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\begin{center}
\vskip 1cm{\LARGE\bf 
Exact Divisibility by Powers of the  \\
\vskip .1in
Fibonacci and Lucas Numbers 
}
\vskip 1cm
\large
Prapanpong Pongsriiam\\
Department of Mathematics, Faculty of Science\\
Silpakorn University\\
Ratchamankanai Rd, Nakhon Pathom\\
Thailand, 73000\\
\href{mailto:prapanpong@gmail.com}{\tt prapanpong@gmail.com}
\end{center}

\vskip .2 in

\begin{abstract}
We give new results on exact divisibility by powers of the Fibonacci and Lucas numbers. For example, we prove that if $F_n^{k}$ exactly divides $m$ and $n$ is not congruent to $3$ modulo $6$, then $F_n^{k+1}$ exactly divides $F_{nm}$. We also provide some examples and open questions.
\end{abstract}

\section{Introduction}
Let $(F_n)_{n\geq 0}$ be the Fibonacci sequence defined by $F_0 = 0$, $F_1 = 1$, and $F_n = F_{n-1}+F_{n-2}$ for $n\geq 2$, and let $(L_n)_{n\geq 0}$ be the Lucas sequence given by $L_0=2$, $L_1=1$ with the same recursive pattern as the Fibonacci sequence. 

The divisibility property of Fibonacci and Lucas numbers has always been a popular area of research. In particular, the divisibility by powers of the Fibonacci numbers became a popular topic when Matijasevi\u{c} \cite{mati1, mati2, mati3} proved in 1970 that
\begin{equation}\label{Matequation}
\text{$F_n^2\mid F_{nm}$ if and only if $F_n\mid m$}
\end{equation}
which led to the solution to Hilbert's 10th Problem. Since \eqref{Matequation} appeared, there have been several results \cite{BR, Ho, Ma, OP, TW1, TW} on divisibility by powers of the Fibonacci numbers in the literature. However, exact divisibility by powers of the Fibonacci and Lucas numbers seems to be new. (Recall that for integers $a$, $d\geq 2$ and $k\geq 1$, we say that $d^k$ exactly divides $a$ and write $d^k\parallel a$ if $d^k\mid a$ and $d^{k+1}\nmid a$). Now let us consider the following divisibility:
\begin{equation}\label{lostthmeq1}
F_n^{k+1}\mid F_{nF_n^k}.
\end{equation}
Marques \cite[p.\ 241]{Ma} mentioned that to the best of his knowledge,
\eqref{lostthmeq1} was first proved by Benjamin and Rouse \cite{BR},
using a combinatorial approach, and a second proof of
\eqref{lostthmeq1} is due to Seibert and Trojovsky \cite{ST} by using
mathematical induction together with an identity for
$\frac{F_{nm}}{F_m}$. Marques \cite{Ma} himself also gave another proof
of \eqref{lostthmeq1} by applying Lengyel's theorem \cite{Le} on the
$p$-adic order of Fibonacci and Lucas numbers. However, there was an
older result which implied \eqref{lostthmeq1}; it was proved in 1977 by
Hoggatt and Bicknell-Johnson \cite{Ho}, as follows:

\begin{theorem}\label{hoggatthmmontly}
\cite{BR, Ho} Let $k$, $m$, and $n$ be positive integers. If $F_n^{k}\mid m$, then $F_n^{k+1}\mid F_{nm}$.
\end{theorem} 
Note that, from Theorem \ref{hoggatthmmontly}, we can easily obtain
\eqref{lostthmeq1} by substituting $m = F_n^k$, and obtain one direction
of \eqref{Matequation} by substituting $k=1$. Now it is natural to ask
if $k+1$ in Theorem \ref{hoggatthmmontly} is the largest exponent of
$F_n$ such that $F_n^{k+1}$ divides $F_{nm}$. Our purpose is to give
another proof of Theorem \ref{hoggatthmmontly}, and generalize the
result to include the divisibility and exact divisibility by powers of
the Fibonacci and Lucas numbers. Our results are as follows:

\begin{theorem}\label{fibothm3}
Let $k, m, n$ be positive integers and $n\geq 3$. Then
\begin{itemize}
\item[(i)] if $F_{n}^{k} \parallel m$ and $n\not\equiv 3\pmod 6$, then $F_n^{k+1}\parallel F_{nm}$;
\item[(ii)] if $F_n^{k}\parallel m$, $n\equiv 3\pmod 6$, and $\frac{F_n^{k+1}}{2}\nmid m$, then $F_n^{k+1}\parallel F_{nm}$;
\item[(iii)] if $F_n^{k}\parallel m$, $n\equiv 3\pmod 6$, and $\frac{F_n^{k+1}}{2}\mid m$, then $F_n^{k+2}\parallel F_{nm}$.
\end{itemize}
\end{theorem}

\begin{theorem}\label{thm1.4fiboandlucas}
Let $k, m, n$ be positive integers and $m$ is odd. Then
\begin{itemize}
\item[(i)] if $L_n^{k}\mid m$, then $L_n^{k+1}\mid L_{nm}$;
\item[(ii)] if $n\geq 2$ and $L_n^{k}\parallel m$, then $L_n^{k+1}\parallel L_{nm}$.
\end{itemize}
\end{theorem}

