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\begin{center}
\vskip 1cm{\Large\bf 
A Simple Proof of Skula's Theorem on Prime \\
\vskip .1in
Power Divisors of Mersenne Numbers
}
\vskip 1cm
Ji\v{r}\'{\i} Kla\v{s}ka\\
Institute of Mathematics\\
Faculty of Mechanical Engineering\\
Brno University of Technology\\
Technick\'{a} 2\\
CZ -- 616 69 Brno\\
Czech Republic \\
\href{mailto:klaska@fme.vutbr.cz}{\tt klaska@fme.vutbr.cz}
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\begin{abstract}
Recently, Skula proved an interesting result concerning the
equivalence between higher order Wieferich primes 
and prime power divisors of Mersenne numbers. In the present paper, we
provide a simple  proof of Skula's result. 
\end{abstract}

\section{Introduction}

The Mersenne numbers $M_n=2^n-1$, $n\in \mathbb{N}$ have been studied by
many authors over  past 
centuries, with a number of remarkable statements discovered and proved
since the time 
of Marin Mersenne (1588--1648).  See sequences
\seqnum{A000225} and \seqnum{A001348} in the {\it On-Line Encyclopedia
of Integer Sequences} \cite{6}. 

The present paper is concerned with the connection between higher-order
Wieferich primes and prime power divisors of Mersenne numbers.
First, however, we recall certain definitions and facts
relating to this topic. 
In 1997, Agoh, Dilcher, and Skula published the well-known
paper \cite{1} in which 
the concepts of Fermat quotient and Wieferich prime were extended
substantially. 

Let $a,m\in \mathbb{Z}$, $m\ge 2$ and let $\gcd(a,m)=1$. By
\cite[Definition 1.2]{1}, 
$$q(a,m)=\frac{a^{\varphi (m)}-1}{m}$$
is called the Euler quotient of $m$ with base $a$.  From Euler's
theorem it is clear 
(see for example \cite[p.\ 33]{2} or \cite[p.\ 42]{4}) that  $q(a,m)\in \mathbb{Z}$.
By \cite[Definition 1.2]{1}, $m$ is called a Wieferich number with  base $a$ if
$q(a,m)\equiv 0$ (mod $m$). See also \cite[\seqnum{A077816}]{6}.
Specifically, if $p$ is a prime satisfying 
$q(2,p)\equiv 0 $ (mod $p$), we obtain the well-known definition of a Wieferich
prime:  $p$  is a  Wieferich prime if $2^{p-1}\equiv 1 $ (mod $p^2$).
See \cite[\seqnum{A001220}]{6}.  
By \cite[Definition 1.4]{5} a Wieferich prime $p$ is called
a~Wieferich prime of order $n\in \mathbb{N}$ if  
$$
q(2,p^n)\equiv 0\pmod{p^{n}}\quad\text{or,
  equivalently,}\quad 2^{p^{n-1}(p-1)}\equiv 1\pmod{p^{2n}}.
$$ 
Hence, a prime $p$ is a Wieferich one if and only if $p$ is Wieferich
prime of order $1$ and, 
$p$ is a Wieferich prime of order $n$ if and only if $p^n$ is a
Wieferich number with base $2$. 

Finally, let us recall that the smallest positive integer $h$ for
which $a^h\equiv 1$ (mod $m$) 
is called the multiplicative order of $a$ modulo $m$, which we 
write as $h=\ord_{m}(a)$. 
Now we can formulate the main result presented in \cite{5}.


\begin{theorem}[Skula, 2019]
\label{thm:1}
Let  $n\in \mathbb{N}$  and let
$p,q$ be odd primes. 
If $p^n|M_q$, then the following statements are equivalent:
\begin{itemize}
\item[$(\mathrm{A})$] $p^{n+1}|M_q$.
\item[$(\mathrm{B})$]  $p$ is a Wieferich prime of order $n$.
\item[$(\mathrm{C})$]  $\ord_{p^{n+1}}(2)=q$.
\end{itemize}
\end{theorem}

