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\begin{center}
\vskip 1cm{\LARGE\bf 
 On Some Magnified Fibonacci Numbers  \\
 \vskip .12in
 Modulo a Lucas Number} \\
\vskip 1cm
\large
Ram Krishna Pandey \\
Department of Mathematics\\
Indian Institute of Technology Patna \\
Patliputra Colony, Patna - 800013\\
India \\
\href{mailto:ram@iitp.ac.in}{\tt ram@iitp.ac.in}
\end{center}

\vskip .2in

\begin{abstract}
Let, as usual, $F_n$ and $L_n$ denote the $n$th Fibonacci number and
the $n$th Lucas number, respectively. In this paper, we consider the
Fibonacci numbers $F_2, F_3, \ldots, F_{t}.$ Let $n \geq 1$ be an
integer such that $4n+2 \leq t \leq 4n+5$ and $m = F_{2n+2} +
F_{2n+4} = L_{2n+3}.$ We prove that the integers $F_2 F_{2n+2},
F_3 F_{2n+2}, \ldots, F_{t} F_{2n+2}$ modulo $m$ all belong to the
interval $[F_{2n+1}, 3F_{2n+2}].$ Furthermore, the endpoints of
the interval $[F_{2n+1}, 3F_{2n+2}]$ are obtained only by the
integers $F_4 F_{2n+2}$ and $F_{4n+2} F_{2n+2}$, respectively.
\end{abstract}


\section{Introduction}


Let $m_1, m_2, \ldots $ be positive integers. Cusick and Pomerance
\cite{cusickpom} discuss the quantity $\kappa$ where
\[\kappa := \sup_{x\in(0,1)}\min_{i}\|xm_{i}\|.\]
Here, for $x \in \mathbb{R}$, $\|x\|$ is distance to the nearest
integer. Observe that if we have a finite number of integers $m_1,
m_2, \ldots, m_n$ and
\[\kappa_1 := \displaystyle\max_{\stackrel{m = m_j + m_l}{1 \leq k \leq m/2}}
\ \frac{1}{m}\min_{i}|km_{i}|_m,\]
where $ |x|_m $ denotes the absolute value of the absolutely least
remainder of $x$ mod $m$, then by a remark of Haralambis \cite{har}
we have that $\kappa = \kappa_1$. The quantity $\kappa$ has become
an important quantity now; it is involved in the
well-known {\em Lonely Runner conjecture}. This conjecture,
due to Bienia et al.\ \cite{bienia}, is stated as follows:
\begin{quote}
Suppose $n$ runners having nonzero distinct constant speeds run laps
on a unit-length circular track. Then there is a time at which all
the $n$ runners are simultaneously at least $1/(n+1)$ units from
their common starting point.
\end{quote}

The original form of this conjecture,
as given by Wills
\cite{wills} and Cusick \cite{cusick}, is as follows:

\begin{quote}
Suppose $m_1, m_2, \ldots, m_n$ be $n$ positive integers. Then
$\kappa \geq 1/(n+1).$
\end{quote}

This is an interpretation also due to Liu and Zhu \cite{liuzhu}. In
the present paper we show that if the integers $m_1, m_2, \ldots,
m_n$ in the definition of $\kappa$ are $F_2, F_3, \ldots, F_t$ and
$n \geq 1$ be an integer such that $4n+2 \leq t \leq 4n+5$, then
\[\kappa \geq \frac{F_{2n+1}}{F_{2n+2}+F_{2n+4}}.\] This also
confirms the lonely runner conjecture in the case where the speeds of the
runners are $F_2, F_3, \ldots, F_t$.