\begin{theorem}\label{addtheorem1.4}
Let $k, m$, and $n$ be positive integers, $m$ is even, and $n\geq 2$. 
Then the following statements hold.
\begin{itemize}
\item[(i)] If $L_n^k\mid m$, then $L_n^{k+1}\mid F_{nm}$;
\item[(ii)] If $L_n^k\parallel m$ and $n\not\equiv 0\pmod 3$, then $L_n^{k+1}\parallel F_{nm}$;
\item[(iii)] If $L_n^k\parallel m$, $n\equiv 0\pmod 6$, and $\frac{L_n^{k+1}}{2}\nmid m$, then $L_n^{k+1}\parallel F_{nm}$;
\item[(iv)] If $L_n^k\parallel m$, $n\equiv 0\pmod 6$, and $\frac{L_n^{k+1}}{2}\mid m$, then $L_n^{k+2}\mid F_{nm}$;
\item[(v)] If $L_n^k\parallel m$, $n\equiv 3\pmod 6$, and $\frac{L_n^{k+1}}{4}\nmid m$, then $L_n^{k+1}\parallel F_{nm}$;
\item[(vi)] If $L_n^k\parallel m$, $n\equiv 3\pmod 6$, and $\frac{L_n^{k+1}}{4}\mid m$, then $L_n^{k+2}\parallel 4F_{nm}$.
\end{itemize}
\end{theorem}

To prove the above theorems, we will need a number of lemmas given in the next section.

\section{Preliminaries and lemmas}
In this section, we will give some auxiliary results that will be used in this article. Let $m, n$, and $r$ be positive integers. Then the following
statements hold. 
\begin{equation}\label{pre1}
\gcd(F_m,F_n)=F_{\gcd(m,n)}.
\end{equation}

\begin{equation}\label{pre2}
F_{nm}=\sum_{j=1}^{m}{m\choose j}F_n^jF_{n-1}^{m-j}F_j .
\end{equation}

\begin{equation}\label{pre3}
\text{If $m\geq3$, then $F_m\mid F_n$ if and only if $m\mid n$} .
\end{equation}

\begin{equation}\label{pre7.5}
L_n = F_{n-1}+F_{n+1} .
\end{equation}

\begin{equation}\label{pre8}
\gcd(L_n,F_n) = \begin{cases}
1, &amp;\text{if $3\nmid n$;}\\
2, &amp;\text{if $3\mid n$} .
\end{cases}
\end{equation}

For the reader's convenience, let us give references for the above
statements. Properties \eqref{pre1} and \eqref{pre3} can be found in
\cite[pp.\ 196--198]{Ko}. Property~\eqref{pre2} can be found in several
articles such as \cite{BR, Ha, Ho}. Property~\eqref{pre7.5} appears in
\cite[p.\ 80]{Ko}, and \eqref{pre8} can be proved by using
\eqref{pre7.5}, \eqref{pre1}, and \eqref{pre3}. We refer the reader to
\cite{BQ, GK, KM, Ko, R, Va} for more details and additional
references.

Next we give the following results, similar to \eqref{pre2}. 

\begin{equation}\label{pre7}
L_{nm}\cdot 2^{m-1} = \sum_{j=0}^{\left\lfloor \frac{m}{2}\right\rfloor} {m\choose m-2j} L_n^{m-2j} F_n^{2j} 5^j.
\end{equation}

\begin{equation}\label{pre11}
F_{nm}\cdot 2^{m-1} = \sum_{j=0}^{\left\lfloor \frac{m-1}{2}\right\rfloor} {m\choose m-2j-1} L_n^{m-(2j+1)} F_n^{2j+1} 5^j.
\end{equation}

These are probably less well known, so we will give a proof for completeness.

\begin{proof}
Recall Binet's formula that $F_n = \frac{\alpha^n-\beta^n}{\alpha-\beta}$ and $L_n = \alpha^n+\beta^n$, where $\alpha = \frac{1+\sqrt 5}{2}$ and $\beta = \frac{1-\sqrt 5}{2}$. Then $\alpha^n = \frac{1}{2}\left(L_n+(\alpha-\beta)F_n\right) = \frac{1}{2}\left(L_n+\sqrt 5 F_n\right)$, and $\beta^n = \frac{1}{2}\left(L_n-(\alpha-\beta)F_n\right) = \frac{1}{2}\left(L_n-\sqrt 5F_n\right)$. Then 
\begin{align*}
L_{nm} &amp;= \alpha^{nm}+\beta^{nm} = \frac{1}{2^m}\left(L_n+\sqrt 5F_n\right)^m+\frac{1}{2^m}\left(L_n-\sqrt 5F_n\right)^m\\
&amp;=\frac{1}{2^m}\sum_{k=0}^m{m\choose m-k} L_n^{m-k}(\sqrt 5F_n)^k\left(1+(-1)^k\right).
\end{align*}
When $k$ is odd, $1+(-1)^k = 0$, so we replace $k$ by $2j$ and obtain
$$
L_{nm} = \frac{1}{2^{m-1}}\sum_{j=0}^{\left\lfloor \frac{m}{2}\right\rfloor}{m\choose m-2j}L_n^{m-2j}F_n^{2j}5^j,
$$
which gives \eqref{pre7}. For \eqref{pre11}, we write
\begin{align*}
\sqrt 5F_{nm} = (\alpha-\beta)F_{nm} &amp;= \alpha^{nm}-\beta^{nm} = \frac{1}{2^m}(L_n+\sqrt 5F_n)^m-\frac{1}{2^m}(L_n-\sqrt 5F_n)^m\\
&amp;=\frac{1}{2^m}\sum_{k=0}^m{m\choose m-k} L_n^{m-k}(\sqrt 5F_n)^k\left(1-(-1)^k\right).
\end{align*}
As in the previous case, we replace $k$ by $2j+1$ to obtain \eqref{pre11}. 
\end{proof}

