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\markboth{U.\,K.\,Dutta and P.\,K.\,Ray}{On arithmetic functions of balancing and Lucas-balancing numbers}

\title{On arithmetic functions of balancing and Lucas-balancing numbers}

\author[U.\,K.\,Dutta and P.\,K.\,Ray]{Utkal Keshari Dutta and Prasanta Kumar Ray\corrauth}

\address{Department of Mathematics, Sambalpur University, Jyoti Vihar, Burla 768\,019, Sambalpur, Odisha, India}

\emails{{\tt utkaldutta@gmail.com}\,\,(U.\,K.\,Dutta), {\tt prasantamath@suniv.ac.in}\,\,(P.\,K.\,Ray)}

\begin{abstract}
For any integers $n\geq1$ and $k\geq0,$ let $\phi(n)$ and $\sigma_{k}(n)$ denote the Euler phi function and the sum of the $k$-th powers of the divisors of $n$, respectively. In this article, the solutions to some Diophantine equations about these functions of balancing and Lucas-balancing numbers are  discussed.
\end{abstract}

\keywords{Euler phi function, divisor function, balancing numbers, Lucas-balancing numbers}

\ams{11B37, 11B39, 11B83}

\maketitle

\section{Introduction}
A balancing number $n$ and a balancer $r$ are the solutions to a simple Diophantine equation $1+2+\cdot \cdot \cdot +(n-1)=(n+1)+(n+2)+ \cdot \cdot \cdot +(n+r)$ \cite{Behera1999}. The sequence of balancing numbers $\{B_n\}$ satisfies the recurrence relation $B_{n+1}=6B_n-B_{n-1},$ $n\geq1$  with initials $(B_0,B_1)=(0,1).$ The companion of $\{B_n\}$ is the sequence of Lucas-balancing numbers $\{C_n\}$ that satisfies the same recurrence relation as that of balancing numbers but with different initials $(C_0,C_1)=(1,3)$ \cite{Panda2009}.
Further, the closed forms known as Binet formulas for both of these sequences are given by
\begin{eqnarray*}
B_n =\frac{\lambda_1^n-\lambda_2^n}{\lambda_1-\lambda_2}, ~~C_n =\frac{\lambda_1^n+\lambda_2^n}{2},
\end{eqnarray*}
where $\lambda_1=3+2\sqrt{2}$ and $\lambda_2=3-2\sqrt{2}$ are the roots of the equation $x^2-6x+1=0.$
Some more recent developments of balancing and Lucas-balancing numbers can be seen in \cite{Irmak2018, Komatsu2014, Patel2017, Ray2017}.

Let $\phi(n)$ and $\sigma_{k}(n)$ denote the Euler phi function and the divisor function of $n$, respectively. In this study, we examine the solutions to some Diophantine equations relating to these functions.


%\underline{This is a template article, which you can use as a skeleton for your own article}.
\section{Auxiliary results}
In this section, we discuss some results which are used subsequently.

The following result is found in \cite{Panda2014}.
\begin{lemma}
For any positive integer $k\geq1$, the period of ${\{B_{n}\}}_{n\geq0}$ modulo ${2}^{k}$ is ${2}^{k}.$
\label{15}
\end{lemma}
The following results are given in \cite{ Ray2012}.
\begin{lemma}
Let $m$ and $n$ be positive integers; then
$\Big( B_m, B_n\Big)=B_{(m, n)},$ where $(x, y)$ denotes the greatest common divisor of $x$ and $y.$
\label{a}
\end{lemma}
\begin{lemma}
Let $m$ and $n$ be positive integers; then
$B_m\big|B_n \Leftrightarrow m|n.$
\label{b}
\end{lemma}
\begin{lemma}
For any odd prime $p$,
$B_{p-1}\!\equiv\!3\!\left(\left(\frac{p}{8}\right)\!-\!1\right) \pmod p$
and
$B_{p+1}\!\equiv\!3\!\left(\left(\frac{p}{8}\right)\!+\!1\right)\pmod p,$ where $(\frac{m}{n})$ denotes the Legendre symbol.
\label{10}
\end{lemma}
\begin{lemma}
Let $p$ be a prime; then
$p|B_{p-\left(\frac{8}{p}\right)}.$
\label{c}
\end{lemma}

We now discuss some congruence properties for balancing and Lucas-balancing numbers that are useful when proving subsequent results.