\section {Main Results}

Let $M = \{F_2, F_3, \ldots, F_t\}$. Let $n \geq 1$ be an integer
such that $4n+2 \leq t \leq 4n+5$ and $m = F_{2n+2} + F_{2n+4} =
L_{2n+3}.$ The above bound about $\kappa$ may be obtained using
Theorem \ref{t1} and $\kappa_1.$ The lonely runner conjecture is
also satisfied in the special case $n=0$. In this special case the
set of speeds may be $\{F_2\}$, $\{F_2, F_3\}$, $\{F_2, F_3, F_4\}$,
and $\{F_2, F_3,F_4,F_5\}$. Taking $x = \frac{1}{2}$, $\frac{1}{3}$,
$\frac{1}{4}$, and $\frac{1}{4}$ in the definition of $\kappa$ for
the sets $\{F_2\}$, $\{F_2, F_3\}$, $\{F_2, F_3, F_4\}$, and $\{F_2,
F_3,F_4,F_5\}$, respectively, we see that the lonely runner
conjecture is satisfied.

In this section, we use the following identities and these
identities may be found in Koshy \cite{koshy}.
\begin{enumerate}
\item Cassini's identity:   $F^2_n - F_{n+1}F_{n-1} = (-1)^{n-1}$.
\item $F^2_n + F^2_{n+1} = F_{2n+1}$.
\item d'Ocagne's identity:   $F_m F_{n+1} - F_{m+1} F_n = (-1)^n
F_{m-n}$.
\item $F_{2n} = \sum_{i=0}^{n-1} F_{2i+1}$.
\item $F_{2n+1}-1 = \sum_{i=1}^{n} F_{2i}$.
\end{enumerate}

\begin{lemma}\label{f1}
\begin{itemize}
\item [(a)] $F_2 F_{2n+2} \equiv F_{4n+5}F_{2n+2} \pmod m$.
\item [(b)] $F_2 F_{2n+2} \equiv - F_{4n+4}F_{2n+2} \pmod m$.
\item [(c)] $F_3 F_{2n+2} \equiv F_{4n+3}F_{2n+2} \pmod m$.
\item [(d)] $F_4 F_{2n+2} \equiv -F_{4n+2}F_{2n+2} \pmod m$.
\end{itemize}
\end{lemma}

\begin{proof}
\begin{itemize}
\item [(a)] We have
\begin{eqnarray*}
m F_{2n+2} &amp;=&amp; (F_{2n+2} + F_{2n+4})F_{2n+2}\\
           &amp;=&amp; F^2_{2n+2} + F_{2n+4}F_{2n+2}\\
           &amp;=&amp; F^2_{2n+2} + (F_{2n+3} + F_{2n+2})(F_{2n+3} - F_{2n+1})\\
           &amp;=&amp; F^2_{2n+2} + F^2_{2n+3} + F_{2n+3}F_{2n+2} - F_{2n+3}F_{2n+1} - F_{2n+2}F_{2n+1}\\
           &amp;=&amp; F^2_{2n+2} + F^2_{2n+3} + (F_{2n+2} + F_{2n+1})F_{2n+2} - F_{2n+3}F_{2n+1} - F_{2n+2}F_{2n+1}\\
           &amp;=&amp; F^2_{2n+2} + F^2_{2n+3} + F^2_{2n+2} - F_{2n+3}F_{2n+1}\\
           &amp;=&amp; F^2_{2n+2} + F^2_{2n+3} + (-1)^{2n+1} \quad \left(\text {using Cassini's identity}\right)\\
           &amp;=&amp; F_{4n+5} - 1 \quad \left(\text {using the identity} F^2_n + F^2_{n+1} = F_{2n+1}\right).
\end{eqnarray*}
Hence we get
\[F_{4n+5}F_{2n+2} = (1+ m F_{2n+2})F_{2n+2} \equiv F_{2n+2} = F_{2}F_{2n+2} \pmod m.\]

