n\in \mathbb{N}$ $\cup \{0\}$ and $k,m\in \mathbb{Z}$.
Then
\begin{equation}
F_{2mn+k}\equiv \left( -1\right) ^{mn}F_{k}\left( \func{mod}F_{m}\right)
\label{2.3}
\end{equation}%
and%
\begin{equation}
L_{2mn+k}\equiv \left( -1\right) ^{mn}L_{k}\left( \func{mod}F_{m}\right) .
\label{2.4}
\end{equation}
\end{theorem}

\begin{theorem}
\label{t:2.4}Let $n\in \mathbb{N\cup }\{0\}$ and $k,m\in \mathbb{Z}$. Then%
\begin{equation}
L_{2mn+k}\equiv \left( -1\right) ^{\left( m+1\right) n}L_{k}\left( \func{mod}%
L_{m}\right)  \label{2.5}
\end{equation}%
and%
\begin{equation}
F_{2mn+k}\equiv \left( -1\right) ^{\left( m+1\right) n}F_{k}\left( \func{mod}%
L_{m}\right) .  \label{2.6}
\end{equation}
\end{theorem}

From the identity (\ref{2.4}), it follows that $8\nmid L_{n}$ for any
natural number $n.$

Now we give two lemmas and a corollary, which will be needed later. The
proofs of the lemmas can be achieved by induction. For the proof of the
corollary, one can consult \cite{NE} or \cite{ZCK}.

\begin{lemma}
\label{L.1}$L_{2^{k}}\equiv 3\ (\func{mod}4)$ for the all positive integers $k$
with $k\geq 1$.
\end{lemma}

\begin{lemma}
\label{L.2}If $r\geq 3,$ then $L_{2^{r}}\equiv 2\ (\func{mod}3).$
\end{lemma}

\begin{corollary}
\label{c.1}If $k\geq 1$, then there is no integer $x$ such that $x^{2}\equiv
-1\ (\func{mod}L_{2^{k}})$.
\end{corollary}

The following lemma can be proved by induction.

\begin{lemma}
\label{L.3}If $r\geq 2,$ then $L_{2^{r}}\equiv 7\ \left( \func{mod}8\right) .$
\end{lemma}

The proofs of\ the following theorems can be found in \cite{CARLITZ}, \cite%
{VAJDA} or \cite{RK}.

\begin{theorem}
\label{T1}Let $m,n\in \mathbb{N}$ and $m\geq 2$. Then $L_{m}|L_{n}$ if and
only if $m|n$ and $\dfrac{n}{m}$ is an odd integer.
\end{theorem}

\begin{theorem}
\label{T2}Let $m,n\in \mathbb{N}$ and $m\geq 3$. Then $F_{m}|F_{n}$ if and
only if $m|n.$
\end{theorem}

\begin{theorem}
\label{T3}Let $m,n\in \mathbb{N}$ and $m\geq 2.$ Then $L_{m}|F_{n}$ if and
only if $m|n$ and $\dfrac{n}{m}$ is an even integer.
\end{theorem}

Also we give some identities about Fibonacci and Lucas numbers which will be
needed in the sequel:%
\begin{equation}
L_{2n}=L_{n}^{2}-2(-1)^{n}  \label{C}
\end{equation}%
\begin{equation}
L_{3n}=L_{n}(L_{n}^{2}-3(-1)^{n})  \label{d}
\end{equation}%
\begin{equation}
F_{2n}=F_{n}L_{n}  \label{f}
\end{equation}%
\begin{equation}
F_{3n}=F_{n}(5F_{n}^{2}+3(-1)^{n})  \label{2.8}
\end{equation}%
\begin{equation}
L_{n}^{2}-5F_{n}^{2}=4(-1)^{n}  \label{2.11}
\end{equation}%
\begin{equation}
2|F_{n}\text{ }\Leftrightarrow 2|L_{n}\Leftrightarrow 3|n  \label{x}
\end{equation}%
\begin{equation}
(F_{n},L_{n})=1\text{ or }(F_{n},L_{n})=2  \label{y}
\end{equation}

Let $\left( \frac{a}{p}\right) $ represent the Legendre symbol. Then we have
\begin{equation}
\left( \frac{2}{p}\right) =1\text{ if and only if }p\equiv \pm 1\ \left( \func{%
mod}8\right)  \label{2.12}
\end{equation}%
and%
\begin{equation}
\left( \frac{-2}{p}\right) =1\text{ if and only if }p\equiv 1,~3\ \left( \func{%
mod}8\right) .  \label{2.13}
\end{equation}

For the proof of (\ref{2.12}) and (\ref{2.13}), one can consult \cite{NE} or
\cite{ZCK}.

