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\begin{center}
\vskip 1cm{\LARGE\bf Some Remarks On the Equation $F_n=kF_m$
In Fibonacci Numbers} \vskip 1cm \large
M. Farrokhi D. G. \\
Faculty of Mathematical Sciences\\
Ferdowski University of Mashhad\\
Iran \\
\href{mailto:m.farrokhi.d.g@gmail.com.}{\tt m.farrokhi.d.g@gmail.com} \\

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\vskip .2 in
\begin{abstract}
Let $\{F_n\}_{n=1}^{\infty}=\{1,1,2,3,\ldots\}$ be the sequence of
Fibonacci numbers. In this paper we give some sufficient
conditions on a natural number $k$ such that the equation
$F_n=kF_m$ is solvable with respect to the unknowns $n$ and $m$.
We also show that for $k&gt;1$ the equation $F_n = k F_m$ has at most one
solution $(n,m)$.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}{Remark}
\newtheorem*{definition}{Definition}

\newcommand{\lcm}{\mbox{lcm}}

\bigskip

%===============================================================================================
\section{Preliminaries}\label{s-1}
Let $F_n$ be the $n$th Fibonacci number, i.e.,
\[F_1=F_2=1\ ,\ F_{n+2}=F_n+F_{n+1},\ \forall n\in\mathbb{N}.\]
It is known that these numbers have the following properties :

(1)\ \ $F_{m+n}=F_{m-1}F_n+F_mF_{n+1}$;

(2)\ \ $\gcd(F_m,F_n)=F_{\gcd(m,n)}$;

(3)\ \ if $m|n$, then $F_m|F_n$;

(4)\ \ if $F_m|F_n$ and $m&gt;2$, then $m|n$.
\\
Now, put
\begin{eqnarray*}
\mathcal{P}&amp;=&amp;\{k\in\mathbb{N}\ :\ \exists m,n\in\mathbb{N}, F_n=kF_m\},\\
\mathcal{Q}&amp;=&amp;\{k\in\mathbb{N}\ :\ \nexists m,n\in\mathbb{N},
F_n=kF_m\}.
\end{eqnarray*}
A simple computations show that the natural numbers which satisfy
in $\mathcal{P}$, less than $100$, are as follows:
\[1,2,3,4,5,7,8,11,13,17,18,21,29,34,47,48,55,72,76,89.\]
By definition of $\mathcal{P}$ and the properties (3) and (4), for
each $k\in\mathcal{P}$ there exist $m,n\in\mathbb{N}$ such that
$k=\frac{F_{mn}}{F_n}$. However, it seems that the elements of
$\mathcal{Q}$ do not have any special form.

Using a theorem of R. D. Carmichael \cite{rdc}, it can be shown
that the product of Fibonacci numbers and their quotients belong
to $\mathcal{Q}$ except for some cases (see Theorem \ref{t-7}).

In this paper, we use elementary methods to prove our claim. In
section \ref{s-3}, we obtain some more properties of
$\mathcal{P}$. For example, we show that for every element
$k$($&gt;1$) of $\mathcal{P}$, the equation $F_n=kF_m$ has a unique
solution $(n,m)$. Moreover, we give a necessary and sufficient
condition for which the product of two elements of $\mathcal{P}$
is again in $\mathcal{P}$.
%===============================================================================================
\section{The Main Theorem}\label{s-2}
In this section, we introduce some elements $k$ in $\mathcal{Q}$,
so that for each fixed $n\in\mathbb{N}$,
\[k=F_{a_1}F_{a_2}\cdots F_{a_n}\]
belongs to $\mathcal{Q}$, for all natural numbers $a_1,\ldots,a_n$
but a finite number.