The next lemma is about the $p$-adic orders of Fibonacci and Lucas
numbers given by Lengyel \cite{Le}. Recall that the order of appearance
of $m$ and the period modulo $m$ of the Fibonacci sequence,
respectively, is  the smallest positive integer $k$ such that $m\mid
F_k$ and the smallest positive integer $k$ such that $F_{n+k}\equiv
F_n\pmod{m}$ for all $n\geq 1$.

\begin{lemma}\label{lemmaofLengyel}
\cite{Le} For each $m\in \mathbb N$, let $\nu_p(m)$ be the $p$-adic order of $m$, $z(m)$ the order of appearance of $m$, and $s(m)$ the period modulo $m$ of the Fibonacci sequence. For $n\geq 1$, we have
\begin{align*}
\nu_2(F_n) &amp;= 
\begin{cases}
0, &amp;\text{if $n\equiv 1, 2\pmod 3$};\\
1, &amp;\text{if $n\equiv 3\pmod 6$};\\
\nu_2(n)+2, &amp;\text{if $n\equiv 0\pmod 6$},
\end{cases}\\
\nu_2(L_n) &amp;= 
\begin{cases}
0, &amp;\text{if $n\equiv 1, 2\pmod 3$};\\
2, &amp;\text{if $n\equiv 3\pmod 6$};\\
1, &amp;\text{if $n\equiv 0\pmod 6$},
\end{cases}
\end{align*}
$\nu_5(F_n) =\nu_5(n)$, $\nu_5(L_n) = 0$, and if $p$ is a prime, $p\neq 2$ and $p\neq 5$, then
$$
\nu_p(F_n) = 
\begin{cases}
\nu_p(n)+\nu_p(F_{z(p)}), &amp;\text{if $n\equiv 0\pmod{z(p)}$};\\
0, &amp;\text{otherwise}.
\end{cases}
$$
$$
\nu_p(L_n) = 
\begin{cases}
\nu_p(n)+\nu_p(F_{z(p)}), &amp;\text{if $s(p)\neq 4z(p)$ and $n\equiv \frac{z(p)}{2}\pmod{z(p)}$};\\
0, &amp;\text{otherwise}.
\end{cases}
$$
\end{lemma}

The next result given by Onphaeng and Pongsriiam \cite{OP} is an important tool for obtaining the main results of this article. 

\begin{lemma}\label{OPnewlemma2}
\cite{OP} Let $k, \ell, m, s$ be positive integers and $s^k\mid m$. Then $s^{k+\ell}\mid {m\choose j}s^j $ for all $1\leq j\leq m$ satisfying $2^{j-\ell+1} &gt; j$. In particular, $s^{k+1}\mid {m \choose j}s^j$ for all $1 \leq j \leq m$, and $s^{k+2}\mid {m \choose j}s^j$ for all $3 \leq j \leq m$. 
\end{lemma}

\section{Proof of the main results}
In this section, we will give the proof of Theorems
\ref{hoggatthmmontly}, \ref{fibothm3}, \ref{thm1.4fiboandlucas}, and
\ref{addtheorem1.4}.

\begin{proof}[Another proof of Theorem \ref{hoggatthmmontly}.]
Assume that $F_n^k\mid m$. Then by Lemma \ref{OPnewlemma2}, $F_n^{k+1}\mid {m\choose j}F_n^j$ for all $1\leq j\leq m$. Therefore $F_n^{k+1}\mid F_{nm}$ by \eqref{pre2}. 
\end{proof}

\begin{proof}[Proof of Theorem \ref{fibothm3}.]
Assume that $F_n^{k}\parallel m$. By Theorem \ref{hoggatthmmontly}, $F_n^{k+1}\mid F_{nm}$. So to show that $F_n^{k+1}\parallel F_{nm}$, it is enough to show that $F_n^{k+2}\nmid F_{nm}$. First we write 
\begin{equation}\label{thm3eq10}
m = F_n^{k}c\quad\text{where $c\geq 1$ and $F_n\nmid c$}.
\end{equation}
By Lemma~\ref{OPnewlemma2}, $F_n^{k+2}\mid {m\choose j}F_n^j$ for all $3\leq j\leq m$. Then we obtain by \eqref{pre2} that 
\begin{equation}\label{thm3eq11}
F_{nm}\equiv mF_nF_{n-1}^{m-1}+\frac{m(m-1)}{2}F_n^2F_{n-1}^{m-2}\pmod{F_n^{k+2}}.
\end{equation} 

\medskip

Proof of (i):

