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\markboth{B.\,Cohen}{An analog of Wolstenholme's theorem}

\title[An analog of Wolstenholme's theorem]{An analog of Wolstenholme's theorem}

\author[Boaz Cohen]{Boaz Cohen\corrauth}

\address{Department of Computer Science, The Academic College of Tel-Aviv, Rabenu Yeruham St., P.O.B. 8\,401 Yaffo, 681\,821\,1, Israel}

\email{{\tt arctanx@gmail.com} (Boaz Cohen)}

\begin{abstract}
In this paper, we shall prove an analogous version of Wolstenholme's theorem, namely,
given a prime number $p\geqslant2$ and positive integers $a,b,m$ such that $p\nmid m$, we shall determine the maximal prime power $p^e$, which divides the numerator of the fraction
\[
\frac{1}{m}+\frac{1}{m+p^b}+\frac{1}{m+2p^b}+\ldots+\frac{1}{m+(p^a-1)p^b},
\]
when written in reduced form, with the exception of one case, where $p=2$, $b=1$, $m>1$ and $2^a\|m-1$. In this exceptional case, a lower bound for $e$ is given.
\end{abstract}

\keywords{Wolstenholme's theorem, Bauer's theorem, congruences, primes}

\ams{11A05, 11A07, 11A41}

\maketitle

\section{Introduction}
The classical Wolstenholme's theorem states that if $p>3$ is a prime number, then the numerator of the fraction
\[
1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{p-1}
\]
is divisible by $p^2$.
%
The proof can be found in~\cite[pp. 112--114]{Hardy&Wright}. Let us
illustrate this result. For $p=13$, we have
\[
1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}
 +\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}=
 \frac{86021}{27720},
\]
and here indeed $13^2\mid 86021$. We remark that for some primes, the numerator in Wolstenholme's theorem can be divisible by $p^3$, even if the fraction is written in reduced form. These prime numbers, which satisfy a stronger version of Wolstenholme's theorem, are called {Wolstenholme primes}. The only Wolstenholme primes known so far are $16843$ and $2124679$, though it is conjectured that their number is infinite. Wolstenholme's theorem has many generalizations and extensions. One of them is due to L. Carlitz~\cite{Carlitz} and it asserts as follows: if $m$ is an arbitrary integer, then for each prime $p>3$, the numerator of the fraction
\[
\frac{1}{mp+1}+\frac{1}{mp+2}+\ldots+\frac{1}{mp+(p-1)}
\]
is divisible by $p^2$. Further generalization can be found in~\cite{Mestrovic}.

In this paper, we shall prove the following analog of Wolstenholme's theorem: if $p$ is an odd prime number and $a,b,m$ are positive integers such that $p\nmid m$, then the numerator of the fraction
\[
\frac{1}{m}+\frac{1}{m+p^b}+\frac{1}{m+2p^b}+\ldots+\frac{1}{m+(p^a-1)p^b},
\]
when written in reduced form, is divisible by $p^a$ but not by $p^{a+1}$. For example, for $p=3$, $a=2$, $b=1$ and $m=2$, we obtain that
\[
\frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}+\frac{1}{17}+
\frac{1}{20}+\frac{1}{23}+\frac{1}{26}=\frac{3688785}{3131128},
\]
and here indeed $3^2\| 3688785$.

In the case of the prime $p=2$ one needs to distinguish cases according to whether $b\geqslant2$ or $b=1$: Suppose that $a,b,m$ are positive integers such that $2\nmid m$. If $b\geqslant2$, then the numerator of the fraction
\[
\frac{1}{m}+\frac{1}{m+2^b}+\frac{1}{m+2\cdot2^b}+\frac{1}{m+3\cdot2^b}
+\ldots+\frac{1}{m+(2^a-1)2^b},
\]
when written in reduced form, is divisible by $2^a$ but not by $2^{a+1}$. For example, for $a=4$, $b=2$ and $m=1$, we obtain that
\[
1+\frac{1}{5}+\frac{1}{9}+\frac{1}{13}+\frac{1}{17}+\ldots+\frac{1}{61}=
\frac{42155877944972752}{24142663793423175},
\]
and here indeed $2^4\|42155877944972752$.

When $b=1$, the results depend upon the value of $m$. If $m=1$, then the numerator of the fraction
\[
1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots+\frac{1}{1+2(2^a-1)}
\]
when written in reduced form, is divisible by $2^{2a}$ but not by $2^{2a+1}$. For example, for $a=4$, we obtain that
\[
1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\ldots+\frac{1}{31}=
\frac{10686452707072}{4512611027925},
\]
and here indeed $2^8\|10686452707072$.