In order to prove the above claim, we need the following
elementary properties of Fibonacci numbers.
%-----------------------------------------------------
\begin{lemma}\label{l-1}
For all $a,b,c,a_1,a_2,\ldots,a_n\in\mathbb{N}$, the following
conditions hold

a) $F_{a+b-1}=F_aF_b+F_{a-1}F_{b-1}$;

b) $F_{a+b-2}=F_aF_b-F_{a-2}F_{b-2}$;

c)
$F_{a+b+c-3}=F_aF_bF_c+F_{a-1}F_{b-1}F_{c-1}-F_{a-2}F_{b-2}F_{c-2}$;

d) if $n\geq 3$, then $F_{a_1+\cdots+a_n-n}\geq
F_{a_1}F_{a_2}\cdots F_{a_n}$.
\end{lemma}
\begin{proof}
Parts (a) and (b) are easily verified.

(c) Using (1), we obtain
\begin{eqnarray*}
F_{a+b+c-3}&amp;=&amp;F_{a-1}F_{b+c-3}+F_aF_{b+c-2}\\
&amp;=&amp;F_{a-1}(F_{b-2}F_{c-2}+F_{b-1}F_{c-1})+F_a(F_bF_c-F_{b-2}F_{c-2})\\
&amp;=&amp;F_aF_bF_c+F_{a-1}F_{b-1}F_{c-1}-(F_a-F_{a-1})F_{b-2}F_{c-2}\\
&amp;=&amp;F_aF_bF_c+F_{a-1}F_{b-1}F_{c-1}-F_{a-2}F_{b-2}F_{c-2}.
\end{eqnarray*}