\section{Main Theorems}

Many authors investigated Fibonacci and Lucas numbers of the form $cx^{2}$.
In \cite{CHNN}, Cohn solved $F_{n}=cx^{2}$ and $L_{n}=cx^{2}$ for $c=1,2$.
In \cite{NRBX}, Robbins considered Fibonacci numbers of the form $px^{2}$.
Robbins solved the equation $F_{n}=px^{2}$ for all $p$ such that $p\equiv 3\ (%
\func{mod}4)$ or $p&lt;10000$. Later, in \cite{RBB} Robbins considered
Fibonacci numbers of the form $cx^{2}$. The author obtained all solutions of
$F_{n}=cx^{2}$ for composite values of $c\leq 1000$. After that, in \cite%
{NRBB}, the same author solved $L_{n}=px^{2},$ where $p$ is an odd prime and
$\ p&lt;1000$. Moreover, in \cite{CHZ}, Zhou dealt with Lucas numbers of the
form $L_{n}=px^{2},$ where $p$ is a prime number, and he gave solutions for $%
1000&lt;p&lt;60000$. In this section, we consider the equations $%
L_{n}=2L_{m}x^{2},~F_{n}=2F_{m}x^{2},$ $L_{n}=6L_{m}x^{2},$ $%
F_{n}=3F_{m}x^{2},$ and $F_{n}=6F_{m}x^{2}.$

In \cite{RIBEN}, Ribenboim considers square-classes of Fibonacci numbers. $%
F_{m},~F_{n}$ are in the same square-class if there exist non-zero integers $%
x,~y$ such that $F_{m}x^{2}=F_{n}y^{2}$; or equivalently, when $F_{m}F_{n}$
is a square. In a similar way, he considers square-classes of Lucas numbers.
A square-class will be called trivial if it consists of only one number.
Ribenboim showed that the square-class of $L_{m}$ is trivial when $m\neq
0,~1,~3,~$and $6.$ He also showed that the square-class of $F_{m}$ is
trivial when $m\neq 1,~2,~3,~6,~12.$ Now, we can give following two
theorems, which can be obtained from Proposition $1$ and Proposition $2$
given in \cite{RIBEN}.

From now on, we will assume that $n$ and $m$ are positive integers.

\begin{theorem}
\label{t:2.5}Let $m&gt;3$ be an integer and $F_{n}=F_{m}x^{2}$ for some $x\in
\mathbb{Z}.$ Then $n=m$.
\end{theorem}

\begin{theorem}
\label{t:2.6}Let $m\geq 2$ be an integer and $L_{n}=L_{m}x^{2}$ for some $%
x\in \mathbb{Z}.$ Then $n=m$.
\end{theorem}

The proofs of the following two theorems can be obtained from Theorem 6 and
Theorem 12 given in \cite{COHN1}, but we will give a different proof.

\begin{theorem}
\label{T3.1}There is no integer $x$ such that $L_{n}=2L_{m}x^{2}$ for $m&gt;1.$
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\proof}}%
%BeginExpansion
\proof%
%EndExpansion
Assume that $L_{n}=2L_{m}x^{2}.$ Then $L_{m}|L_{n}$ and therefore $n=mk$ for
some odd natural number $k$ by Theorem \ref{T1}. Firstly assume that $m$ is an
odd integer. Since $2|L_{n},$ we get $3|n$ by (\ref{x}). Thus we see that $%
3\nmid m$. For if $3|m$, then $L_{3}|L_{m},$ i.e., $4|L_{m}$ by Theorem \ref%
{T1}. This implies that $8|L_{n}$, which is impossible. Since $3\nmid m,$ it
follows that $3|k$. That is, $k=3t$ for some odd positive integer $t$. Thus $%
n=mk=3mt$ and $mt$ is an odd integer. Therefore, since $3|n,$ it follows
that $L_{3}|L_{n},$ i.e., $4|2L_{m}x^{2}$ by Theorem \ref{T1}.\ Since $%
3\nmid m,$ $L_{m}$ is an odd integer. Therefore $2|x^{2},$ i.e., $x$ is an
even integer. This implies that $8|L_{n}$, which is impossible.