If $m>1$ and $c$ is the positive integer such that
$2^c\|m-1$, then the maximal prime power $2^e$ which divides the numerator of the fraction
\[
\frac{1}{m}+\frac{1}{m+2}+\frac{1}{m+4}+\frac{1}{m+6}
+\ldots+\frac{1}{m+2(2^a-1)},
\]
when written in reduced form, is either $2^{a+\min\{a,c\}}$ if $a\neq c$, or at least $2^{2a+1}$ if $a=c$. As an example, for $a=4$ and $m=9$, we obtain that
\[
\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+\ldots+\frac{1}{39}=\frac{134154786738304}{166966608033225}.
\]
In this case, $2^3\|m-1$, so $c=3$, and indeed $2^7\|134154786738304$.
%==================================================================================================
\section{Preliminaries}
We begin with a brief overview of several concepts concerning elementary symmetric polynomials. Recall that the {kth elementary symmetric function} in $n$ variables is defined by
\[
e_k(x_1,x_2,\ldots,x_n)=\hspace{-0.9cm}\sum_{1\leqslant i_1<i_2<\cdots<i_k\leqslant n}\hspace{-0.8cm}x_{i_1}x_{i_2}\cdots x_{i_k}.
\]
For example, the elementary symmetric functions in $4$ variables are:
\[
\begin{split}
e_1(x_1,x_2,x_3,x_4)&=x_1+x_2+x_3+x_4 \\
e_2(x_1,x_2,x_3,x_4)&=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 \\
e_3(x_1,x_2,x_3,x_4)&=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 \\
e_4(x_1,x_2,x_3,x_4)&=x_1x_2x_3x_4.
\end{split}
\]
It is convenient to define $e_0(x_1,x_2,\ldots,x_n)=1$. Note that the $k$th elementary symmetric function is homogenous of degree $k$, that is,
\[
e_k(\lambda x_1,\lambda x_2,\ldots,\lambda x_n)=\lambda^ke_k(x_1,x_2,\ldots,x_n)
\]
for every $\lambda$.

