\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \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 $p<10000$. 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 $\ p<1000$. 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 $% 10003$ 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 $m>1.$ \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 $m>3$ 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 $m>2.$ 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 $m>1$. 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} \begin{thebibliography}{99} \bibitem{Alf} B. U. Alfred, \textrm{On~square Lucas~numbers,} \textit{% Fibonacci Quart.}\emph{\ }\textbf{2} (1964), 11--12. \bibitem{NE} D. M. Burton, \textrm{Elementary Number Theory,} McGraw-Hill, 1998. \bibitem{CARLITZ} L. Carlitz, \textrm{A~note~on~Fibonacci~numbers}$\mathrm{,} $ \textit{Fibonacci Quart.}\emph{\ }\textbf{1} (1964), 15--28. \bibitem{Chn1} J. H. E. Cohn, \textrm{On square Fibonacci numbers, etc.,} \textit{J. London Math. Soc.} \textbf{39} (1964), 537--540. \bibitem{CHNN} J. H. E. Cohn, \textrm{Square Fibonacci numbers, etc.,} \textit{Fibonacci Quart.} \textbf{2.2} (1964), 109--113. \bibitem{COHN} J. H. E. Cohn, \textrm{Lucas and Fibonacci numbers and some Diophantine equations,} \textit{Proc. Glasgow Math. Assoc.} \textbf{7} (1965), 24--28. \bibitem{COHN1} J. H. E. Cohn, \textrm{Squares in some recurrent sequences,}% \emph{\ }\textit{Pacific J. Math.}\emph{\ }\textbf{41} (1972), 631--646. \bibitem{RK} R. Keskin and B. Demirt\"{u}rk, \textrm{Fibonacci and Lucas congruences and their applications,}\emph{\ }\textit{Acta Math. Sin.} (Engl. Ser.) \textbf{27} (2011), 725--736. \bibitem{CH} T. Koshy, \textrm{Fibonacci and Lucas Numbers With Applications,% } John Wiley and Sons, Proc., New York-Toronto, 2001. \bibitem{Mos} L. Moser and L. Carlitz, \textrm{Advanced problem H-2,} \textit{Fibonacci Quart.} \textbf{1} (1963), 46. \bibitem{ZCK} I. Niven, H. S. Zuckerman, and H. L. Montgomery, \textit{An Introduction to The Theory of Numbers,} John Wiley \& Sons, 1991. \bibitem{Ogi} S. C. Ogilvy, \textit{Tomorrow's Math: Unsolved Problems for the Amateur,}\emph{\ }\textit{Oxford University Press}, 1962. \bibitem{RIBEN} P. Ribenboim, \textrm{Square classes of Fibonacci and Lucas numbers,}\emph{\ }\textit{Portugal. Math.} \textbf{46} (1989), 159--175. \bibitem{NRBX} N. Robbins, \textrm{On Fibonacci numbers of the form }$px^{2}$% \textrm{, where }$p$\textrm{\ is prime,} \textit{Fibonacci Quart.} \textbf{21% } (1983), 266--271. \bibitem{RBB} N. Robbins, \textrm{Fibonacci numbers of the form }$cx^{2}$% \textrm{, where }$1\leq c\leq 1000$\textrm{,} \textit{Fibonacci Quart.} \textbf{28} (1990), 306--315. \bibitem{NRBB} N. Robbins, \textrm{Lucas numbers of the form }$px^{2}$% \textrm{, where }$p$\textrm{\ is prime,} \textit{Internat. J. Math. Math. Sci.} \textbf{14} (1991), 697--703. \bibitem{Rol} A. P. Rollet, \textrm{Advanced problem 5080,} \textit{Amer. Math. Monthly} \textbf{70} (1963), 216. \bibitem{VAJDA} S. Vajda, \textit{Fibonacci and Lucas Numbers and The Golden Section,} Ellis Horwood, 1989. \bibitem{Wy} O. Wyler, \textrm{In the Fibonacci series }$F_{1}=1,$\textrm{\ }% $F_{2}=1,~F_{n+1}=F_{n}+F_{n-1}$\textrm{\ the first, second and twelfth terms are squares,} \textit{Amer. Math. Monthly} \textbf{71} (1964), 221--222. \bibitem{CHZ} C. Zhou, \textrm{A general conclusion on Lucas numbers of the form }$px^{2},$\textrm{\ where }$p$\textrm{\ is prime,} \textit{Fibonacci Quart. }\textbf{37} (1999), 39--45. \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}