\begin{lemma}
For any natural number $k\geq2$, $B_{2^{k-1}}\equiv2^{k-1}\cdot 3\pmod {2^{k+1}}.$
\label{12}
\end{lemma}
\begin{proof}
The method of induction is used to prove this result. For $k=2,$ the result is obvious. Assume that $B_{2^{k-1}}\equiv 3\cdot2^{k-1}\pmod{2^{k+1}}$ holds for $k\geq3$. It follows that $B_{2^{k-1}}=2^{k-1}\cdot3u,$ where $u\equiv1 \pmod 4.$
From the identity $B_{2m}=B_m(B_{m+1}-B_{m-1}),$
\[
B_{2^{k}}=B_{2\cdot2^{k-1}}=B_{2^{k-1}}\cdot(B_{2^{k-1}+1}-B_{2^{k-1}-1})
=3\cdot2^{k-1}u\cdot 2v=3\cdot2^{k}uv,
\]
where $B_{2^{k-1}+1}-B_{2^{k-1}-1}=2v$ for any positive integer $v.$ This completes the proof.
\end{proof}
\begin{lemma}
For any odd integer $n$ and $k\geq1,$ if $B_{n}\equiv1 \pmod {2^{k}}$, then
$n\equiv1\pmod {2^{k}}.$
\label{100}
\end{lemma}
\begin{proof}
Consider $B_{n}\equiv1 \pmod {2^{k}}$. Then from Lemma \ref{15}, we can write $B_{2^{k}+1}\equiv1 \pmod {2^{k}}$. It follows that $2^{k}+1\equiv1\pmod {2^{k}}$. Since $B_1=1$ and $n$ is odd, $n\equiv1\pmod {2^{k}}.$
\end{proof}
\begin{lemma}
For any positive integer $k\geq2,$ $C_{2^{k}}\equiv1 \pmod {2^{k+4}}.$
\end{lemma}
\begin{proof}
The proof of this result is analogous to Lemma \ref{12}.
\end{proof}
\section{Main result}
In this section, we prove our main result.
\begin{theorem} The following statements hold:
\begin{itemize}
\item[$($i$)$] The ~only ~solutions ~to ~the ~equation $\phi(
|{B_{n}}|)=2^{m}$ ~are ~obtained ~for ~$n=\pm1,\pm2,\pm4.$
\item[$($ii$)$] The ~only ~solutions ~to ~the ~equation $\phi(
|{C_{n}}|)=2^{m}$ ~are ~obtained ~for ~$n=0, \pm1,\pm2.$
\item[$($iii$)$] The ~only ~solutions ~to ~the ~equation $\sigma(|{B_{n}}|)=2^{m}$ ~are ~obtained ~for ~$n=\pm1.$
\item[$($iv$)$] The ~only ~solutions ~to ~the ~equation ~$\sigma(|{C_{n}}|)=2^{m}$ ~are ~obtained ~for ~$n=0, \pm1.$
\end{itemize}
\end{theorem}

\begin{proof}
In order to prove ${(i)}$, we first show that if $\phi(
{B_{n}})=2^{m}$; then $2$ is the only prime factor of $n.$ Assume that the above statement does not hold, that is, there exists a prime $p>2$ such that $p|n$ and hence using Lemma \ref{b}, $B_{p}|B_{n}$, it follows that
\[
\phi(B_{p})|\phi(B_{n})=2^{m}.
\]
Therefore, $\phi{(B_{p})}=2^{m_{1}}$ for some $m_{1} \leq m,$ and it follows that
\begin{eqnarray}
B_{p}=2^{t}p_{1}p_{2}\cdots p_{s}, ~\rm{for}~t\geq1, ~p\geq1,
\end{eqnarray}
where $p_{1}, p_{2},\cdots ,p_{s}$ are distinct Fermat primes. Since $B_{p}$ is co-prime to $B_{1}$ and $B_{2}$ for $p>2,$ then $2\nmid B_{p}$ and $3\nmid B_{p}.$ This forces $t=0$ and $p_{1}>3$ and hence $p_{i}>3$ for all $i=1, 2, \dots , s.$

Let us consider $p_{i}=2^{2^{e_{i}}}+1$ for $e_{i}\geq1.$ Since $p_{1}\equiv 5\pmod8,$ from Lemma \ref{10}, it follows that $p_{1}$ divides $B_{p_{1}-1}$ and hence $p_{1}|(B_{p}, B_{p_{1}-1}).$ Using Lemma \ref{a}, $p_{1}|B_{(p,~p_{1}-1)}.$ It follows that $p|p_{1}-1=2^{2^{e_{1}}}$ forces $p=2$ and hence $e_{1}=0,$ which is a contradiction.

Assume that $n=2^{u}3^{v}.$ To show $u \leq 2,$ assume that $u > 2$. Then
\[
235416=B_{8}|B_{n},
\]
which concludes $3|73728=\phi(235416)|\phi(B_{n})=2^{m},$ which is a contradiction. To show that $v=0,$ assuming $v>0$ we get $35=B_{3}|B_{n}$; therefore
\[
3|24=\phi(35)|\phi(B_{n})=2^{m},
\]
which is again a contradiction. The above discussion concludes $n|2^{2}$ and the result follows.

In order to prove ${(ii)},$ we proceed as follows. Since $\phi(1)=1=2^{0},$ $\phi(3)=2$ and $\phi(17)=16=2^{4},$ it follows that the solution to ${(ii)}$ are the elements from the set $n=\{0, 1, 2\}$. In order to prove
identity $(ii)$ completely, we need to show that these are the only solutions. If possible, let $\phi (C_{n})=2^{m}$ for $n\geq3$; it follows that
\begin{eqnarray*}
C_{n}=2^{l}p_{1}\ldots p_{k},
\end{eqnarray*}
where $l\geq0$ and $p_{1}<\ldots<p_{k}$ are Fermat primes. Since $C_{n}^2=8B_{n}^2+1,$ so $C_n$ are odd, which forces $l$ to be zero.