(d) We use induction on $n$. By part (c), the result holds for
$n=3$. Now assume it is true for $n\geq 3$. Clearly
\begin{eqnarray*}
F_{a_1+\cdots+a_{n+1}-(n+1)}&amp;=&amp;F_{a_{n+1}-1}F_{a_1+\cdots+a_n-(n+1)}+F_{a_{n+1}}F_{a_1+\cdots+a_n-n}\\
&amp;\geq&amp;F_{a_{n+1}}F_{a_1+\cdots+a_n-n}\\
&amp;\geq&amp;F_{a_1}F_{a_2}\cdots F_{a_{n+1}},
\end{eqnarray*}
which gives the assertion.
\end{proof}
%-----------------------------------------------------
\begin{remark}
In Lemma \ref{l-1}(d), if $a_1=\cdots=a_n=1$, then
$a_1+\cdots+a_n-(n+1)=-1$ and by generalizing the recursive
relation for negative numbers, we get $F_{-1}=F_1-F_0=1$.
\end{remark}
%-----------------------------------------------------
\begin{remark}
Note that all the formulas in Lemma \ref{l-1} can be also deduced
from Binet's formula
\[F_n=\frac{\alpha^n-\beta^n}{\sqrt{5}},\]
where
\[\alpha=\frac{1+\sqrt{5}}{2},\ \beta=\frac{1-\sqrt{5}}{2}.\]
\end{remark}
%-----------------------------------------------------
\begin{lemma}\label{l-2}
Suppose $m,n$ and $k$ are any natural numbers with $k|n$, then
\[\frac{F_{mn}}{F_n}\stackrel{F_k}{\equiv}mF_{n-1}^{m-1}.\]
\end{lemma}
\begin{proof}
We proceed by induction on $m$. Clearly, the result is true for
$m=1$. Assume it is true for $m$. Now, using (1) and (3), we have
\begin{eqnarray*}
\frac{F_{(m+1)n}}{F_n}&amp;\stackrel{F_k}{\equiv}&amp;F_{n-1}\frac{F_{mn}}{F_n}+F_{mn+1}\\
&amp;\stackrel{F_k}{\equiv}&amp;mF_{n-1}^m+F_{mn+1}\\
&amp;\stackrel{F_k}{\equiv}&amp;mF_{n-1}^m+F_{n-1}F_{(m-1)n+1}+F_nF_{(m-1)n+2}\\
&amp;\stackrel{F_k}{\equiv}&amp;mF_{n-1}^m+F_{n-1}F_{(m-1)n+1}\\
&amp;\vdots&amp;\\
&amp;\stackrel{F_k}{\equiv}&amp;mF_{n-1}^m+F_{n-1}^m\\
&amp;\stackrel{F_k}{\equiv}&amp;(m+1)F_{n-1}^m.
\end{eqnarray*}
\end{proof}
%-----------------------------------------------------
\begin{lemma}\label{l-3}
Let $a_1,\ldots,a_n,n\geq 3$ and $F_{a_1}F_{a_2}\cdots
F_{a_n}=F_b$, then
\[b+n\leq a_1+\cdots+a_n\leq b+2n-2.\]
\end{lemma}
\begin{proof}
By Lemma \ref{l-1}, $F_b=F_{a_1}F_{a_2}\cdots F_{a_n}\leq
F_{a_1+a_2+\cdots+a_n-n}$ and hence $b\leq a_1+a_2+\cdots+a_n-n$.
This gives the left hand side of the inequality. By repeated
application of Lemma \ref{l-1} we have
\begin{eqnarray*}
F_b&amp;=&amp;F_{a_1}F_{a_2}\cdots F_{a_n}\\
&amp;\geq&amp;F_{a_1+a_2-2}F_{a_3}\cdots F_{a_n}\\
&amp;\geq&amp;F_{a_1+a_2+a_3-4}F_{a_4}\cdots F_{a_n}\\
&amp;\vdots&amp;\\
&amp;\geq&amp;F_{a_1+\cdots+a_n-2(n-1)},
\end{eqnarray*}
and so $b\geq a_1+\cdots+a_n-2(n-1)$, which completes the proof.
\end{proof}
%-----------------------------------------------------
\begin{remark}
Note that using Binet's formula, for $n&gt;2$, one obtains
\[(1-\beta^8)\alpha^n\leq\sqrt{5}F_n\leq(1+\beta^6)\alpha^n,\]
which implies the following inequalities
\[v n - u \le a_1+\cdots +a_n - b \le u n - v,\]
where
\[u = -{\log((1-\beta^8)/\sqrt 5) \over \log \alpha}=1.717\ldots\]
and
\[v = -{\log((1+\beta^6)/\sqrt 5) \over \log \alpha}=1.559\ldots.\]
One observes that the above inequalities are sharper than Lemma
\ref{l-3}.
\end{remark}
%-----------------------------------------------------
\begin{definition}
A solution of the equation $F_{a_1}F_{a_2}\cdots F_{a_n}=F_b$ is
said to be nontrivial, whenever $a_1,\ldots,a_n\geq 3$ or
equivalently $F_{a_1},\ldots,F_{a_n}&gt;1$.
\end{definition}
%-----------------------------------------------------
\begin{lemma}\label{l-4}
The equation $F_aF_b=F_c $ has no nontrivial solution, for any
natural numbers $a,b$ and $c$.
\end{lemma}
\begin{proof}
We may assume $a\leq b$ and the triple $(a,b,c)$ is a nontrivial
solution of the equation, i.e., $a,b\geq 3$. Clearly, $F_b|F_c$
and hence $b|c$. Now put $c=kb$ which gives $k\geq 2$ and
therefore $F_aF_b=F_{kb}\geq
F_{2b}=F_b(F_{b-1}+F_{b+1})&gt;F_b^2\geq F_aF_b$, which is
impossible.
\end{proof}