The proof of Theorem~\ref{thm:1} given in \cite{5} is mainly based on some
special properties of Euler quotient, 
such as the logarithm property presented in \cite[Proposition 1.1]{5}
or the proposition on the Euler 
quotient for two bases \cite[Proposition 2.1]{5}. Furthermore, the
proof in \cite{5} uses the results 
\cite[Lemma 5.1]{1} and \cite[Corollary 5.2]{1} on the order of the
number $q(2,p)$ modulo $p$ 
within the  meaning of definition  \cite[p.~3]{2}. Hence,  the proof
of Theorem~\ref{thm:1} in \cite{5} 
is the result of resourcefully applying a series of findings. 

The main purpose of this paper is to show that Theorem~\ref{thm:1} can be
proved in a way simpler than the one in \cite{5}, 
that is, without using special propositions on the Euler quotient. 

\section{Proof of Theorem~\ref{thm:1}}

We begin by recalling  some of the known properties of $\ord_m(a)$,
which are needed 
in our proof of Theorem~\ref{thm:1}.

\begin{lemma} Let $a,m\in \mathbb{Z}$, $m\ge 2$ and let
  $\gcd(a,m)=1$. Then $(\mathrm{i})$--$(\mathrm{vii})$ holds. 
\vskip 2mm

\begin{itemize}

\item[$(\mathrm{i})$] Let $k\in \mathbb{N}$. Then
$a^k\equiv 1$ \textup{(mod $m$)}  if and only if $\ord_m(a)|k$.  

\item[$(\mathrm{ii})$] $\ord_m(a)|\varphi(m)$. Consequently, if $p$ is an odd prime, then $\ord_p(2)|p-1$.

\item[$(\mathrm{iii})$] Let $a,k,n\in \mathbb{N}$ and let $p$ be an odd prime satisfying $p\nmid a$.
Further, let $\ord_p(a)=h$ and let $p^k\| a^h-1$.  Then
\begin{displaymath} \ord_{p^n}(a)=
\begin{cases} h, &amp; \text{for $n\le k$;}\\
       p^{n-k}h, &amp; \text{for $n&gt; k$.}
\end{cases}
\end{displaymath}
Here, $p^k\|a^h-1$ means that $p^k|a^h-1$ but $p^{k+1}\nmid a^h-1$.


\item[$(\mathrm{iv})$]  Let $a,n\in \mathbb{N}$ and let $p$ be an odd prime satisfying $p\nmid a$. If
$\ord_{p^{n}}(a)=h$, then $\ord_{p^{n+1}}(a)\in \{h,ph\}$.
Consequently, $\ord_{p^n}(a) | \ord_{p^{n+1}}(a)$.


\item[$(\mathrm{v})$] Let $k,n\in \mathbb{N}$ and let $p$ be an odd prime.
If $\ord_{p}(2)=\cdots=\ord_{p^n}(2)\ne \ord_{p^{n+1}}(2)$, then
$\ord_{p^{n+k}}(2)=p^k\ord_p(2)$.

\item[$(\mathrm{vi})$]  Let $n\in \mathbb{N}$ and let $p$ be an  odd prime.
If $\ord_{p^n}(2)|p-1$, then $\ord_{p^k}(2)=\ord_{p^n}(2)$ for any
$k\in\{1,\ldots,n\}$. 

\item[$(\mathrm{vii})$]  Let $n\in \mathbb{N}$ and let $p,q$ be odd primes.
If $\ord_{p^n}(2)|q$, then $\ord_{p^k}(2)=q$ for any $k\in\{1,\ldots,n\}$.
\end{itemize}
\end{lemma}

The proofs of (i) and (ii)  can be found in \cite[p.\ 43]{4}. Part (iii)
is  Theorem 4.4 proved by LeVeque 
in \cite[pp.\ 80--81]{3}. See also Theorem 4--6 in \cite[pp.~52--53]{4}.
Part (iv) immediately follows from (iii). Part (v) is a direct
consequence of (iii) for $p=2$. Finally, (vi) and (vii)  follow from
(iv). 

\begin{proposition}
\label{prop:1}
Let $n\in  \mathbb{N}$ and let $p$ be a prime. Then,
\begin{equation} \label{eq:21}
2^{p^{n-1}(p-1)}\equiv 1\pmod{p^{2n}}\quad\text{if and only if}\quad
2^{p-1}\equiv 1\pmod{p^{n+1}}.
\end{equation}
\end{proposition}

\begin{proof} Let us first assume  that $p$ is an odd prime.