\medskip

\noindent Case 1 of (i): $n \equiv 1, 2, 4, 5\pmod 6$. Then by Lemma \ref{lemmaofLengyel}, $F_n$ is odd. Since $F_n^{k}\mid 2\left(\frac{m(m-1)}{2}\right)$ and $F_n$ is odd, we obtain that $F_n^{k}\mid \frac{m(m-1)}{2}$. Therefore
\begin{equation}\label{thm1.4case2eqstar}
F_n^{k+2}\left|\frac{m(m-1)}{2}F_n^2F_{n-1}^{m-2}\right..
\end{equation}
By \eqref{thm3eq11} and \eqref{thm1.4case2eqstar}, we obtain $F_{nm} \equiv mF_nF_{n-1}^{m-1}\pmod{F_n^{k+2}}$. Since $\gcd(F_n,F_{n-1})=1$ with respect to \eqref{pre1}, we see that 
\begin{equation}\label{thm1.4case2eqstar3.5.5}
F_n^{k+2}\mid F_{nm} \;\Leftrightarrow\; F_n^{k+2}\mid mF_nF_{n-1}^{m-1} \;\Leftrightarrow\; F_n^{k+1}\mid m.
\end{equation}
Since $F_n^{k}\parallel m$, $F_n^{k+1}\nmid m$. So we obtain by \eqref{thm1.4case2eqstar3.5.5} that $F_n^{k+2}\nmid F_{nm}$.

\medskip

\noindent Case 2 of (i): $n\equiv 0\pmod 6$. Then by Lemma \ref{lemmaofLengyel} 
we have $F_n\equiv 0\pmod 4$. Let 
$$
d = F_{n-1}+(m-1)\frac{F_n}{2}.
$$
 Since $F_n$ is even, 
\begin{equation}\label{thm3eq12}
\text{$d$ is an integer and $\left(\frac{F_n}{2},d\right) = 1$}
\end{equation}
By \eqref{thm3eq11} and \eqref{thm3eq10}, we obtain
$$
F_{nm} \equiv mF_nF_{n-1}^{m-2}\left(F_{n-1}+\frac{m-1}{2}F_n\right) \equiv cF_n^{k+1}F_{n-1}^{m-2}d\pmod{F_n^{k+2}}.
$$
Suppose, in order to get a contradiction, that $F_n^{k+2}\mid F_{nm}$. Then $F_n^{k+2} \mid cF_n^{k+1}F_{n-1}^{m-2}d$ which implies that $F_n\mid cd$. By $F_n\equiv 0\pmod 4$, $F_n\mid cd$, and \eqref{thm3eq12} we obtain that $d$ is odd, $c$ is even, and $\frac{F_n}{2}\mid \frac{c}{2}d$. This and \eqref{thm3eq12} imply that $F_n\mid c$, which contradicts \eqref{thm3eq10}. Hence $F_n^{k+2}\nmid F_{nm}$.

\medskip

Proof of (ii):

\medskip

Assume that $n\equiv 3\pmod 6$ and $\frac{F_n^{k+1}}{2}\nmid m$. Then by Lemma \ref{lemmaofLengyel} and \eqref{thm3eq10}, respectively, we obtain $\nu_2(F_n) = 1$ and $\frac{F_n}{2}\nmid c$. We can follow the argument in Case 2 of (i) to see that $F_{nm} \equiv cF_n^{k+1}F_{n-1}^{m-2}d\pmod{F_n^{k+2}}$, and that $F_n^{k+2}\mid F_{nm}$ if and only if $F_n\mid cd$. But if $F_n\mid cd$, then $\frac{F_n}{2}\mid cd$ which implies by \eqref{thm3eq12} that $\frac{F_n}{2}\mid c$, a contradiction. So $F_n\nmid cd$ and therefore $F_n^{k+2}\nmid F_{nm}$. Hence $F_n^{k+1}\parallel F_{nm}$. 

\medskip

Proof of (iii):

\medskip

Assume that $n\equiv 3\pmod 6$ and $\frac{F_n^{k+1}}{2}\mid m$. In this case, we will show that $F_n^{k+2}\parallel F_{nm}$. We can still follow the argument in Case 2 of (i) and the proof of (ii) to obtain $F_{nm} \equiv cF_n^{k+1}F_{n-1}^{m-2}d\pmod{F_n^{k+2}}$ and $F_n^{k+2}\mid F_{nm}$ if and only if $F_n\mid cd$. Since $n\equiv 3\pmod 6$, $F_n$ is even and $\frac{F_n}{2}$ is odd. Therefore $F_{n-1}$ and $m-1$ are odd and $d$ is even. Since $m = F_n^kc$ and $\frac{F_n^{k+1}}{2}\mid m$, we obtain $F_n\mid 2c$. Since $2c\mid cd$, we obtain $F_n\mid cd$ and therefore $F_n^{k+2}\mid F_{nm}$. It remains to show that $F_n^{k+3}\nmid F_{nm}$. Suppose for a contradiction that $F_n^{k+3}\mid F_{nm}$. So for every prime $p$ dividing $F_n$, $\nu_p(F_n^{k+3})\leq \nu_p(F_{nm})$. By Lemma \ref{lemmaofLengyel}, we obtain the following inequalities:
\begin{align*}
2+\nu_2(F_n^{k+1}) &amp;= \nu_2(F_n^{k+3}) \leq \nu_2(F_{nm}) = \nu_2(nm)+2 = \nu_2(m)+2,\\
2\nu_5(n)+\nu_5(F_n^{k+1}) &amp;= \nu_5(F_n^{k+3}) \leq \nu_5(F_{nm}) = \nu_5(n)+\nu_5(m),
\end{align*}
and for every prime $p\notin \{2, 5\}$ and $p\mid F_n$, we have
\begin{align*}
2\nu_p(F_n)+\nu_p(F_n^{k+1}) &amp;= \nu_p(F_n^{k+3}) \leq \nu_p(F_{nm}) \\
&amp;= \nu_p(nm)+\nu_p(F_{z(p)}) = \nu_p(m)+\nu_p(n)+\nu_p(F_{z(p)})\\
&amp;= \nu_p(m)+\nu_p(F_n).
\end{align*}
From the above inequalities, we obtain that $\nu_p(m)\geq \nu_p(F_n^{k+1})$ for every prime $p$ dividing $F_n$. Therefore $F_n^{k+1}\mid m$ which contradicts $F_n^k\parallel m$. This gives (iii). Hence the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.4fiboandlucas}.]   