We are now able to prove the main theorem of this section.
%-----------------------------------------------------
\begin{theorem}\label{t-1}
For each fixed $n\geq2$, the equation $F_{a_1}F_{a_2}\cdots
F_{a_n}=F_b$ has at most finitely many nontrivial solutions.
\end{theorem}
\begin{proof}
By Lemma \ref{l-4}, the result follows for $n=2$. Assume, $n\geq
3$ and let $(a_1,\ldots,a_n;b)$ be a nontrivial solution of the
equation. Without loss of generality, we may assume $3\leq a_1\leq
a_2\leq\cdots\leq a_n$. Put $a_1+\cdots+a_n=b+k$. Clearly, by
Lemma \ref{l-3} there are only finitely many natural numbers $k$,
which can satisfy the latter equation. As $F_{a_n}|F_b$ and
$a_n\geq 3$, we have $a_n|b$ and so $b=k'a_n$ for some
$k'\in\mathbb{N}$. Similarly, $F_{a_{n-1}}|F_b=F_{k'a_n}$ and
$a_{n-1}\geq 3$, which implies that $a_{n-1}|k'a_n$ and so
$a_{n-1}=k''k'''$ with $k''|k'$ and $k'''|a_n$. Now since
$F_{k'''}|F_{a_{n-1}}|\frac{F_{k'a_n}}{F_{a_n}}$, Lemma \ref{l-2}
implies that $F_{k'''}|k'$. By Lemma \ref{l-3}, there are only
finitely many $k,k',k'',k'''$ satisfying these equations. Thus
there are only finitely many choices for $a_{n-1}$ and
consequently for $a_1,\ldots,a_{n-2}$. Finally, there are only
finitely many choices for $a_n$ and $b$ satisfying the equation.
\end{proof}
%-----------------------------------------------------
\begin{remark}
The above theorem shows that except finitely many cases if
$k=F_{a_1}\cdots F_{a_n}$, where $a_1,\ldots,a_n$ $\geq 3$ the
equation $F_t=kF_s$ has no solution.
\end{remark}
%===============================================================================================
\section{Some More Results}\label{s-3}
In this section, we consider some more properties of the elements
of $\mathcal{P}$ and $\mathcal{Q}$. For instant, it is shown that
every element $k&gt;1$ of $\mathcal{P}$ satisfies a unique equation
of the form $F_n=kF_m$.
%-----------------------------------------------------
\begin{theorem}\label{t-2}
The equation $F_aF_b=F_cF_d$ holds for natural numbers $a,b,c,d$
if and only if $F_a=F_c$ and $F_b=F_d$, or $F_a=F_d$ and
$F_b=F_c$.
\end{theorem}
\begin{proof}
Clearly, if one the numbers $a,b,c$ or $d$, ($a$, say), is less
than $3$ then $F_b=F_cF_d$ and Lemma \ref{l-4} implies that either
$F_c=F_a=1$ and $F_b=F_d$, or $F_d=F_a=1$ and $F_b=F_c$.
Therefore, we assume that $a,b,c,d\geq 3$ and by symmetry we may
assume that $3\leq a\leq b,c,d$. Using Lemma \ref{l-1}, we have
\[F_{a+b-2}&lt;F_aF_b=F_cF_d&lt;F_{c+d-1},\]
which implies that $a+b-2&lt;c+d-1$ and hence $a+b\leq c+d$.
Similarly $c+d\leq a+b$ and so $a+b=c+d$. By repeated application
of Lemma \ref{l-1}, we obtain
\[\begin{array}{crcl}
&amp;F_aF_b&amp;=&amp;F_cF_d\\
\Rightarrow&amp;F_{a-1}F_{b-1}&amp;=&amp;F_{c-1}F_{d-1}\\
&amp;&amp;\vdots&amp;\\
\Rightarrow&amp;F_2F_{b-a+2}&amp;=&amp;F_{c-a+2}F_{d-a+2}\\
\Rightarrow&amp;F_{b-a+2}&amp;=&amp;F_{c-a+2}F_{d-a+2}.
\end{array}\]
Now by Lemma \ref{l-4}, $F_{c-a+2}=1$ or $F_{d-a+2}=1$, which
implies that either $a=c$ and $b=d$, or $a=d$ and $b=c$.
\end{proof}