Now assume that $m$ is an even integer. If $x$ is an even integer, then we
see that $8|L_{n}$, which is impossible. Therefore $x$ is an odd integer.
Assume that $3|m.$ Then $L_{m}$ is an even integer. Therefore $L_{3}|L_{n}$
by Theorem \ref{T1}. It follows that $n=3b$ for some odd integer $b$ by
Theorem \ref{T1}. That is, $n$ is an odd integer. But this is impossible.
Because since $m$ is an even integer, $n$ is also an even integer. Assume
that $3\nmid m.$ Then since $n=mk$ and $3|n,$ we get $3|k,$ i.e., $k=3t$ for
some odd integer $t$. Since $t$ is an odd integer, $t=4q\pm 1$ for some
nonnegative integer $t.$ Thus $n=mk=3m(4q\pm 1)=2\cdot 6mq\pm 3m.$ Then
\begin{equation*}
L_{n}=L_{2\cdot 6mq\pm 3m}\equiv L_{\pm 3m} \ (\func{mod}F_{6})
\end{equation*}%
and therefore%
\begin{equation*}
2L_{m}x^{2}\equiv L_{3m}\ (\func{mod}8)
\end{equation*}%
by (\ref{2.4}). Since $x^{2}\equiv 1 \ (\func{mod}8)$ and $m$ is even integer,
we get
\begin{equation*}
2L_{m}\equiv L_{m}(L_{m}^{2}-3)\ (\func{mod}8)
\end{equation*}%
by (\ref{d}). Moreover, since $3\nmid m,$ $L_{m}$ is odd integer. Therefore
we get
\begin{equation*}
2\equiv L_{m}^{2}-3 \ (\func{mod}8).
\end{equation*}%
Thus%
\begin{equation*}
2\equiv -2\ (\func{mod}8),
\end{equation*}%
which is impossible. This completes the proof.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}%
%BeginExpansion
\endproof%
%EndExpansion

In \cite{CHNN}, it is shown that, for $m=1,2,$ the equation $%
F_{n}=2F_{m}x^{2}=$ $2x^{2}$ has solution only for $n=3,6.$ More generally,
we can give the following theorem.

\begin{theorem}
\label{T3.2}If $F_{n}=2F_{m}x^{2}$ and $m\geq 3,$ then $m=3,$ $x^{2}=36,$
and $n=12$ or $m=6,$ $x^{2}=9,$ and $n=12.$
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\proof}}%
%BeginExpansion
\proof%
%EndExpansion
If $m=3,$ then $F_{n}=2F_{3}x^{2}=(2x)^{2}.$ Thus it can be seen that $n=12,$
$x^{2}=36$ by Theorem \ref{t:2.1}.\ Assume that $m&gt;3$ and $%
F_{n}=2F_{m}x^{2}. $ Then $F_{m}|F_{n}$ and therefore $n=mk$ for some
natural number $k$ by Theorem \ref{T2}.

Firstly, assume that $k$ is an even integer. Then $k=2t$ for some integer $%
t. $ Therefore $n=mk=2mt.$ Thus
\begin{equation*}
F_{n}=F_{2mt}=F_{mt}L_{mt}=2F_{m}x^{2}
\end{equation*}%
by (\ref{f}). This shows that $\left( F_{mt}/F_{m}\right) L_{mt}=2x^{2}.$ It
can be easily seen that if $\left( F_{mt}/F_{m},L_{mt}\right) =d,$ then $d=1$
or $d=2$ by (\ref{y}). Thus we have the following equations:

\begin{equation}
\frac{F_{mt}}{F_{m}}=u^{2},\text{ }L_{mt}=2v^{2},  \label{3.1}
\end{equation}

\begin{equation}
\frac{F_{mt}}{F_{m}}=2u^{2},\text{ }L_{mt}=v^{2},  \label{3.2}
\end{equation}