Proof of (i):

\medskip

Assume that $L_n^{k}\mid m$. Since $m$ is odd and $L_n\mid m$, $L_n$ is odd too. By Lemma \ref{OPnewlemma2}, $L_n^{k+1}\mid {m\choose j}L_n^j$ for each $1\leq j\leq m$. Then $L_n^{k+1}\mid {m\choose m-2j}L_n^{m-2j}$ for all $0\leq j\leq \frac{m-1}{2}$. Then by \eqref{pre7}, we see that $L_n^{k+1}\mid L_{nm}\cdot 2^{m-1}$.  Hence $L_n^{k+1}\mid L_{nm}$. This proves (i).

\medskip

Proof of (ii):

\medskip

Next assume that $n\geq 2$ and $L_n^{k}\parallel m$. Then $m\geq 3$ and by Lemma \ref{OPnewlemma2}, $L_n^{k+2}\mid {m\choose j}L_n^j$ for all $3\leq j\leq m$. This implies that $L_n^{k+2}\mid {m\choose m-2j}L_n^{m-2j}$ for all $0\leq j\leq \frac{m-3}{2}$. Then by \eqref{pre7}, we see that 
\begin{equation}\label{pfofth1.4eq1}
L_{nm}\cdot 2^{m-1}\equiv mL_nF_n^{m-1}5^{\frac{m-1}{2}}\pmod{L_n^{k+2}}
\end{equation}
Since $L_n$ is odd, we obtain
\begin{align*}
L_n^{k+2}\mid L_{nm} &amp;\Leftrightarrow L_n^{k+2}\mid L_{nm}\cdot 2^{m-1}\\
&amp;\Leftrightarrow L_n^{k+1}\mid mF_n^{m-1}5^{\frac{m-1}{2}},\quad\text{by \eqref{pfofth1.4eq1}}\\
&amp;\Leftrightarrow L_n^{k+1}\mid m,\quad\text{by \eqref{pre8} and Lemma \ref{lemmaofLengyel}}.
\end{align*}
Now $L_n^{k}\parallel m$, so $L_n^{k+1}\nmid m$ and thus $L_n^{k+2}\nmid L_{nm}$. Hence $L_n^{k+1}\parallel L_{nm}$ as desired.
\end{proof}

\begin{proof}[Proof of Theorem \ref{addtheorem1.4}.]

Assume that $L_n^k\mid m$. Since $n\geq 2$, we have $m\geq 4$. Then by Lemma \ref{OPnewlemma2}, $L_n^{k+1}\mid {m\choose j}L_n^j$ for every $1\leq j\leq m$. Therefore $L_n^{k+1}\mid{m\choose m-2j-1}L_n^{m-(2j+1)}$ for every $0\leq j\leq \frac{m-2}{2}$. Then by \eqref{pre11}, we obtain $F_{nm}\cdot 2^{m-1}\equiv 0\pmod{L_n^{k+1}}$. From this point on, $p$-adic orders will be used and we sometimes apply Lemma \ref{lemmaofLengyel} without referring to it explicitly.

\medskip

Proof of (i):

\medskip

\noindent Case 1 of (i): $n\equiv 1, 2\pmod{3}$. Then $\gcd(2, L_n) = 1$. Since $L_n^{k+1}\mid F_{nm}\cdot 2^{m-1}$, $L_n^{k+1}\mid F_{nm}$. 

\medskip

\noindent Case 2 of (i): $n\equiv 0\pmod 6$. Then $\nu_2(L_n) = 1$. Since $L_n^k\mid m$, $\nu_2(m)\geq \nu_2(L_n^k) = k$. Since $n\equiv 0\pmod 6$, $nm\equiv 0\pmod 6$ and therefore 
$$
\nu_2(F_{nm}) = \nu_2(nm)+2 = \nu_2(n)+\nu_2(m)+2\geq k+2.
$$
Since $L_n^{k+1}\mid F_{nm}\cdot 2^{m-1}$ and $\nu_2(L_n) = 1$, we obtain $\left(\frac{L_n}{2}\right)^{k+1}\mid \frac{F_{nm}}{2^{k+1}}\cdot 2^{m-1}$, which implies $\left(\frac{L_n}{2}\right)^{k+1}\mid \frac{F_{nm}}{2^{k+1}}$. Therefore $L_n^{k+1}\mid F_{nm}$.