The following corollaries follow immediately.
%-----------------------------------------------------
\begin{corollary}\label{c-1}
Suppose $\frac{F_a}{F_b}=\frac{F_c}{F_d}\neq 1$, then $F_a=F_c$
and $F_b=F_d$.
\end{corollary}
%-----------------------------------------------------
\begin{corollary}\label{c-2}
Every element $k&gt;1$ of $\mathcal{P}$ satisfies a unique equation
of the form $F_n=kF_m$, for some natural numbers $m$ and $n$.
\end{corollary}
%-----------------------------------------------------
\begin{corollary}\label{c-3}
The least common multiple of two Fibonacci numbers is again a
Fibonacci number if and only if one divides the other.
\end{corollary}
\begin{proof}
Suppose $\lcm(F_m,F_n)=F_k$, for some natural numbers $m$ and $n$.
Then clearly
\[F_mF_n=\gcd(F_m,F_n)\lcm(F_m,F_n)=F_{\gcd(m,n)}F_k\]
and so $\gcd(F_m,F_n)=F_{\gcd(m,n)}$ is either $F_m$ or $F_n$.
Hence either $F_m|F_n$ or $F_n|F_m$.
\end{proof}
%-----------------------------------------------------
\begin{theorem}\label{t-3}
For any natural numbers $a,b,c,d$ and $e$, the equation
$F_aF_bF_c=F_dF_e$ has no nontrivial solution.
\end{theorem}
\begin{proof}
Assume $(a,b,c;d,e)$ is a nontrivial solution of the equation
$F_aF_bF_c=F_dF_e$. Hence $a,b,c,d,e\geq 3$. By Lemma \ref{l-1},
we have
\[F_{a+b+c-4}&lt;F_aF_bF_c=F_dF_e&lt;F_{d+e-1}\]
and
\[F_{d+e-2}&lt;F_dF_e=F_aF_bF_c\leq F_{a+b+c-3},\]
which imply that $a+b+c=d+e+2$. Using Lemma \ref{l-1} once more
and noting the identity $a+b+c-3=d+e-1$, we obtain
\begin{eqnarray*}
F_{d+e-4}&amp;\leq&amp;F_{d-1}F_{e-1}\\
&amp;=&amp;F_{a-1}F_{b-1}F_{c-1}-F_{a-2}F_{b-2}F_{c-2}\\
&amp;&lt;&amp;F_{a-1}F_{b-1}F_{c-1}\\
&amp;\leq&amp;F_{a+b+c-6}.
\end{eqnarray*}
Thus $d+e+2&lt;a+b+c$, which is impossible.
\end{proof}
%-----------------------------------------------------
\begin{theorem}\label{t-4}
Let $(a,b,c;d,e,f)$ be a nontrivial solution of the equation
$F_aF_bF_c=F_dF_eF_f$, then $a,b,c$ are equal to $d,e,f$, in some
order.
\end{theorem}
\begin{proof}
Without loss of generality, we may assume that $a\leq d$, $3\leq
a\leq b\leq c$ and $3\leq d\leq e\leq f$. If $a=d$, the result
follows immediately by Theorem \ref{t-2}. Now assume that $a&lt;d$.
Using Lemma \ref{l-1}, we have
\[F_{a+b+c-4}&lt;F_aF_bF_c=F_dF_eF_f\leq F_{d+e+f-3}\]
and
\[F_{d+e+f-4}&lt;F_dF_eF_f=F_aF_bF_c\leq F_{a+b+c-3}.\]
Thus $a+b+c=d+e+f$, and so by Lemma \ref{l-1} we obtain
\begin{eqnarray*}
F_{a-1}F_{b-1}F_{c-1}-F_{a-2}F_{b-2}F_{c-2}&amp;=&amp;F_{d-1}F_{e-1}F_{f-1}-F_{d-2}F_{e-2}F_{f-2}\\
2F_{a-2}F_{b-2}F_{c-2}-F_{a-3}F_{b-3}F_{c-3}&amp;=&amp;2F_{d-2}F_{e-2}F_{f-2}-F_{d-3}F_{e-3}F_{f-3}\\
&amp;\vdots&amp;
\end{eqnarray*}
Hence for each $i\geq 1$
\begin{eqnarray*}
F_{i+1}F_{a-i}F_{b-i}F_{c-i}-F_iF_{a-i-1}F_{b-i-1}F_{c-i-1}&amp;=&amp;F_{i+1}F_{d-i}F_{e-i}F_{f-i}-F_iF_{d-i-1}F_{e-i-1}F_{f-i-1}.
\end{eqnarray*}
By replacing $i$ by $a$ in the above equality, we obtain
\[0\geq -F_aF_{b-a-1}F_{c-a-1}=F_{a+1}F_{d-a}F_{e-a}F_{f-a}-F_aF_{d-a-1}F_{e-a-1}F_{f-a-1}\geq 0.\]
Then
\[F_{a+1}F_{d-a}F_{e-a}F_{f-a}-F_aF_{d-a-1}F_{e-a-1}F_{f-a-1}=0,\]
which is impossible, since otherwise we must have
\[F_{d-a}F_{e-a}F_{f-a}=F_{d-a-1}F_{e-a-1}F_{f-a-1}=0,\]
which implies that $d=a$.
\end{proof}