\begin{equation}
\frac{F_{mt}}{F_{m}}=2u^{2},\text{ }L_{mt}=(2v)^{2},  \label{3.3}
\end{equation}%
and

\begin{equation}
\frac{F_{mt}}{F_{m}}=(2u)^{2},\text{ }L_{mt}=2v^{2}.  \label{3.4}
\end{equation}
Assume that (\ref{3.1}) is satisfied. Then $mt=m,$ i.e., $t=1$ by Theorem %
\ref{t:2.5}. Therefore $L_{m}=2v^{2}$ and this implies that $m=6$ by Theorem %
\ref{t:2.1}. Thus we get $m=6,$ $x^{2}=9,$ and $n=12.$\ By using Theorem \ref%
{t:2.1} and Theorem\ \ref{t:2.5}, it can be seen that the other three cases
are impossible.

Secondly, assume that $k$ is an odd integer. Suppose that $m$ is an even
integer, i.e., $m=2r$ for some natural number $r.$ Then we can write $%
n=mk=2kr.$ Thus
\begin{equation*}
F_{n}=F_{2kr}=F_{kr}L_{kr}=2F_{r}L_{r}x^{2}
\end{equation*}%
by (\ref{f}). This shows that $\left( F_{kr}/F_{r}\right)
(L_{kr}/L_{r})=2x^{2}.$ A similar argument shows that the equation $\left(
F_{kr}/F_{r}\right) (L_{kr}/L_{r})=2x^{2}$ has no solution. Now assume that $%
m$ is an odd integer. Firstly, suppose that $3\nmid k.$ Since $k$ is an odd
integer, we can write $k=6q\pm 1$ for some nonnegative integer $q.$
Therefore $n=mk=m(6q\pm 1)=2\cdot 3mq\pm m.$ Thus we get
\begin{equation*}
F_{n}=F_{2\cdot 3mq\pm m}\equiv F_{\pm m} \ (\func{mod}L_{3}),
\end{equation*}%
i.e.,%
\begin{equation*}
F_{n}\equiv F_{m} \ (\func{mod}4)
\end{equation*}%
by (\ref{2.6}). Since $F_{n}$ is even integer, $F_{m}$ is also an even
integer. Thus $3|m,$ and therefore $m=3a$ for some integer $a$ by (\ref{x}).
On the other hand, since $F_{m}$ is even integer, $4|F_{n},$ and thus $6|n$
by Theorem \ref{T2}. Since $n=mk=3ak,$ we get $6|3ak,$ i.e., $2|ak.$
Moreover, since $k$ is odd integer, it is seen that $2|a.$ This implies that
$2|m,$ which is impossible. Because $m$ is an odd integer. Assume that $3|k.$
Then $k=3s$ for some odd integer $s$. Therefore $n=mk=3ms.$ Thus since $ms$
is odd integer, we get
\begin{equation*}
F_{n}=F_{3ms}=F_{ms}(5F_{ms}^{2}-3)=2F_{m}x^{2}
\end{equation*}%
by (\ref{2.8}). This shows that $\left( F_{ms}/F_{m}\right) \left(
5F_{ms}^{2}-3\right) =2x^{2}.$ It can be easily seen that if $d=\left(
F_{ms}/F_{m},5F_{ms}^{2}-3\right) ,$ then $d=1$ or $d=3$. Assume that $d=3.$
Then $3|F_{ms},$ and \ thus $4|ms$ by Theorem \ref{T2}. But this is
impossible, since $ms$ is odd integer. Therefore $d=1.$ Then we get

\begin{equation}
\frac{F_{ms}}{F_{m}}=u^{2},\text{ }5F_{ms}^{2}-3=2v^{2}  \label{3.9}
\end{equation}%
or