\medskip

\noindent Case 3 of (i): $n\equiv 3\pmod 6$. Then $\nu_2(L_n) = 2$ and $\nu_2(F_n) = 1$. Since $L_n^k\mid m$, $\nu_2(m) \geq \nu_2(L_n^k) = 2k$. Therefore $nm \equiv 0\pmod 6$ and $\nu_2(F_{nm}) = \nu_2(nm)+2 = \nu_2(n)+\nu_2(m)+2\geq 2k+2$. Since $L_n^{k+1}\mid F_{nm}\cdot 2^{m-1}$, $\nu_2(L_n) = 2$, and $\nu_2(F_{nm}) \geq 2k+2$, we obtain 
$$
\left.\frac{L_n^{k+1}}{(2^2)^{k+1}}\right| \frac{F_{nm}}{2^{2k+2}}\cdot 2^{m-1},
$$
which implies that $\frac{L_n^{k+1}}{(2^2)^{k+1}}\mid \frac{F_{nm}}{2^{2k+2}}$. Therefore $L_n^{k+1}\mid F_{nm}$.

In any case, we obtain that $L_n^{k+1}\mid F_{nm}$. This proves (i). 

\medskip

Proof of (ii):

\medskip

Next assume that $L_n^{k}\parallel m$. Recall that $m\geq 4$. By Lemma \ref{OPnewlemma2}, $L_n^{k+2}\mid {m\choose j}L_n^j$ for every $3\leq j\leq m$. Therefore $L_n^{k+2}\mid {m\choose m-2j-1}L_n^{m-(2j+1)}$ for every $0\leq j\leq \frac{m-4}{2}$. Then by \eqref{pre11},
\begin{equation}\label{pfofthm1.4(ii)eq1}
F_{nm}\cdot 2^{m-1} \equiv mL_nF_n^{m-1}5^{\frac{m-2}{2}}\pmod{L_n^{k+2}}.
\end{equation}

Assume that $n\equiv 1, 2\pmod{3}$. Then by Lemma \ref{lemmaofLengyel}, \eqref{pfofthm1.4(ii)eq1}, and \eqref{pre8}, we obtain that 
\begin{align*}
L_n^{k+2}\mid F_{nm} \Leftrightarrow L_n^{k+2}\mid F_{nm}\cdot 2^{m-1} \Leftrightarrow L_n^{k+2}\mid mL_nF_n^{m-1}5^{\frac{m-2}{2}}\Leftrightarrow L_n^{k+1}\mid m.
\end{align*}
But $L_n^k\parallel m$, so $L_n^{k+1}\nmid m$ and thus $L_n^{k+2}\nmid F_{nm}$. Then $L_n^{k+1}\parallel F_{nm}$. This proves (ii).

\medskip

Proof of (iii) and (iv):

\medskip

Assume that $L_n^k\parallel m$ and $n\equiv 0\pmod 6$. In analogy with the proof of Case 2 of (i), we see that 
\begin{align}\label{pfofthm1.4(ii)eq2}
L_n^{k+2}\mid F_{nm}\cdot 2^{m-1} &amp;\Leftrightarrow \frac{L_n^{k+2}}{2^{k+2}}\mid \frac{F_{nm}}{2^{k+2}}\cdot 2^{m-1} \Leftrightarrow \frac{L_n^{k+2}}{2^{k+2}}\mid \frac{F_{nm}}{2^{k+2}}\Leftrightarrow L_n^{k+2}\mid F_{nm}.
\end{align}
Let $m = L_n^kc$ where $c\geq 1$ and $L_n\nmid c$. Then by \eqref{pfofthm1.4(ii)eq2}, \eqref{pfofthm1.4(ii)eq1}, and Lemma \ref{lemmaofLengyel}, we obtain that 
\begin{align}\label{pfofthm1.4(ii)eq3}
L_n^{k+2}\mid F_{nm} \Leftrightarrow L_n^{k+2}\mid mL_nF_n^{m-1}5^{\frac{m-2}{2}}\Leftrightarrow L_n\mid cF_n^{m-1}. 
\end{align}
Now $\gcd(L_n,F_n) = 2$ and $\nu_2(L_n) = 1$, so $\gcd\left(\frac{L_n}{2},F_n\right) = 1$, and therefore 
\begin{align}\label{pfofthm1.4(ii)eq4}
L_n\mid cF_n^{m-1} \Leftrightarrow \frac{L_n}{2}\mid c\frac{F_n^{m-1}}{2} \Leftrightarrow \frac{L_n}{2}\mid c\Leftrightarrow \frac{L_n^{k+1}}{2}\mid m.
\end{align}
By \eqref{pfofthm1.4(ii)eq3} and \eqref{pfofthm1.4(ii)eq4}, we see that $L_n^{k+2}\mid F_{nm} \Leftrightarrow \frac{L_n^{k+1}}{2}\mid m$. This proves (iii) and (iv).