If we expand the product $(z-x_1)(z-x_2)\cdots(z-x_n)$ into a polynomial in the variable $z$, we get the following relation:
\[
(z-x_1)(z-x_2)\cdots(z-x_n)=\sum_{k=0}^{n}(-1)^ke_k(x_1,x_2,\ldots,x_n)z^{n-k}.
\]
%
We continue with the following result, which will helpful in the sequel.
%
\begin{proposition}\label{prop_2_1}
Suppose that $p$ is a prime number and let $t$ be a positive integer
such that $p\nmid t$. If $a,b$ are integers such that $0\leqslant
b\leqslant a$ and $t\leqslant p^{a-b}$,  then
\[
p^{a-b}\|\binom{p^a}{p^bt}.
\]
\end{proposition}
%
\begin{proof}
If $a=b$, then $t=1$ and the result certainly holds. So assume that
$a>b$. For a positive number $n$ define
$\nu_p(n)$ to be the multiplicity of $p$ in $n$, namely $\nu_p(n)$
is the non-negative integer such that $p^{\nu_p(n)}\|n$.
By~\cite[pp. 90--91]{Leveq}, we know that
\[
\nu_{p}(n!)=\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor
    +\left\lfloor\frac{n}{p^3}\right\rfloor+\ldots
\]
Here the lower square-bracket notation $\lfloor x\rfloor$ denotes
the largest integer not exceeding the real number $x$. Hence
\[
\begin{split}
\nu_p((p^bt)!)&=\left\lfloor\frac{p^bt}{p}\right\rfloor+\left\lfloor\frac{p^bt}{p^2}\right\rfloor
                          +\ldots+\left\lfloor\frac{p^bt}{p^b}\right\rfloor+\left\lfloor\frac{p^bt}{p^{b+1}}\right\rfloor
                          +\left\lfloor\frac{p^bt}{p^{b+2}}\right\rfloor+\ldots \\
                       &=(p^{b-1}+p^{b-2}+\ldots+p+1)t
                          +\left\lfloor\frac{t}{p}\right\rfloor+\left\lfloor\frac{t}{p^2}\right\rfloor+\ldots \\
                       &=t\frac{p^b-1}{p-1}+\nu_p(t!).
\end{split}
\]
Now, for a real number $x$ and an integer $n$, we have $\lfloor
n+x\rfloor=n+\lfloor x\rfloor$ and $\lfloor-x\rfloor=-1-\lfloor
x\rfloor$ if $x$ is not an integer (see~\cite[p. 90]{Leveq}).
Therefore, as $a>b$, we get
\[
\begin{split}
\nu_p((p^{a-b}-t)!)&=\left\lfloor\frac{p^{a-b}-t}{p}\right\rfloor+\left\lfloor\frac{p^{a-b}-t}{p^2}\right\rfloor
                               +\ldots+\left\lfloor\frac{p^{a-b}-t}{p^{a-b}}\right\rfloor \\
                           % &=&(p^{\al-\bt-1}+\left[-\frac{t}{p}\right])+(p^{\al-\bt-2}+\left[-\frac{t}{p^{2}}\right])
                           %   +\ldots+(1+\left[-\frac{t}{p^{\al-\bt}}\right]) \\
                            &=(p^{a-b-1}+\ldots+p+1)+\left\lfloor-\frac{t}{p}\right\rfloor+\left\lfloor-\frac{t}{p^2}\right\rfloor
                               +\ldots+\left\lfloor-\frac{t}{p^{a-b}}\right\rfloor \\
                            &=\frac{p^{a-b}-1}{p-1}-\underbrace{(1+1+\ldots+1)}_{a-b\ \mathrm{times}}-\left(\left\lfloor\frac{t}{p}\right\rfloor+\left\lfloor\frac{t}{p^2}\right\rfloor
                               +\ldots+\left\lfloor\frac{t}{p^{a-b}}\right\rfloor\right) \\
                            &=\frac{p^{a-b}-1}{p-1}-(a-b)-\nu_p(t!).
\end{split}
\]
Observe that the third equality follows from the fact that $p\nmid
t$. From the first formula we deduce that
\[