\item [(b)] We have
\begin{eqnarray*}
F_{2}F_{2n+2} &amp;\equiv&amp; F_{4n+5}F_{2n+2} \pmod m \quad \left(\text {by (a)}\right)\\
           &amp;=&amp; (F_{4n+6} - F_{4n+4})F_{2n+2}\\
           &amp;=&amp; F_{4n+6}F_{2n+2} - F_{4n+4}F_{2n+2}\\
           &amp;=&amp; (L_{2n+3}F_{2n+3})F_{2n+2} - F_{4n+4}F_{2n+2} \quad \left(\text{since } L_n F_n = F_{2n}\right)\\
           &amp;\equiv&amp; - F_{4n+4}F_{2n+2} \pmod m \quad  \left(\text{as } m = L_{2n+3}\right).
\end{eqnarray*}

\item [(c)] $F_3 F_{2n+2} = (F_2 + F_1)F_{2n+2} \equiv (F_{4n+5} -F_{4n+4})F_{2n+2}  \equiv F_{4n+3}F_{2n+2} \pmod m$.
\item [(d)] $F_4 F_{2n+2} = (F_3 + F_2)F_{2n+2} \equiv (F_{4n+3} -F_{4n+4})F_{2n+2} \equiv - F_{4n+2}F_{2n+2} \pmod m$.
\end{itemize}
\end{proof}


\begin{lemma}
$F_{r+5} F_{2n+2} = (F_{r+4} + F_{r+3})F_{2n+2} \equiv \epsilon
F_{4n+1-r} F_{2n+2} \pmod m$ for each $0 \leq r \leq 2n-2,$ where
\[
\epsilon =
\left\{%
\begin{array}{ll}
    +1, &amp; \text{if $r$ is even;} \\
    -1, &amp; \text{if $r$ is odd.} \\
\end{array}%
\right.
\]
\end{lemma}

\begin{proof}
Using the recurrence relation $F_{n+1} = F_{n} + F_{n-1}$ for $n
\geq 1$ and Lemma \ref{f1}, the proof follows by induction on $r$.
\end{proof}

Thus in order to examine the integers $F_2 F_{2n+2}, F_3 F_{2n+2},
\ldots, F_{4n+5} F_{2n+2}$ modulo $m$, it is sufficient to examine
only the integers $F_2 F_{2n+2}, F_3 F_{2n+2}, \ldots, F_{2n+3}
F_{2n+2}$ modulo $m$.

\begin{lemma}\label{f2}
\begin{itemize}
\item [(a)] $F_2 F_{2n+2} \equiv F_{2n+2} \pmod m$.
\item [(b)] $F_3 F_{2n+2} \equiv -F_{2n+3} \pmod m$.
\item [(c)] $F_4 F_{2n+2} \equiv -F_{2n+1} \pmod m$.
\item [(d)] $F_5 F_{2n+2} \equiv 2F_{2n+2} - F_{2n+1} \pmod m$.
\end{itemize}
\end{lemma}

\begin{proof}
\begin{itemize}
\item [(a)] $F_2 F_{2n+2} \equiv F_{2n+2} \pmod m$ (as $F_2 = 1$).
\item [(b)] $F_3 F_{2n+2} = 2F_{2n+2} \equiv - (F_{2n+1}+F_{2n+2}) = -F_{2n+3} \pmod m$.
\item [(c)] $F_4 F_{2n+2} = (F_2+F_3)F_{2n+2} \equiv F_{2n+2} - F_{2n+3} = -F_{2n+1} \pmod m$.
\item [(d)] $F_5 F_{2n+2} = 5F_{2n+2} \equiv 2F_{2n+2} - F_{2n+1} \pmod m$.
\end{itemize}
\end{proof}

We observe that in all above four congruences if $R$ is the
remainder then we have, $F_{2n+1} \leq |R| \leq \frac{m}{2}$, i.e.,
we have that the positive remainder is always in the interval
$[F_{2n+1}, 3F_{2n+2}]$.