The following corollary is an immediate consequence of the above
theorem.
%-----------------------------------------------------
\begin{corollary}\label{c-4}
Let $x=\frac{F_a}{F_b}, y=\frac{F_c}{F_d}$ be in $\mathcal{P}$.
Then $xy\in\mathcal{P}$ if and only if one of the following occurs

i) $x=1$;

ii) $y=1$;

iii) $x=y=2$;

iv) $F_a=F_d$, or

v) $F_b=F_c$.
\end{corollary}

Now we turn to the equation $F_{a_1}F_{a_2}\cdots F_{a_n}=F_b$.
The special case when $a_i$'s are equal follows easily from the
following theorem. We are not aware of its proof so we prove it
here (see \cite{sv}).
%-----------------------------------------------------
\begin{theorem}\label{t-5}
Let $p$ be a prime and let $m$ and $n$ be natural numbers such
that $p\nmid m$ and $p^\alpha\|F_n$, for $\alpha&gt;0$. Then

i) $p^{\alpha+1}\|F_{nmp}$, if $(p,\alpha)\neq(2,1)$;

ii) $p^{\alpha+2}\|F_{nmp}$, if $(p,\alpha)=(2,1)$.
\end{theorem}
\begin{proof}
By the assumption and Lemma \ref{l-2},
\[\frac{F_{nm}}{F_n}\stackrel{p}{\equiv}mF_{n-1}^{m-1}.\]
Thus if $p\nmid m$ then $p^{\alpha}\|F_{nm}$ and hence it is
enough to show that $p^{\alpha+1}\|F_{np}$. By repeated
applications of (1), we have
\begin{eqnarray*}
\frac{F_{pn}}{F_n}&amp;=&amp;F_{n-1}\frac{F_{(p-1)n}}{F_n}+F_{(p-1)n+1}\\
&amp;=&amp;F_{n-1}\left(F_{n-1}\frac{F_{(p-2)n}}{F_n}+F_{(p-2)n+1}\right)+F_{(p-1)n+1}\\
&amp;\vdots&amp;\\
&amp;=&amp;F_{n-1}^{p-1}+F_{n-1}^{p-2}F_{n+1}+F_{n-1}^{p-3}F_{2n+1}+\cdots+F_{n-1}F_{(p-2)n+1}+F_{(p-1)n+1}.
\end{eqnarray*}
Now, for each $k\in\mathbb{N}$
\begin{eqnarray*}
F_{kn+1}&amp;=&amp;F_nF_{(k-1)n}+F_{n+1}F_{(k-1)n+1}\\
&amp;\stackrel{p^{2\alpha}}{\equiv}&amp;F_{n+1}F_{(k-1)n+1}\\
&amp;\vdots&amp;\\
&amp;\stackrel{p^{2\alpha}}{\equiv}&amp;F_{n+1}^k\\
&amp;\stackrel{p^{2\alpha}}{\equiv}&amp;(F_n+F_{n-1})^k\\
&amp;\stackrel{p^{2\alpha}}{\equiv}&amp;kF_nF_{n-1}^{k-1}+F_{n-1}^k.
\end{eqnarray*}
Hence
\begin{eqnarray*}
\frac{F_{pn}}{F_n}&amp;\stackrel{p^{2\alpha}}{\equiv}&amp;F_{n-1}^{p-1}+F_{n-1}^{p-2}F_{n+1}+\cdots+F_{n-1}F_{(p-2)n+1}+F_{(p-1)n+1}\\
&amp;\stackrel{p^{2\alpha}}{\equiv}&amp;F_{n-1}^{p-1}+F_{n-1}^{p-2}(F_n+F_{n-1})+\cdots+F_{n-1}\left((p-2)F_nF_{n-1}^{p-3}+F_{n-1}^{p-1}\right)\\
&amp;&amp;+\left((p-1)F_nF_{n-1}^{p-2}+F_{n-1}^{p-1}\right)\\
&amp;\stackrel{p^{2\alpha}}{\equiv}&amp;\frac{p(p-1)}{2}F_nF_{n-1}^{p-2}+pF_{n-1}^{p-1},
\end{eqnarray*}
which implies that $p^{\alpha+1}\|F_{np}$ whenever
$(p,\alpha)\neq (2,1)$. This proves (i).