\medskip

Proof of (v) and (vi):

\medskip

Assume that $L_n^k\parallel m$ and $n\equiv 3\pmod 6$. In analogy with the proof of Case 3 of (i), we obtain that 
\begin{align}\label{pfofthm1.4(ii)eq5}
L_n^{k+2}\mid F_{nm}\cdot 2^{m-1} &amp;\Leftrightarrow \frac{L_n^{k+2}}{2^{2(k+2)}}\mid \frac{F_{nm}}{2^{2k+2}}\cdot \frac{2^{m-1}}{2^2}\notag\\
&amp;\Leftrightarrow \frac{L_n^{k+2}}{2^{2(k+2)}}\mid \frac{F_{nm}}{2^{2k+2}}\notag\\
&amp;\Leftrightarrow L_n^{k+2}\mid 4F_{nm}.
\end{align}
Next, we write $m = L_n^kc$, use \eqref{pfofthm1.4(ii)eq1}, and use the similar idea in the proof of (iii) and (iv) to obtain that 
\begin{align}\label{pfofthm1.4(ii)eq6}
L_n^{k+2}\mid F_{nm}\cdot 2^{m-1} &amp;\Leftrightarrow L_n^{k+2}\mid mL_nF_n^{m-1}5^{\frac{m-2}{2}}\Leftrightarrow L_n\mid cF_n^{m-1}\Leftrightarrow \frac{L_n}{4}\mid c\frac{F_n^{m-1}}{4}\notag\\
&amp;\Leftrightarrow \frac{L_n}{4}\mid c\Leftrightarrow \frac{L_n^{k+1}}{4}\mid m. 
\end{align}
From \eqref{pfofthm1.4(ii)eq5} and \eqref{pfofthm1.4(ii)eq6}, we conclude that $L_n^{k+2}\mid 4F_{nm} \Leftrightarrow \frac{L_n^{k+1}}{4}\mid m$. This gives (v) and to prove (vi), it suffices to show that $L_n^{k+3}\nmid 4F_{nm}$. Suppose for a contradiction that $L_n^{k+3}\mid 4F_{nm}$. Then for every prime $p$ dividing $L_n$, $\nu_p(L_n^{k+3}) \leq \nu_p(4F_{nm})$. Consider the following inequalities: 
$$
4+\nu_2(L_n^{k+1}) =\nu_2(L_n^{k+3}) \leq \nu_2(4F_{nm}) = 2+\nu_2(nm)+2 = 4+\nu_2(m),
$$
where we use that $mn\equiv 0\pmod 6$, and for every prime $p\notin \{2,5\}$ and $p\mid L_n$, we have
\begin{align*}
\nu_p(L_n^{k+1})+2\nu_p(L_n) &amp;= \nu_p(L_n^{k+3}) \leq \nu_p(4F_{nm}) \leq \nu_p(nm)+\nu_p(F_{z(p)})\\
&amp; = \nu_p(m)+\nu_p(n)+\nu_p(F_{z(p)}) = \nu_p(m)+\nu_p(L_n).
\end{align*}
From the above inequalities, we obtain that $\nu_p(m) \geq \nu_p(L_n^{k+1})$ for every prime $p$ dividing $L_n$. Therefore $L_n^{k+1}\mid m$ which contradicts the fact that $L_n^k\parallel m$. Hence $L_n^{k+3}\nmid 4F_{nm}$. This proves (vi) and the proof is complete.  
\end{proof}

\section{Examples}
In this section, we give some examples to clarify the results in Theorems \ref{thm1.4fiboandlucas} and \ref{addtheorem1.4}. Then we give some open questions at the end of this article. First we will show that the assumption that $m$ is odd in Theorem \ref{thm1.4fiboandlucas} and that $m$ is even in Theorem \ref{addtheorem1.4} cannot be omitted.

\begin{example}\label{exam8add}
Let $n\equiv 0\pmod 3$, $k\geq 1$, and $m = L_n^k$. Then $m$ is even, $L_n^k\mid m$, but $\nu_2(L_n^{k+1}) \geq k+1 &gt; 1 = \nu_2(L_{nm})$. Therefore $L_n^{k+1}\nmid L_{nm}$. This shows that Theorem \ref{thm1.4fiboandlucas} does not hold if $m$ is not odd.
\end{example}

\begin{example}
Let $n\geq 2$, $n\equiv 1, 2 \pmod 3$, $k\geq 1$, and $m = L_n^k$. Then $m$ is odd and $L_n^k\mid m$. As in the argument used in beginning of the proof of Theorem \ref{addtheorem1.4}(i), we see that $L_n^{k+1}\mid {m\choose m-2j-1}L_n^{m-(2j+1)}$ for every $0\leq j\leq \frac{m-3}{2}$ and $F_{nm}\cdot 2^{m-1}\equiv F_n^m5^{\frac{m-1}{2}}\pmod{L_n^{k+1}}$. Since $n\equiv 1, 2\pmod 3$, we obtain by \eqref{pre8} that $\gcd(L_n,F_n) = 1$. As $L_n\geq 3$ and $\gcd(L_n,5)=1$ with respect to Lemma \ref{lemmaofLengyel}, we obtain $L_n^{k+1}\nmid F_n^m5^{\frac{m-1}{2}}$. Hence $L_n^{k+1}\nmid F_{nm}$. This shows that Theorem \ref{addtheorem1.4} does not hold if $m$ is not even.
\end{example}