Inductively, we can prove that
\[F_{2n+3-k} = F_{2n-1-k}F_{5} + F_{2n-2-k}F_{4}\]
for each $0 \leq k \leq 2n-3$. Hence we have
\begin{eqnarray*}
F_{2n+3-k}F_{2n+2} &amp;=&amp; (F_{2n-1-k}F_{5} + F_{2n-2-k}F_{4})F_{2n+2}\\
                   &amp;\equiv&amp; F_{2n-1-k}(2F_{2n+2}-F_{2n+1}) - F_{2n-2-k}F_{2n+1}\pmod m  \quad \left(\text{using (c), (d) of Lemma \ref{f2}}\right)\\
                   &amp;=&amp; F_{2n-1-k}F_{2n+2} + F_{2n-1-k}(F_{2n+1} + F_{2n}) - F_{2n-1-k}F_{2n+1} - F_{2n-2-k}F_{2n+1}\\
                   &amp;=&amp; F_{2n-1-k}F_{2n+2} + F_{2n-1-k}F_{2n} - F_{2n-2-k}F_{2n+1}\\
                   &amp;=&amp; F_{2n-1-k}F_{2n+2} + (-1)^{2n-2-k}F_{2+k} \quad \left(\text {using d'Ocagne's identity}\right).
\end{eqnarray*}
Thus we get
\begin{equation}
F_{2n+3-k}F_{2n+2} \equiv F_{2n-1-k}F_{2n+2} + \epsilon F_{2+k} \pmod m, \hfill \ldots 
\label{star}
\end{equation}
where
\[
\epsilon =
\left\{%
\begin{array}{ll}
    +1, &amp; \text{if $k$ is even;} \\
    -1, &amp; \text{if $k$ is odd.} \\
\end{array}%
\right.
\]

\begin{lemma}\label{f3}
Let $R_k = F_{2n-1-k}F_{2n+2} + \epsilon F_{2+k}$ for each $0 \leq k
\leq 2n-3$, where
\[
\epsilon =
\left\{%
\begin{array}{ll}
    +1, &amp; \text{if $k$ is even;} \\
    -1, &amp; \text{if $k$ is odd.} \\
\end{array}%
\right.
\]
Then $F_{2n+1} &lt; |R_k|_m &lt; \frac{m}{2}$.
\end{lemma}

\begin{proof}
We partition the set $\{0,1,2, \ldots, 2n-3\}$ into the sets $A_2,
A_3, A_4,$ and $A_5$ where
\[A_i = \left\{2n-1-i-4t : 0 \leq t \leq \left\lfloor \frac{2n-1-i}{4} \right\rfloor\right\},\]
for each $2 \leq i \leq 5.$ Notice that
\[
\epsilon =
\left\{%
\begin{array}{ll}
    +1, &amp; \text{if $k \in A_3 \cup A_5$;} \\
    -1, &amp; \text{if $k \in A_2 \cup A_4$.} \\
\end{array}%
\right.
\]
We consider the following cases:

\medskip

{\bf Case I:} ($k \in A_2)$. Clearly (\ref{star}) implies that
\[F_{6+4t}F_{2n+2} \equiv F_{2+4t}F_{2n+2} - F_{2n-1-4t} \pmod m.\]

$t =0 \Rightarrow F_{6}F_{2n+2} \equiv F_{2}F_{2n+2} - F_{2n-1}
\equiv F_{2n+2} - F_{2n-1} \pmod m.$ Observe that $F_{2n+1} &lt;
F_{2n+2} - F_{2n-1} &lt; m/2.$

$t =1 \Rightarrow F_{10}F_{2n+2} \equiv F_{6}F_{2n+2} - F_{2n-5}
\equiv F_{2n+2} - F_{2n-1} - F_{2n-5}\pmod m.$ Observe that
$F_{2n+1} &lt; F_{2n+2} - F_{2n-1} - F_{2n-5} &lt; m/2.$

Inductively, we let $t = \lfloor \frac{2n-3}{4} \rfloor$. In this
case $k = 1$ or $k = 3$ depending on $n$ is even or odd. If $k =1$,
then we have

$F_{2n+2}F_{2n+2} \equiv F_{2n-2}F_{2n+2} - F_{3} \equiv F_{2n+2} -
(F_{2n-1} + F_{2n-5} + \dots + F_3) \pmod m.$ Observe that $F_{2n+1}
&lt; F_{2n+2} - (F_{2n-1} + F_{2n-5} + \dots + F_3) &lt; m/2$ as we have
the identity $F_{2n} = \sum_{i=0}^{n-1} F_{2i+1}$.