Let $2^{p^{n-1}(p-1)}\equiv 1$ (mod $p^{2n}$). Applying (i), we obtain
\begin{equation}
\label{eq:22} \ord_{p^{2n}} (2) | p^{n-1}(p-1).
\end{equation}
Suppose that $2^{p-1}\not\equiv 1$ (mod $p^{n+1}$).
From (i), it now  follows that  $\ord_{p^{n+1}}(2)\nmid p-1$ and, by
(ii), $\ord_p(2)|p-1$. 
This means that $\ord_{p^{n+1}} (2)\ne \ord_{p} (2)$.
From (v), it follows that there exists an $s\in \{1,\ldots,n\}$ such
that $\ord_{p^{n+1}} (2)=p^s \ord_{p} (2)$. Hence, 
\begin{equation} \label{eq:23}
\ord_{p^{2n}} (2)=p^n \ord_{p^{n+1}} (2)= p^{n+s} \ord_{p} (2).
\end{equation}
Combining  \eqref{eq:23} and  \eqref{eq:22}, we  obtain  $p^{n+s}
\ord_{p} (2)|p^{n-1}(p-1)$. 
Since $s\in \{1,\ldots,n\}$ and $\ord_{p} (2)|p-1$, we have a
contradiction. 

Conversely, let $2^{p-1}\equiv 1$ (mod $p^{n+1}$). Then, by (i), we
have $\ord_{p^{n+1}} (2)|p-1$ and, using (vi), 
we obtain  $\ord_{p^{n+1}} (2)=\ord_{p} (2)$. Further, by (v), there
exist a $t\in \{0,\ldots,n-1\}$  such that 
\begin{equation}
\label{eq:24}
\ord_{p^{2n}} (2)=p^t \ord_{p^{n+1}} (2)= p^t \ord_{p} (2).
\end{equation} 
Since $t\le n-1$,  \eqref{eq:24} implies  $\ord_{p^{2n}} (2) | p^{n-1}
\ord_{p} (2)$, which, together with 
$\ord_{p} (2)|p-1$, yields $\ord_{p^{2n}} (2) | p^{n-1}(p-1)$. Hence,
$2^{p^{n-1}(p-1)}\equiv 1 $ (mod $p^{2n}$). 

Finally, let $p=2$. Then, for any $n\in \mathbb{N}$,
$ 2^{2^{n-1}}\not\equiv 1$ (mod $2^{2n}$)  and $2\not\equiv 1$ (mod $2^{n+1}$).
This means that, \eqref{eq:21} holds for $p=2$ as well. The proof is
now complete. 
\end{proof}

\begin{proposition}
\label{prop:2}
Let $n\in  \mathbb{N}\cup \{0\}$ and let $p$ be an odd prime. Then,
$$
2^{p-1}\equiv 1 \pmod{p^{n+1}}\quad\text{if and only if}\quad
\ord_{p^{n+1}}(2)=\ord_{p}(2).
$$
\end{proposition}

\begin{proof}
Let $2^{p-1}\equiv 1$ (mod $p^{n+1}$). Then, by  (i),
$\ord_{p^{n+1}}(2)|p-1$. 
Next, applying (iii), we obtain  $\ord_{p^{n+1}}(2)=p^s\ord_{p}(2)$
for some $s\in\{0,\ldots,n\}$. Suppose that $s\ne 0$. 
Then, $p^s\ord_{p}(2)|p-1$, which is a contradiction. Hence,
$\ord_{p^{n+1}}(2)=\ord_{p}(2)$. 