\nu_p((p^a)!)=1\cdot\frac{p^a-1}{p-1}+\nu_p(1)=\frac{p^a-1}{p-1},
\]
and the first and the second formula yield:
\[
\begin{split}
\nu_p((p^a-p^bt)!)&=\nu_p((p^b(p^{a-b}-t))!) \\
                               &=(p^{a-b}-t)\frac{p^b-1}{p-1}+\nu_{p}((p^{a-b}-t)!) \\
                               &=(p^{a-b}-t)\frac{p^b-1}{p-1}+\frac{p^{a-b}-1}{p-1}-(a-b)-\nu_p(t!).
\end{split}
\]
Hence
\[
\nu_p\left(\binom{p^a}{p^bt}\right)
        =\nu_p((p^a)!)-\nu_p((p^bt)!)-\nu_p((p^a-p^bt)!)
        =a-b,
\]
and the proof is complete.
\end{proof}
%==================================================================================================
\section{The value of $e_k(0,1,2,\ldots,p^a-1)$ modulo $p^a$}
Let $p$ be a prime number and $a$ a positive integer. Our aim in this section is to compute the values of the expressions $e_k(0,1,2,\ldots,p^a-1)$ modulo $p^a$, where $e_k$ denotes the $k$th elementary symmetric function. As we shall see, the values of these expressions will play an important role in our analysis. To do so, we shall use a specific generalization of Lagrange's Indeterminate Congruence Theorem. Recall that Lagrange's Indeterminate Congruence Theorem claims that
\[
x(x-1)(x-2)\cdots(x-p+1)\equiv x^p-x\pmod{p},
\]
where the congruence indicates that the corresponding coefficients
of the polynomials are congruent modulo $p$. A generalization of Lagrange's Indeterminate
Congruence Theorem was given by Bauer (see~theorems~126 and 127 of \cite{Hardy&Wright}) and later by Vandiver. Equations~(8') and~(9') of~\cite{Vandiver}, listed below, are the ones we need for our analysis:
%
\begin{theorem}[Vandiver \cite{Vandiver}]\label{tm_3_1}
Let $p$ be a prime number and let $a$ be a positive integer.
\begin{enumerate}
  \item[\textup{(a)}] If $p>2$, then
  \[
x(x-1)(x-2)\cdots(x-(p^a-1))\equiv\left(x^p-x\right)^{p^{a-1}}\pmod{p^a}.
\]
  \item[\textup{(b)}] If $p=2$ and $a\geqslant2$ , then
\[
x(x-1)(x-2)\cdots(x-(2^a-1))\equiv \big((x^2-x)^2-2(x^2-x)\big)^{2^{a-2}}\pmod{2^a}.
\]
\end{enumerate}
\end{theorem}
%
\begin{proposition}\label{prop_3_2}
Suppose that $p$ is an odd prime number, $a$ is a positive integer and $0\leqslant k\leqslant p^a$ is an integer. Then
\[
e_k(0,1,2,\ldots,p^a-1)\equiv(-1)^k(-1)^{k/(p-1)}\binom{p^{a-1}}{\frac{k}{p-1}}\pmod{p^a}
\]
if $p-1\mid k$ and $0\leqslant k\leqslant p^a-p^{a-1}$, and $e_k(0,1,2,\ldots,p^a-1)\equiv0\pmod{p^a}$ otherwise.
\end{proposition}
%
\begin{proof}
On the one hand, by Theorem~\ref{tm_3_1}(a)
\[
\begin{split}
x(x-1)(x-2)\cdots(x-(p^a-1))
&\equiv (x^p-x)^{p^{a-1}} \\
&=\sum_{i=0}^{p^{a-1}}\binom{p^{a-1}}{i}\left(x^p\right)^{p^{a-1}-i}(-x)^{i} \\
&=\sum_{i=0}^{p^{a-1}}\binom{p^{a-1}}{i}(-1)^{i}x^{p^a-i(p-1)}\pmod{p^a},
\end{split}
\]
so the powers of $x$ in the sum are of the form $x^{p^a-k}$, where $p-1\mid k$ and $0\leqslant k\leqslant(p-1)p^{a-1}=p^a-p^{a-1}$. On the other hand,
\[
x(x-1)(x-2)\cdots(x-(p^a-1))=\sum_{k=0}^{p^a}(-1)^ke_k(0,1,2,\ldots,p^a-1)x^{p^a-k}.
\]
Now, comparing the corresponding exponents yields
\[