Now, if $(p,\alpha)=(2,1)$ then $F_n$ is even, $3|n$ and
$\frac{n}{3}$ is odd. On the other hand, $8\|F_6$ and by the proof
of part (i), $8\|F_{2n}$ which completes the proof of part (ii).
\end{proof}
%-----------------------------------------------------
\begin{theorem}\label{t-6}
For all $k&gt;1$, the equation $F_n=F_m^k$ has only the solutions
$F_m=F_n=1$, or $k=3$, $m=3$ and $n=6$.
\end{theorem}
\begin{proof}
Let $k&gt;1$, $n\geq m\geq 3$ and $F_n=F_m^k$. As $F_m|F_n$, we have
$m|n$ and so $n=dm$, for some $d\in\mathbb{N}$. Also, by Lemma
\ref{l-2}, $F_m|d$. Now, if $p$ is a prime divisor of $F_m$ such
that $p^a\|F_m$, where $(p,a)\neq (2,1)$, then $p$ is also a
divisor of $d$ and by Theorem \ref{t-5}, $p^{a+b}\|F_n$, where
$p^b\|d$. On the other hand, $p^{ka}\|F_n$ and so $a+b=ka$, i.e.,
$b=(k-1)a$. Now, we have
\[k-1\geq d=p^bd'\geq p^b=p^{(k-1)a}\geq p^{k-1}\geq k,\]
which is impossible and hence $F_m=2$. If $p&gt;3$ and $p$ divides
$n$, then $F_p|2^k$, which is also impossible. Hence $n=2^s3^t$
and as $F_4,F_9\nmid 2^k$, we must have $n=6$.
\end{proof}

R. D. Carmichael \cite{rdc} showed that if $n&gt;2$ and $n\neq6,12$
then $F_n$ has a prime divisor $p$, which does not divide the
Fibonacci numbers $F_m$, for all $1\leq m&lt;n$. Applying this
result one can obtain the general solutions of the equation
$F_{a_1}\cdots F_{a_m}=F_b$ and more generally the solutions of
the equation $F_{a_1}\cdots F_{a_m}=F_{b_1}\cdots F_{b_n}$. For
some applications of this beautiful theorem, see \cite{ybflmmss}.