\begin{equation}
\frac{F_{ms}}{F_{m}}=2u^{2},\text{ }5F_{ms}^{2}-3=v^{2}  \label{3.10}
\end{equation}%
for some integers $u$ and $v$. Assume that (\ref{3.9}) is satisfied. Then $%
ms=m,$ i.e., $s=1$ by Theorem \ref{t:2.5}. Therefore $5F_{m}^{2}-3=2v^{2}$
and this shows that $2v^{2}=5F_{m}^{2}-3=L_{m}^{2}+1=L_{2m}-1$ by (\ref{2.11}%
) and (\ref{C}). This implies that $L_{2m}=2v^{2}+1.$ Since $%
L_{2m}=2v^{2}+1, $ we get $3\nmid m.$ Thus we can write $m=6q\pm 1=3\cdot
2^{r+1}b\pm 1,$ where $q=2^{r}b$ for some odd integer $b$ with $r\geq 0.$
This shows that
\begin{equation*}
L_{2m}=L_{2\cdot 2^{r+1}3b\pm 2}\equiv -L_{\pm 2}\ (\func{mod}L_{2^{r+1}})
\end{equation*}%
and therefore%
\begin{equation*}
2v^{2}+1\equiv -3 \ (\func{mod}L_{2^{r+1}}),
\end{equation*}%
i.e.,%
\begin{equation*}
2v^{2}\equiv -4 \ (\func{mod}L_{2^{r+1}})
\end{equation*}%
by (\ref{2.5}). Since $L_{2^{r+1}}$ is an odd integer, we get
\begin{equation*}
v^{2}\equiv -2\ (\func{mod}L_{2^{r+1}}).
\end{equation*}%
This shows that $\left( \dfrac{-2}{p}\right) =1$ for every prime divisor of $%
L_{2^{r+1}}$. Then it follows that%
\begin{equation*}
p\equiv 1,3\ (\func{mod}8)
\end{equation*}%
by (\ref{2.13}) and therefore\ \ \ \ \ \ \ \ \
\begin{equation*}
L_{2^{r+1}}\equiv 1,3\ (\func{mod}8).
\end{equation*}%
This shows that $r=0$ by Lemma \ref{L.3}. Consequently, $q$ is an odd
integer. Therefore it can be easily seen that $m=12c+5$ or $m=12c+7$ for
some integer $c.$ Thus we get \ \ \ \ \ \ \ \ \
\begin{equation*}
L_{m}\equiv 3\ (\func{mod}8)
\end{equation*}%
or \ \ \ \ \ \ \ \ \
\begin{equation*}
L_{m}\equiv 5\ (\func{mod}8)
\end{equation*}%
by (\ref{2.4}). On the other hand, since

\begin{equation*}
2v^{2}=L_{m}^{2}+1,
\end{equation*}%
we get%
\begin{equation*}
2v^{2}\equiv 1\ (\func{mod}L_{m}),
\end{equation*}%
and therefore
\begin{equation*}
(2v)^{2}\equiv 2\ (\func{mod}L_{m}).
\end{equation*}%
This shows that $\left( \dfrac{2}{p}\right) =1$ for every prime divisor $p$
of $L_{m}$. Then it follows that%
\begin{equation*}
p\equiv \pm 1\ (\func{mod}8)
\end{equation*}%
by (\ref{2.12}) and therefore
\begin{equation*}
L_{m}\equiv \pm 1\ (\func{mod}8).
\end{equation*}%
But this contradicts with the fact that $L_{m}\equiv 3,5\ (\func{mod}8).$
Assume that (\ref{3.10}) is satisfied. Then we get $%
v^{2}=5F_{ms}^{2}-3=L_{ms}^{2}+1$ by (\ref{2.11}). This implies that $%
L_{ms}=0,$ which is impossible. This completes the proof.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}%
%BeginExpansion
\endproof%
%EndExpansion