e_k(0,1,2,\ldots,p^a-1)\equiv(-1)^k(-1)^{k/(p-1)}\binom{p^{a-1}}{\frac{k}{p-1}}\pmod{p^a}
\]
if $p-1\mid k$ and $0\leqslant k\leqslant p^a-p^{a-1}$, and $e_k(0,1,2,\ldots,p^a-1)\equiv0\pmod{p^a}$ otherwise, as required.
\end{proof}
%
\begin{proposition}\label{prop_3_3}
Suppose that $a\geqslant3$ and $0\leqslant k\leqslant 2^a$ are integers. Then
\[
e_k(0,1,2,\ldots,2^a-1)\equiv
\begin{cases}
  1, & \hbox{if $k=0$}\\
  2^{a-1}, & \hbox{if $k=1$}\\
  2^{a-2}, & \hbox{if $k=2$}\\
  0, & \hbox{if $2<k\leqslant 2^{a-1}$ and $k$ is odd}\\
  \binom{2^{a-1}}{k} & \hbox{if $2<k\leqslant 2^{a-1}$ and $k$ is even}\\
  2^{a-1}, & \hbox{if $2^{a-1}<k\leqslant2^{a-1}+2$}\\
  0, & \hbox{if $2^{a-1}+2<k\leqslant 2^a$}\\
\end{cases}\pmod{2^a}.
\]
\end{proposition}
%
\begin{proof}
Set $f(x)=x(x-1)(x-2)\cdots(x-(2^a-1))$. Note that by Theorem~\ref{tm_3_1}(b)
\[
  f(x)
  \equiv\left((x^2-x)^2-2(x^2-x)\right)^{2^{a-2}}
  =\sum_{k=0}^{2^{a-2}}\binom{2^{a-2}}{k}(-2)^k(x^2-x)^{2^{a-1}-k}\pmod{2^a}.
\]
We claim that
\[
2^a\mid2^k\binom{2^{a-2}}{k}
\]
for every $3\leqslant k\leqslant 2^a$. Let $t$ be the non-negative integer such that $2^t\|k$. By Proposition~\ref{prop_2_1},
\[
2^{a-2-t}\|\binom{2^{a-2}}{k}.
\]
Hence, in order to prove our claim, it suffices to prove that $a\leqslant k+a-2-t$, that is, $t+2\leqslant k$. Indeed, if $0\leqslant t\leqslant1$, then $t+2\leqslant3\leqslant k$, and if $t\geqslant2$, then $t+2\leqslant 2^t\leqslant k$, as required. Since $a\geqslant 3$, it follows that
\[
\begin{split}
  f(x)&\equiv(x^2-x)^{2^{a-1}}-2\binom{2^{a-2}}{1}(x^2-x)^{2^{a-1}-1}+4\binom{2^{a-2}}{2}(x^2-x)^{2^{a-1}-2}\\
      &=(x^2-x)^{2^{a-1}}-2^{a-1}(x^2-x)^{2^{a-1}-1}+2^{a-1}(2^{a-2}-1)(x^2-x)^{2^{a-1}-2}\\
      &\equiv(x^2-x)^{2^{a-1}}+2^{a-1}(x^2-x)^{2^{a-1}-1}+2^{a-1}(x^2-x)^{2^{a-1}-2}\\
      &=(x^2-x)^{2^{a-1}}+2^{a-1}
      \big(\underbrace{(x^2-x)^{2^{a-1}-1}+(x^2-x)^{2^{a-1}-2}}_{g(x)}\big)\pmod{2^a}.\\
\end{split}
\]
We shall prove that
\[
g(x)\equiv
x^{2^a-2}+\sum_{j=1}^{2^{a-2}}x^{2^a-(2j+1)}+x^{2^{a-1}-2}
\pmod{2}
\]
for every $a\geqslant3$. Set $A=2^{a-1}$. Note that $A\geqslant4$ and
\[
\begin{split}
  g(x)&=\sum_{i=0}^{A-1}(-1)^i\binom{A-1}{i}x^{2A-2-i}
      +\sum_{j=0}^{A-2}(-1)^j\binom{A-2}{j}x^{2A-4-j}\\
      &=x^{2A-2}-(A-1)x^{2A-3}+\sum_{i=2}^{A-1}(-1)^i\binom{A-1}{i}x^{2A-2-i}\\
      &\quad+\sum_{j=0}^{A-3}(-1)^j\binom{A-2}{j}x^{2A-4-j}+x^{A-2}\\
      &=x^{2A-2}-(A-1)x^{2A-3}+
      \sum_{j=0}^{A-3}(-1)^j\left(\binom{A-1}{j+2}+\binom{A-2}{j}\right)x^{2A-4-j}+x^{A-2}\\
      &\equiv
      x^{2A-2}+x^{2A-3}+
      \sum_{j=0}^{A-3}\left(\binom{A-1}{j+2}+\binom{A-2}{j}\right)x^{2A-4-j}+x^{A-2}\pmod{2}.
\end{split}
\]
We shall prove now that
\[
\binom{A-1}{j+2}+\binom{A-2}{j}\equiv
\begin{cases}
  0, & \hbox{if $j$ is even}\\
  1, & \hbox{if $j$ is odd}
\end{cases}\pmod{2},
\]
for every $0\leqslant j\leqslant A-3$. To do so, we shall use Lucas's theorem~\cite{Fine}, which claims, in particular for the prime $p=2$, that if $m,n$ are non-negative integers and if
\[