Now,  write $p_{i}=2^{2^{e_{i}}}+1,$ $i=1, 2, \dots , k.$ For $n\geq3,$ $C_{n}\geq99$ and hence $p_{i}>3.$ It can be observed that $p_{i}\equiv5~{\rm{or}}~1 \pmod 8.$ For $p_{i}\equiv5\pmod 8,$ $\left(\frac{p_{i}}{8}\right)=-1=\left(\frac{8}{p_{i}}\right)$. As $8B_{n}^2=C_{n}^2-1 \equiv -1 \pmod{p_i},$ it follows that $\left(\frac{8}{p_{i}}\right)=1,$ a contradiction.

On the other hand, for $p_{i}\equiv1 \pmod8,$  $\left(\frac{p_{i}}{8}\right)=1=\left(\frac{8}{p_{i}}\right).$
Again, the identity $8B_{n}^2 \equiv -1 \pmod{p_i}$ gives $\left(\frac{8}{p_{i}}\right)=-1,$ also a contradiction and hence the result follows.

Further, clearly $n=1$ is a solution for $\sigma(|{B_{n}}|)=2^{m}$ as $\sigma(B_{1})=1.$
Now it remains to show that there is no other solution except $n=1$. If possible, let there exist a solution to $\sigma(B_{n})=2^{m}$ with $n\geq2$. For $\sigma(B_{n})=2^{m}$, let $B_{n}=q_{1}\ldots q_{k},$ where $q_{1}<q_{2}<\ldots<q_{k}$ are Mersenne primes. Let $q_{i}=2^{p_{i}}-1$ for $p_{i}\geq2.$ In particular, $q_{i}\equiv-1 \pmod 8$ and it follows that $\left(\frac{8}{q_{i}}\right)=\left(\frac{2}{q_{i}}\right)^{3}=-1$. From Lemma \ref{c}, $B_{p-\left(\frac{8}{p}\right)}\equiv0 \pmod p,$  it follows that $q_{i}|B_{q_{i}+1}$ and hence  $q_{i}|(B_{n}, B_{q_{i}+1})$. It is well known that if $q|n$, then $B_{q}|B_{n}.$ Therefore, $q_{i}|(B_{q},B_{q_{i}+1})=B_{(q, q_{i}+1)}$ implies $q_{i}|q_{i}+1=2^{p_{i}},$ again a contradiction and ${(iii)}$ follows.

Since $\sigma(1)=1=2^{0}, \sigma(3)=4=2^{2}$ it follows that $n=0, 1$ are the solutions to $\sigma(|{C_{n}}|)=2^{m}.$ To prove that these are the only ones, assume that there exists a solution with $n\geq2$. For $\sigma(C_{n})=2^{m},$ let $C_{n}=q_{1}\ldots q_{k},$ where $q_{1}<q_{2}<\ldots<q_{k}$ are Mersenne primes. Let $q_{i}=2^{p_{i}}-1$ for  $p_{i}\geq2.$ Assume that $p_{1}>2.$ Since  $C_{p}^{2}\equiv9~({\rm{mod}}~2^{p_1+1})$, then
\[
C_{p}^{2}-9=8(B_{p}^{2}-1)\equiv0~({\rm{mod}}~ 2^{p_1+1}),
\]
which gives
\[
B_{p}^{2}\equiv1~({\rm{mod}}~2^{p_1+1}).
\]
It follows that $B_{p}\equiv\pm1\pmod{2^{p_1}}$ and hence using Lemma \ref{100}, $p\equiv1({\rm{mod}}~2^{p_1}).$ In particular,
\begin{equation}\label{13}
p\geq2^{p_{i}}+1.
\end{equation}
Further, since $q_{1}|C_{p}$, then $8B_{p}^{2}\equiv-1\pmod{q_1},$ which implies that $q_{1}|B_{q_{1}-1}.$ As $B_{2p}=2B_{p}C_{p},$  $q_{1}|B_{2p}.$ Therefore,
\begin{eqnarray*}
 q_{1}|(B_{2p},B_{q_{1}-1})=B_{(2p,q_{1}-1)}.
 \end{eqnarray*}
For $p_{1}>2,$  $q_{1}>3$ and hence $2p|q_{1}-1=2^{p_{1}}-2=2(2^{p_{1}-1}-1).$ In particular,
\begin{eqnarray}
p\leq2^{p_{i}-1}-1,
\end{eqnarray}
which is a contradiction to (\ref{13}). Thus, $p_{i}\leq2$, which follows ${(iv)}.$ This completes the proof of the theorem.
\end{proof}
%%%% Acknowledgment %%%%%%%%
\section*{Acknowledgement}
The authors are grateful to the
referee for his/her useful and helpful remarks which significantly improved
the presentation of this paper.

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\end{thebibliography}
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