\begin{theorem}
\label{t:2.7}If $L_{n}=6L_{m}x^{2}$ and $m\geq 1$, then $m=2$, $x^{2}=1,$
and $n=6$.
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\proof}}%
%BeginExpansion
\proof%
%EndExpansion
Assume that $L_{n}=6L_{m}x^{2}$ for some integer $x$. Then $3|L_{n}$ and
therefore $n=2k_{0}$ for some odd integer $k_{0}$ by Theorem \ref{T1}.
Moreover, since $2|L_{n}$, we get $3|n$ by (\ref{x}). This shows that $%
3|k_{0}$ and then $k_{0}=3t$ for some odd integer $t$. Thus $%
n=6t=6(2u+1)=12u+6$. Therefore
\begin{equation*}
L_{n}=L_{12u+6}\equiv L_{6}\ (\func{mod}8)
\end{equation*}%
by (\ref{2.4}). That is,
\begin{equation*}
L_{n}\equiv 2\ (\func{mod}8).
\end{equation*}%
Since $8\nmid $ $L_{n}$, it can be seen that $x$ is an odd integer. Therefore%
\begin{equation*}
x^{2}\equiv 1\ (\func{mod}8),
\end{equation*}%
which implies that
\begin{equation*}
6L_{m}x^{2}\equiv 6L_{m}\ (\func{mod}8).
\end{equation*}%
This shows that\ \ \ \
\begin{equation*}
6L_{m}\equiv 2\ (\func{mod}8),
\end{equation*}%
which implies that $m\neq 1.$ Now assume that $m&gt;2.$ Since\ $L_{m}|L_{n}$,
there exists an odd integer $k$ such that $n=mk$ by Theorem \ref{T1}. On the
other hand, since $2|n,$ it is seen that $2|m$. Therefore $m=2r$ for some
odd integer $r$. If $r=6q+3,$ then $m=2r=12q+6$ and therefore\ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \
\begin{equation*}
L_{m}\equiv L_{6}\ (\func{mod}8)
\end{equation*}%
by (\ref{2.4}). That is,%
\begin{equation*}
L_{m}\equiv 2\ (\func{mod}8),
\end{equation*}%
which is impossible since
\begin{equation*}
6L_{m}\equiv 2\ (\func{mod}8).
\end{equation*}%
Therefore $3\nmid $ $r$. Since $n=mk$, $m=2r$ and $3\nmid $ $r$, it follows
that $3|k$ and thus $k=3s$ for some odd integer $s$. Then
\begin{equation*}
L_{n}=L_{mk}=L_{3ms}=L_{ms}(L_{ms}^{2}-3)=6L_{m}x^{2}
\end{equation*}%
by (\ref{d}). It can be seen that $(L_{ms},L_{ms}^{2}-3)=3$. Thus $\left(
L_{ms},\dfrac{L_{ms}^{2}-3}{3}\right) =1$. Then we get
\begin{equation*}
\frac{L_{ms}}{L_{m}}\left( \frac{L_{ms}^{2}-3}{3}\right) =2x^{2}.
\end{equation*}%
This shows that
\begin{equation}
\frac{L_{ms}}{L_{m}}=2u^{2\text{ }}\text{and }\frac{L_{ms}^{2}-3}{3}=v^{2}
\label{g}
\end{equation}%
or
\begin{equation}
\frac{L_{ms}}{L_{m}}=u^{2}\text{ and }\frac{L_{ms}^{2}-3}{3}=2v^{2}
\label{h}
\end{equation}%
for some integers $u$ and $v.$ Assume that (\ref{g}) is satisfied. Then $%
3\left( \dfrac{L_{ms}}{3}\right) ^{2}-1=v^{2}$ and therefore
\begin{equation*}
v^{2}\equiv -1\ (\func{mod}3),
\end{equation*}%
which is a contradiction. Now assume that (\ref{h}) is satisfied. Then $%
L_{ms}=L_{m}u^{2}$, which implies that $ms=m$ by Theorem \ref{t:2.6}. That
is, $s=1$. Thus $L_{m}^{2}-3=6v^{2}$. Since $L_{m}^{2}=L_{2m}+2$ by (\ref{C}%
), we see that $L_{2m}-1=6v^{2}$. Moreover, since $m=2r$, it follows that $%
L_{4r}-1=6v^{2}$. On the other hand, we can write $4r$ as $4r=4(4u\pm
1)=16u\pm 4=2\cdot 2^{k}a\pm 4$ for some odd integer $a$ with $k\geq 3$.
This shows that
\begin{equation*}
L_{4r}=L_{2\cdot 2^{k}a\pm 4}\equiv -L_{\pm 4}\ (\func{mod}L_{2^{k}})
\end{equation*}%
by (\ref{2.5}) and therefore\ \ \ \ \ \ \ \ \ \ \
\begin{equation*}
1+6v^{2}\equiv -7\ (\func{mod}L_{2^{k}}).
\end{equation*}%
Then we get\ \ \ \ \ \ \ \ \ \ \ \
\begin{equation*}
6v^{2}\equiv -8\ (\func{mod}L_{2^{k}}).
\end{equation*}%
That is, \ \ \ \ \ \
\begin{equation*}
3v^{2}\equiv -4\ (\func{mod}L_{2^{k}}).
\end{equation*}%
Thus\ \ \ \ \ \ \ \
\begin{equation*}
(3v)^{2}\equiv -12\ (\func{mod}L_{2^{k}}).
\end{equation*}%
This shows that $\left( \dfrac{-12}{p}\right) =1$ for every prime divisor $p$
of $L_{2^{k}}$. Then it follows that\ \ \ \ \ \ \ \ \ \
\begin{equation*}
p\equiv 1\ (\func{mod}3)
\end{equation*}%
and therefore\ \ \ \ \ \ \ \ \
\begin{equation*}
L_{2^{k}}\equiv 1\ (\func{mod}3).
\end{equation*}%
But this contradicts with Lemma \ref{L.2}. This completes the proof.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}%
%BeginExpansion
\endproof%
%EndExpansion