Conversely, let $\ord_{p^{n+1}}(2)=\ord_{p}(2)=h$. Then, by (ii),
$h|p-1$. 
This means that there exists a $k\in \mathbb{N}$ such that $p-1=hk$. Since
$2^h\equiv 1$ (mod $p^{n+1}$), we have 
$2^{p-1}=2^{hk}=(2^h)^k\equiv 1$ (mod $p^{n+1}$), as required.
\end{proof}

Now we are ready to prove Theorem~\ref{thm:1}.

\begin{proof}
First, we show that (A) implies (B). Let $p^{n+1}|M_q$. Then, $2^q\equiv 1$ (mod $p^{n+1}$) which
yields $\ord_{p^{n+1}}(2)|q$. Applying (vii), we obtain
$\ord_{p^{n+1}}(2)=\ord_{p}(2)=q$. This means, by Proposition~\ref{prop:2}, that $2^{p-1}\equiv 1$ (mod $p^{n+1}$)
and, using Proposition~\ref{prop:1}, we conclude that $p$ is a Wieferich prime of order $n$.

Next, we show that (B) implies (C). Assume that  $p$ is a Wieferich
prime of order $n$. 
Then, by Proposition~\ref{prop:1}, $2^{p-1}\equiv 1$ (mod $p^{n+1}$)
and, using (i), we get 
$\ord_{p^{n+1}}(2)|p-1$.
Next, from the basic assumption $p^n|M_q$, we obtain $2^q\equiv 1$
(mod $p^{n}$) and, by (i), we get $\ord_{p^{n}}(2)|q$.
Since $q$ is an odd prime and $\ord_{p^{n}}(2)\ne 1$, we have $\ord_{p^{n}}(2)=q$.
Hence, by (iv), $\ord_{p^{n+1}}(2)=pq$ or $\ord_{p^{n+1}}(2)=q$.
Suppose that $\ord_{p^{n+1}}(2)=pq$. Since $\ord_{p^{n+1}}(2)|p-1$, we
get $pq|p-1$, a contradiction. 

Finally, we show that (C) implies (A). Let
$\ord_{p^{n+1}}(2)=q$. Then, $2^q\equiv 1$ (mod $p^{n+1}$), 
which yields $p^{n+1}|M_q$. The proof is complete.
\end{proof}

\section{Acknowledgments}
The author thanks the  anonymous referee for carefully reading  the
manuscript. 


\begin{thebibliography}{9}

\bibitem{1}
T. Agoh, K. Dilcher, and L. Skula, Fermat quotients for composite
moduli, \emph{J. Number Theory} \textbf{66} (1997), 29--50. 

\bibitem{2}
K. Ireland and M. Rosen, \emph{A Classical Introduction to Modern Number
Theory}, 2nd edition, Springer-Verlag, 1992.

\bibitem{3}
W. J. LeVeque, \emph{Fundamentals of Number Theory}, Addison-Wesley,
1977.

\bibitem{4}
W. J. LeVeque, \emph{Topics in Number Theory}, Volume I, Dover
Publications, 2002.

\bibitem{5}
L. Skula, Prime power divisors of Mersenne numbers and Wieferich
primes of  higher order, \emph{Integers} \textbf{19} (2019), \#A19.

\bibitem{6}
N. J. A. Sloane et al., \emph{The On-Line Encyclopedia of Integer
  Sequences}, available at \url{https://oeis.org}, 2022.

\end{thebibliography}



\bigskip
\hrule
\bigskip

\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11A41; Secondary 11A07.

\noindent \emph{Keywords: } 
Mersenne number, Wieferich number, Wieferich prime.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000225},
\seqnum{A001220},
\seqnum{A001348}, and
\seqnum{A077816}.)

\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent
Received January 27 2022;
revised version received March 29 2022. 
Published in {\it Journal of Integer Sequences},
March 31 2022.

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