\begin{split}
  m&=m_k2^k+\ldots+m_22^2+m_12+m_0\\
  n&=n_k2^k+\ldots+n_22^2+n_12+n_0\\
\end{split}
\]
are the base-2 expansions of $m$ and $n$, respectively, then the following congruence relation holds:
\[
\binom{m}{n}\equiv\binom{m_k}{n_k}\cdots\binom{m_2}{n_2}\binom{m_1}{n_1}\binom{m_0}{n_0}\pmod{2},
\]
where we use the convention that $\binom{m}{n}=0$ whenever $m<n$. In our case, the base-2 expansions of $A-1$ and of $A-2$ are
\[
A-1=2^{a-1}-1=2^{a-2}+\ldots+2^2+2+1
\]
and
\[
A-2=2^{a-1}-2=2^{a-2}+\ldots+2^2+2.
\]
If
\[
j+2=m_{a-2}2^{a-2}+\ldots+m_22^2+m_12+m_0
\]
and
\[
j=n_{a-2}2^{a-2}+\ldots+n_22^2+n_12+n_0
\]
are the base-2 expansions of $j$ and $j+2$, then
\[
\binom{A-1}{j+2}\equiv\binom{1}{m_{a-2}}\cdots\binom{1}{m_2}\binom{1}{m_1}\binom{1}{m_0}\pmod{2}
\]
and
\[
\binom{A-2}{j}\equiv\binom{1}{n_{a-2}}\cdots\binom{1}{n_2}\binom{1}{n_1}\binom{0}{n_0}\pmod{2}.
\]
Since each   of $m_0,m_1,\ldots,m_{a-2}$ and $n_0,n_1,\ldots,n_{a-2}$ is either $0$ or $1$, it follows that $\binom{A-1}{j+2}\equiv 1\pmod{2}$. If $j$ is odd, then $n_0=1$, so $\binom{A-2}{j}\equiv 0\pmod{2}$, and if $j$ is even, then $n_0=0$, so $\binom{A-2}{j}\equiv 1\pmod{2}$. Therefore,
\[
\binom{A-1}{j+2}+\binom{A-2}{j}\equiv
\begin{cases}
  0, & \hbox{if $j$ is even}\\
  1, & \hbox{if $j$ is odd}
\end{cases}
\pmod{2},
\]
as required. Hence
\[
\begin{split}
  g(x)&\equiv
  x^{2A-2}+x^{2A-3}+
      \sum_{{0\leqslant j\leqslant A-3}\atop\hbox{\tiny $j$ is odd}}x^{2A-4-j}+x^{A-2}\\
  &=x^{2A-2}+x^{2A-3}+
    x^{2A-5}+x^{2A-7}+\cdots+x^{2A-(A+1)}+x^{A-2}\\
  &= x^{2A-2}+
    \sum_{j=1}^{A/2}x^{2A-(1+2j)}+x^{A-2}\\
  &=x^{2^a-2}+\sum_{j=1}^{2^{a-2}}x^{2^a-(2j+1)}+x^{2^{a-1}-2}
\pmod{2},
\end{split}
\]
as claimed. Therefore,
\[
\begin{split}
  f(x)&\equiv
(x^2-x)^{2^{a-1}}+2^{a-1}\left(x^{2^a-2}+\sum_{j=1}^{2^{a-2}}x^{2^a-(2j+1)}+x^{2^{a-1}-2}\right)\\
&\equiv\sum_{i=0}^{2^{a-1}}(-1)^i\binom{2^{a-1}}{i}x^{2^a-i}\\
&\quad+2^{a-1}\left(x^{2^a-2}+\sum_{j=1}^{2^{a-2}}x^{2^a-(2j+1)}+x^{2^a-(2^{a-1}+2)}\right)
\pmod{2^a}.
\end{split}
\]
Now, given $0\leqslant k\leqslant2^a$, let $c_k$ be the coefficient of the term $x^{2^a-k}$ in $f(x)$. By the above expression we deduce that
\[
c_k\equiv
\begin{cases}
  1, & \hbox{if $k=0$}\\
  -\binom{2^{a-1}}{1}, & \hbox{if $k=1$}\\
  \binom{2^{a-1}}{2}+2^{a-1}, & \hbox{if $k=2$}\\
  2^{a-1}-\binom{2^{a-1}}{k}, & \hbox{if $2<k\leqslant 2^{a-1}$ and $k$ is odd}\\
  \binom{2^{a-1}}{k}, & \hbox{if $2<k\leqslant 2^{a-1}$ and $k$ is even}\\
  2^{a-1}, & \hbox{if $2^{a-1}<k\leqslant2^{a-1}+2$}\\
  0, & \hbox{if $2^{a-1}+2<k\leqslant 2^a$}\\
\end{cases}\pmod{2^a}.
\]
Note that
\[
-\binom{2^{a-1}}{1}=-2^{a-1}\equiv 2^{a-1}\pmod{2^a}
\]
and
\[
\binom{2^{a-1}}{2}+2^{a-1}=2^{a-2}(2^{a-1}-1)+2^{a-1}=2^{2a-3}+2^{a-2}\equiv2^{a-2}\pmod{2^a}.
\]
In addition, if $k$ is odd, then by Proposition~\ref{prop_2_1} it follows that $\binom{2^{a-1}}{k}=2^{a-1}b$, where $2\nmid b$. Thus
\[
2^{a-1}-\binom{2^{a-1}}{k}=2^{a-1}(1-b)\equiv 2^a\cdot\frac{1-b}{2}\equiv0\pmod{2^a}.
\]
Therefore,
\[