In \cite{RK}, the authors showed that $L_{n}=3L_{m}x^{2}$ has no solution if
$m&gt;1$. Now we give a similar result for Fibonacci numbers.

\begin{theorem}
\label{t:2.8}Let $m\geq 3$ be an integer and $F_{n}=3F_{m}x^{2}$ for some
integer $x$. Then $m=4$, $x^{2}=16,$ and $n=12$.
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\proof}}%
%BeginExpansion
\proof%
%EndExpansion
Assume that $m\geq 3$ and $F_{n}=3F_{m}x^{2}$ for some integer $x$. Then $%
F_{m}|F_{n}$ and therefore $n=mk$ for some integer $k$ by Theorem \ref{T2}.

Firstly, assume that $k$ is an even integer. Then $k=2s$ for some $s\in
\mathbb{N}$. Therefore $n=mk=2ms$. Thus
\begin{equation*}
F_{n}=F_{2ms}=F_{ms}L_{ms}=3F_{m}x^{2}
\end{equation*}%
by (\ref{f}). This shows that

\begin{equation*}
(F_{ms}/F_{m})L_{ms}=3x^{2}.
\end{equation*}%
By using Theorem \ref{t:2.1}, Theorem \ref{t:2.6}, and Theorem \ref{t:2.7},
it can be shown that the equation $(F_{ms}/F_{m})L_{ms}=3x^{2}$ has no
solution.

Now assume that $k$ is an odd integer. Since $F_{n}=3F_{m}x^{2}$, we get $%
4|n $ by Theorem \ref{T2}. Moreover, since $n=mk$ and $k$ is odd, we get $%
4|m $. Assume that $x$ is an even integer. Then $4|F_{n}$. Thus $L_{3}|F_{n}$
and $3|n$ by Theorem \ref{T3}. Therefore since $4|n$ and $3|n$, we get $12|n$%
. That is, $n=12t$ for some $t\in \mathbb{N}$. On the other hand since $4|m$%
, we get $m=4r$ for some $r\in \mathbb{N}$. Therefore $12t=n=mk=4rk$. Then
it follows that $3t=rk$. Thus
\begin{equation*}
F_{n}=F_{12t}=F_{6t}L_{6t}=3F_{2r}L_{2r}x^{2}
\end{equation*}%
by (\ref{f}). Since $(6t/2r)=k$ and $k$ is odd, we can write

\begin{equation*}
\frac{F_{6t}}{F_{2r}}.\frac{L_{6t}}{L_{2r}}=3x^{2}.
\end{equation*}%
Assume that $3|r$. Then, it can be seen that $\left( \dfrac{F_{6t}}{F_{2r}},%
\dfrac{L_{6t}}{L_{2r}}\right) =1$. Therefore

\begin{equation}
\frac{F_{6t}}{F_{2r}}=u^{2},\text{ }\frac{L_{6t}}{L_{2r}}=3v^{2}  \label{a5}
\end{equation}%
or

\begin{equation}
\frac{F_{6t}}{F_{2r}}=3u^{2},\text{ }\frac{L_{6t}}{L_{2r}}=v^{2}  \label{a6}
\end{equation}%
for some integers $u$ and $v$. A similar argument shows that (\ref{a5}) and (%
\ref{a6}) are impossible. Now assume that $3\nmid r$. Then since $3t=rk,$ it
follows that $3|k$. Thus $k=3s$ for some $s\in \mathbb{N}$. Then $3t=rk=3rs$
and therefore $t=rs$. Also since $3\nmid r$, it can be seen that $\left(
\dfrac{F_{6t}}{F_{2r}},\dfrac{L_{6t}}{L_{2r}}\right) =2$. Therefore