We say a solution of the equation $F_{a_1}\cdots
F_{a_m}=F_{b_1}\cdots F_{b_n}$ is nontrivial, whenever
$a_i,b_j\geq 3$ and $a_i\neq b_j$, for all $i=1,\ldots,m$ and
$j=1,\ldots,n$.
%-----------------------------------------------------
\begin{theorem}\label{t-7}
i) The only nontrivial solutions of the equation
$F_{a_1}F_{a_2}\cdots F_{a_n}=F_b$ with $n&gt;1$ and
$a_1\leq\cdots\leq a_n$ are
\[(3,3,3;6)\ ,\ (3,4,4,6;12)\ ,\ (3,3,3,3,4,4;12)\]
ii) The only nontrivial solutions of the equation $F_{a_1}\cdots
F_{a_m}=F_{b_1}\cdots F_{b_n}$ are
\begin{eqnarray*}
(3,\ldots,3;6,\ldots,6)&amp;,&amp;m=3n\\
(\overbrace{3,\ldots,3}^a,\overbrace{6,\ldots,6}^b,4,\ldots,4;12,\ldots,12)&amp;,&amp;a+3b=4n\\
(\overbrace{3,\ldots,3}^a,4,\ldots,4;\overbrace{6,\ldots,6}^b,12,\ldots,12)&amp;,&amp;a=3b+4n\\
(\overbrace{6,\ldots,6}^a,4,\ldots,4;\overbrace{3,\ldots,3}^b,12,\ldots,12)&amp;,&amp;3a=b+4n
\end{eqnarray*}
\end{theorem}
\begin{proof}
The proofs of both parts follow easily from Carmichael's theorem.
\end{proof}

The following theorem is another consequence of Carmichael's
theorem.
%-----------------------------------------------------
\begin{theorem}\label{t-8}
Suppose $p_1,p_2,\ldots,p_n$ are arbitrary distinct prime numbers.
Then there are only finitely many $n$-tuples $(a_1,\ldots,a_n)$ of
nonnegative integers such that $p_1^{a_1}\cdots
p_n^{a_n}\in\mathcal{P}$.
\end{theorem}
\begin{proof}
Assume $\{(a_{i_1},\ldots,a_{i_n})\}_{i=1}^{\infty}$ is an
infinite sequence of distinct $n$-tuples such that for each $i$
the number $k_i=p_1^{i_1}\cdots p_n^{i_n}$ belongs to
$\mathcal{P}$. Then there exist some natural numbers $m_i$ and
$n_i$ such that $F_{n_i}=k_iF_{m_i}$. Without loss of generality,
we may assume that $n_i\neq m_i$ and $n_i's$ are all distinct and
greater than $12$. Since there are infinitely many $n$-tuples, we
may ignore the prime factors of the equations
$F_{n_i}=k_iF_{m_i}$ so that we obtain an equation of type as in
Theorem \ref{t-7}, which contradicts Theorem \ref{t-7}.
\end{proof}

Although we were able to obtain the general solutions of the
equation $F_{a_1}\cdots F_{a_n}$ $=F_b$ using Carmichael's
theorem, an elementary proof may nevertheless be of interest.
%-----------------------------------------------------

\section{Acknowledgment}

The author would like to thank the referee for some useful
suggestions and corrections.

\begin{thebibliography}{0}

\bibitem{ybflmmss}Y. Bugeaud, F. Luca, M. Mignotte and S. Siksek,
On Fibonacci numbers with few prime divisors,
{\it Proc. Japan Acad. Ser. A} \textbf{81} (2005), 17--20.

\bibitem{rdc}R. D. Carmichael, On the numerical factors of the arithmetic 
forms $\alpha^n\pm\beta^n$, {\it Ann. Math.} (2) \textbf{15} (1913/14),
30--48.

\bibitem{sv}S. Vajda, {\it Fibonacci \&amp; Lucas Numbers, and the Golden Section},
Ellis Horwood Limited, Chichester, England, 1989.
\end{thebibliography}


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\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B39; Secondary 11B50, 11D99.

\noindent \emph{Keywords: } Fibonacci numbers.


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

\bigskip
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\vspace*{+.1in}
\noindent
Received March 28 2007;
revised version received  May 22 2007.
Published in {\it Journal of Integer Sequences}, June 4 2007.

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