If $k =3$, then we have

$F_{2n}F_{2n+2} \equiv F_{2n-4}F_{2n+2} - F_{5} \equiv F_{2n+2} -
(F_{2n-1} + F_{2n-5} + \dots + F_5) \pmod m.$ Observe that $F_{2n+1}
&lt; F_{2n+2} - (F_{2n-1} + F_{2n-5} + \dots + F_5) &lt; m/2$ as we have
the identity $F_{2n} = \sum_{i=0}^{n-1} F_{2i+1}$.

\medskip

{\bf Case II:} ($k \in A_3$). Clearly (\ref{star}) implies that
\[F_{7+4t}F_{2n+2} \equiv F_{3+4t}F_{2n+2} + F_{2n-2-4t} \pmod m.\]

$t =0 \Rightarrow F_{7}F_{2n+2} \equiv F_{3}F_{2n+2} + F_{2n-2}
\equiv -F_{2n+3} + F_{2n-2} \pmod m.$ Observe that $F_{2n+1} &lt;
|-F_{2n+3} + F_{2n-2}| &lt; m/2.$

$t =1 \Rightarrow F_{11}F_{2n+2} \equiv F_{7}F_{2n+2} + F_{2n-6}
\equiv -F_{2n+3} + F_{2n-2} + F_{2n-6}\pmod m.$ Observe that
$F_{2n+1} &lt; |-F_{2n+3} + F_{2n-2} + F_{2n-6}| &lt; m/2.$

Inductively, we let $t = \lfloor \frac{2n-4}{4} \rfloor$. In this
case $k = 0$ or $k = 2$ depending on $n$ is even or odd. If $k =0$,
then we have

$F_{2n+3}F_{2n+2} \equiv F_{2n-1}F_{2n+2} + F_{2} \equiv -F_{2n+3} +
(F_{2n-2} + F_{2n-6} + \dots + F_2) \pmod m.$ Observe that $F_{2n+1}
&lt; |-F_{2n+3} + (F_{2n-2} + F_{2n-6} + \dots + F_2)| &lt; m/2$ as we
have $F_{2n+3} - (F_{2n-2} + F_{2n-6} + \dots + F_2) = F_{2n+1} +
F_{2n+2} - (F_{2n-2} + F_{2n-6} + \dots + F_2)$ and the identity
$F_{2n+1}-1 = \sum_{i=1}^{n} F_{2i}$

If $k =2$, the result follows from the above case $k=0$.

\medskip

{\bf Case III:} ($k \in A_4$). Clearly (\ref{star}) implies that
\[F_{8+4t}F_{2n+2} \equiv F_{4+4t}F_{2n+2} - F_{2n-3-4t} \pmod m.\]

$t =0 \Rightarrow F_{8}F_{2n+2} \equiv F_{4}F_{2n+2} - F_{2n-3}
\equiv -F_{2n+1} - F_{2n-3} \pmod m.$ Observe that $F_{2n+1} &lt;
F_{2n+1} + F_{2n-3} &lt; m/2.$

$t =1 \Rightarrow F_{12}F_{2n+2} \equiv F_{8}F_{2n+2} - F_{2n-7}
\equiv -(F_{2n+1} + F_{2n-3} + F_{2n-7})\pmod m.$ Observe that
$F_{2n+1} &lt; F_{2n+1} + F_{2n-3} + F_{2n-7} &lt; m/2.$

Inductively, we let $t = \lfloor \frac{2n-5}{4} \rfloor$. In this
case $k = 1$ or $k = 3$ depending on $n$ is odd or even. If $k =1$,
then we have

$F_{2n+2}F_{2n+2} \equiv F_{2n-2}F_{2n+2} - F_{3} \equiv -(F_{2n+1}
+ F_{2n-3} + F_{2n-7}+\ldots + F_3) \pmod m.$ Observe that $F_{2n+1}
+ F_{2n-3} + F_{2n-7}+\ldots + F_3 &lt; m/2$ as we have the identity
$F_{2n} = \sum_{i=0}^{n-1} F_{2i+1}.$

If $k =3$. It follows from the above case $k=1$.