\begin{equation}
\frac{F_{6t}}{F_{2r}}=2u^{2},\text{ }\frac{L_{6t}}{L_{2r}}=6v^{2}  \label{a7}
\end{equation}%
or

\begin{equation}
\frac{F_{6t}}{F_{2r}}=6u^{2},\text{ }\frac{L_{6t}}{L_{2r}}=2v^{2}  \label{a8}
\end{equation}%
for some integers $u$ and $v$. Assume that (\ref{a7}) is satisfied. Then $%
2r=2$ by Theorem \ref{t:2.7}. This shows that $r=1$ and $t=s.$ Thus $%
L_{6t}=6L_{2}v^{2}=L_{6}v^{2}$ and this implies that $6t=6$, i.e., $t=1$ by
Theorem \ref{t:2.6}. Therefore $k=3s=3t=3$ and $m=4r=4$. Therefore $n=12$
and $x^{2}=16.$

Now assume that (\ref{a8}) is satisfied. Then it follows that

\begin{equation*}
L_{6t}=2L_{2r}v^{2},
\end{equation*}%
which is impossible by Theorem \ref{T3.1} and Theorem \ref{t:2.7}.

Now assume that $x$ is an odd integer. Then

\begin{equation*}
F_{n}\equiv 3F_{m}\ (\func{mod}8).
\end{equation*}%
Since $4|m$, it follows that $m=12q$ or $m=12q$ $\pm 4$ for some integer $q$%
. If $m=12q$ $\pm 4$, then

\begin{equation*}
F_{m}\equiv F_{12q\pm 4}\equiv F_{\pm 4}\equiv \pm 3\ (\func{mod}8)
\end{equation*}%
by (\ref{2.3}). Therefore

\begin{equation*}
F_{n}\equiv \pm 1\ (\func{mod}8),
\end{equation*}%
which is impossible since $4|n$. Because if $4|n,$ then $n=12r\pm 4$ or $%
n=12r$ for some integer $r,~$and therefore $F_{n}\equiv \pm 3,0\ (\func{mod}8)$
by (\ref{2.3})$.$ If $m=12q$, then $n=mk=12qk$. This shows that $6qk/6q$ is
an odd integer. Then from the identity%
\begin{equation*}
F_{n}=F_{12qk}=F_{6qk}L_{6qk}=3F_{m}x^{2}=3F_{6q}L_{6q}x^{2},
\end{equation*}%
it follows that

\begin{equation*}
\frac{F_{6qk}}{F_{6q}}.\frac{L_{6qk}}{L_{6q}}=3x^{2}.
\end{equation*}%
Since $\left( \dfrac{F_{6qk}}{F_{6q}},\dfrac{L_{6qk}}{L_{6q}}\right) =1$, we
get

\begin{equation}
\frac{F_{6qk}}{F_{6q}}=u^{2},\text{ }\frac{L_{6qk}}{L_{6q}}=3v^{2}
\label{a9}
\end{equation}%
or

\begin{equation}
\frac{F_{6qk}}{F_{6q}}=3u^{2},\text{ }\frac{L_{6qk}}{L_{6q}}=v^{2}
\label{a10}
\end{equation}%
for some integers $u$ and $v$. Similarly, it can be seen that (\ref{a9}) and
(\ref{a10}) are impossible. This completes the proof.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}%
%BeginExpansion
\endproof%
%EndExpansion
\ \

Lastly, we can give the following theorem without proof since its proof is
similar to that of Theorem \ref{t:2.8}.

\begin{theorem}
\label{t:2.9}There is no integer $x$ such that $F_{n}=6F_{m}x^{2}.$
\end{theorem}

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\end{thebibliography}




\bigskip
\hrule
\bigskip

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

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

\bigskip
\hrule
\bigskip

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

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 29 2011;
revised version received  August 18 2011; September 25 2011.
Published in {\it Journal of Integer Sequences}, October 16 2011.

\bigskip
\hrule
\bigskip

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


\end{document}



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