The conclusion $L_n^{k+2}\mid F_{nm}$ in Theorem \ref{addtheorem1.4}(iv) may or may not be $L_n^{k+2}\parallel F_{nm}$ as shown in the next example.

\begin{example}
Let $k\geq 1$, $\ell \geq 2k+2$, $n=6$, and $m = 2^k3^\ell$. Then $m$ is even, $n\equiv 0\pmod 6$, $L_n=18$, $L_n^k\parallel m$, and $\frac{L_n^{k+1}}{2}\mid m$. Then we obtain
\begin{align*}
\nu_2(F_{nm}) &amp;= \nu_2(nm)+2 = k+3 = (k+3)\nu_2(L_n) = \nu_2(L_n^{k+3}),\\
\nu_3(F_{nm}) &amp;= \nu_3(nm)+\nu_3(F_4) = \ell+2,\quad\text{and}\\
\nu_3(L_n^{k+3}) &amp;= \nu_3(18^{k+3}) = 2k+6.
\end{align*}
From this we see that if $\ell \geq 2k+4$, then $L_n^{k+3}\mid F_{nm}$ and if $2k+2\leq \ell &lt; 2k+4$, then $L_n^{k+3}\nmid F_{nm}$. This shows that the conclusion in Theorem \ref{addtheorem1.4}(iv) may or may not be replaced by an exact divisibility.
\end{example}

In view of Theorems \ref{fibothm3}, \ref{thm1.4fiboandlucas}, and \ref{addtheorem1.4}, the reader may expect to see the result concerning the divisibility of Lucas numbers by powers of the Fibonacci numbers. But this is not possible in general. For example, let $n\geq 3$, $n\equiv 0\pmod 3$, $k\geq 1$, $m = F_n^k$, then $\nu_2(F_n^{k+1}) \geq k+1 \geq 2 &gt; \nu_2(L_{nm})$ and therefore $F_n^k\parallel m$ but $F_n^{k+1}\nmid L_{nm}$. Even though we assume $n\not\equiv 0\pmod 3$, the desired result is still false as shown in the next example.

\begin{example}
Let $n\geq 3$ and $n\not\equiv 0\pmod 3$. Let $k, m\geq 1$ and $F_n^k\mid m$. Then by Lemma \ref{OPnewlemma2}, $F_n^{k+1}\mid {m\choose 2j}F_n^{2j}$ for every $1\leq j\leq \left\lfloor \frac{m}{2}\right\rfloor$. By \eqref{pre7}, we obtain 
$$
L_{nm}\cdot 2^{m-1} = \sum_{j=0}^{\left\lfloor \frac{m}{2}\right\rfloor}{m\choose 2j}F_n^{2j}L_n^{m-2j}5^j \equiv L_n^m\pmod{F_n^{k+1}}.
$$
As $\gcd(F_n,L_n) = 1$ by \eqref{pre8}, we have $F_n^{k+1}\nmid L_n^m$. Therefore $F_n^{k+1}\nmid L_{nm}$.
\end{example}

\section{Open questions}

It is natural to ask if the converse of each theorem holds. Marques \cite[Corollary 3]{Ma} shows that a partial converse of Theorem \ref{hoggatthmmontly} holds. More precisely, he shows that if $F_n^{k+1}\mid F_{nm}$ and $m$ is odd, then $F_n^k\mid m$. Whether or not the converse of Theorem \ref{hoggatthmmontly} holds when $m$ is even is still open. Similar questions are the following.
\begin{itemize}
\item[Q1] Assume that $F_n^{k+1}\parallel F_{nm}$ and $m$ is odd. Can we conclude that $F_n^k\parallel m$?
\item[Q2] Assume that $L_n^{k+1}\parallel L_{nm}$ and $m$ is odd (or any other assumption). Can we conclude that $L_n^k\parallel m$?
\item[Q3] Assume that $L_n^{k+1}\parallel F_{nm}$ and $m$ is even (or any other assumption). Can we conclude that $L_n^k\parallel m$? 
\item[Q4] Suppose that Q1, Q2, or Q3 has a negative answer. Can we say something about the divisibility of $m$ by powers of the Fibonacci and Lucas numbers?
\item[Q5] Benjamin and Rouse \cite{BR} show that Theorem \ref{hoggatthmmontly} holds if the Fibonacci numbers are replaced by the generalized Lucas numbers of the first kind (given by $u_0 = 0$, $u_1 = 1$, and $u_n = au_{n-1}+bu_{n-2}$ for $n\geq 2$). Does Theorem \ref{hoggatthmmontly} hold for other sequences? Can we extend the results to include exact divisibility by powers of the generalized Lucas numbers of the first kind?
\end{itemize}

\section{Acknowledgment}
The author receives financial support from Faculty of Science,
Silpakorn University, contract number SRF-PRG-2557-07. The author would
like to thank the referee for his/her careful reading and for his/her
suggestions which improve the presentation of this article.

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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11A07.

\noindent \emph{Keywords: } 
Fibonacci number, Lucas number, exact divisibility.

\bigskip
\hrule
\bigskip

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

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 24 2014;
revised versions received November 5 2014; November 6 2014.
Published in {\it Journal of Integer Sequences}, November 7 2014.

\bigskip
\hrule
\bigskip

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