\medskip

{\bf Case IV:} ($k \in A_5$) Clearly (\ref{star}) implies that
\[F_{9+4t}F_{2n+2} \equiv F_{5+4t}F_{2n+2} + F_{2n-4-4t} \pmod m.\]

$t =0 \Rightarrow F_{9}F_{2n+2} \equiv F_{5}F_{2n+2} + F_{2n-4}
\equiv 2F_{2n+2} - F_{2n+1}+ F_{2n-4} \pmod m.$ Observe that
$F_{2n+1} &lt; 2F_{2n+2} - F_{2n+1}+ F_{2n-4} &lt; m/2.$

$t =1 \Rightarrow F_{13}F_{2n+2} \equiv F_{9}F_{2n+2} + F_{2n-8}
\equiv 2F_{2n+2} - F_{2n+1}+ F_{2n-4} + F_{2n-8} \pmod m.$ Observe
that $F_{2n+1} &lt; 2F_{2n+2} - F_{2n+1}+ F_{2n-4} + F_{2n-8} &lt; m/2.$

Inductively, we let $t = \lfloor \frac{2n-6}{4} \rfloor$. In this
case $k = 0$ or $k = 2$ depending on $n$ is odd or even. If $k =0$,
then we have

$F_{2n+3}F_{2n+2} \equiv F_{2n-1}F_{2n+2} + F_{2} \equiv 2F_{2n+2} -
F_{2n+1}+ (F_{2n-4} + F_{2n-8} + \ldots + F_2) \pmod m.$ Observe
that $F_{2n+1} &lt; 2F_{2n+2} - F_{2n+1}+ (F_{2n-4} + F_{2n-8} + \ldots
+ F_2) &lt; m/2$ as we have $2F_{2n+2} - F_{2n+1}+ (F_{2n-4} + F_{2n-8}
+ \ldots + F_2) = F_{2n+2} + F_{2n}+ (F_{2n-4} + F_{2n-8} + \ldots +
F_2) &lt; F_{2n+2} + F_{2n+1} - 1 &lt; m/2$ using the identity $F_{2n+1}-1
= \sum_{i=1}^{n} F_{2i}$

If $k =2$. It follows from the above case $k=0$.
\end{proof}


\begin{theorem} \label{t1}
Let $F_2, F_3, \ldots, F_{t}$ be the Fibonacci numbers and let $n
\geq 1$ be an integer such that $4n+2 \leq t \leq 4n+5$ and $m =
F_{2n+2} + F_{2n+4} = L_{2n+3}.$ Then the integers $F_2 F_{2n+2},
F_3 F_{2n+2}, \ldots, F_{t} F_{2n+2}$ modulo $m$ all belong to the
interval $[F_{2n+1}, 3F_{2n+2}].$ Moreover, both the endpoints of
the interval $[F_{2n+1}, 3F_{2n+2}]$ are obtained only by the
integers $F_4 F_{2n+2}$ and $F_{4n+2} F_{2n+2}$, respectively.
\end{theorem}

\begin{proof}
Lemma \ref{f2} and Lemma \ref{f3} together prove the theorem.
\end{proof}



\section {Acknowledgements} The author wishes to thank to the
referees for their useful comments.




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D. D.-F. Liu and X. Zhu, Fractional chromatic number for distance
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\end{thebibliography}

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

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

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

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\vspace*{+.1in}
\noindent
Received September 28 2012;
revised version received  January 21 2013.
Published in {\it Journal of Integer Sequences}, January 26 2013.

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