c_k\equiv
\begin{cases}
  1, & \hbox{if $k=0$}\\
  2^{a-1}, & \hbox{if $k=1$}\\
  2^{a-2}, & \hbox{if $k=2$}\\
  0, & \hbox{if $2<k\leqslant 2^{a-1}$ and $k$ is odd}\\
  \binom{2^{a-1}}{k}, & \hbox{if $2<k\leqslant 2^{a-1}$ and $k$ is even}\\
  2^{a-1}, & \hbox{if $2^{a-1}<k\leqslant2^{a-1}+2$}\\
  0, & \hbox{if $2^{a-1}+2<k\leqslant 2^a$}\\
\end{cases}\pmod{2^a}.
\]
Since
\[
f(x)=\sum_{k=0}^{2^a}(-1)^ke_k(0,1,2,\ldots,2^a-1)x^{2^a-k},
\]
it follows that $(-1)^ke_k(0,1,2,\ldots,2^a-1)\equiv c_k\pmod{2^a}$, that is, $e_k(0,1,2,\ldots,2^a-1)\equiv (-1)^kc_k\pmod{2^a}$. By noting that $-2^{a-1}\equiv2^{a-1}\pmod{2^a}$, we deduce that
\[
e_k(0,1,2,\ldots,2^a-1)\equiv
\begin{cases}
  1, & \hbox{if $k=0$}\\
  2^{a-1}, & \hbox{if $k=1$}\\
  2^{a-2}, & \hbox{if $k=2$}\\
  0, & \hbox{if $2<k\leqslant 2^{a-1}$ and $k$ is odd}\\
  \binom{2^{a-1}}{k}, & \hbox{if $2<k\leqslant 2^{a-1}$ and $k$ is even}\\
  2^{a-1}, & \hbox{if $2^{a-1}<k\leqslant2^{a-1}+2$}\\
  0, & \hbox{if $2^{a-1}+2<k\leqslant 2^a$}\\
\end{cases}\pmod{2^a},
\]
as required.
\end{proof}
%==================================================================================================
\section{Our main results}
%
In this section, we shall prove the four main theorems of this paper. All four proofs are based on the following observation: The fraction $s=\frac{1}{a_1}+\ldots+\frac{1}{a_k}$ can be written as $N/D$, where $N=e_{k-1}(a_1,\ldots,a_k)$, $D=a_1\cdots a_k$, and $e_{k-1}$ denotes the $(k-1)$th elementary symmetric function. Now, if $p\nmid D$ and $p^e\|N$, then  $p^e$ is also the maximal prime power which divides the numerator of the fraction $s$, when written in reduced form. Therefore, it suffices to determine the maximal power $e$ for which $p^e\|N$. This goal will be accomplished by using the results proven in Section~3 according to   different cases. We begin with the case where $p$ is an odd prime number.
%
\begin{theorem}\label{tm_4_1}
If $p$ is an odd prime number and $a,b,m$ are positive integers such that $p\nmid m$, then the maximal prime power $p^e$ which divides the numerator of the fraction
\[
\frac{1}{m}+\frac{1}{m+p^b}+\frac{1}{m+2p^b}+\ldots+\frac{1}{m+(p^a-1)p^b},
\]
when written in reduced form, is $p^a$.
\end{theorem}
%
\begin{proof}
Suppose that
\[
\frac{1}{m}+\frac{1}{m+p^b}+\frac{1}{m+2p^b}+\ldots+\frac{1}{m+(p^a-1)p^b}=\frac{n}{d},
\]
where $\gcd(n,d)=1$. We claim that $p^a\|n$. Set
\[
\begin{split}
  N&=e_{p^a-1}(m,m+p^b,m+2p^b,\cdots,m+(p^a-1)p^b)\\
  D&=m(m+p^b)(m+2p^b)\cdots(m+(p^a-1)p^b),
\end{split}
\]
where $e_{p^a-1}$ denotes the $(p^a-1)$th elementary symmetric function. Note that $n/d=N/D$, but the fraction $N/D$ is not necessary in its reduced form. Nevertheless,
since $\gcd(m,p)=1$ and $b\geqslant1$, it follows that $p\nmid D$, so $p^a\|n$ if and only if $p^a\|N$. Hence, it suffices to prove that $p^a\|N$. To do so, consider the polynomial
\[
\begin{split}
  f(x)&=(x-m)(x-(m+p^b))(x-(m+2p^b))\cdots (x-(m+(p^a-1)p^b))\\
      &=(x-m)(x-m-p^b)(x-m-2p^b)\cdots (x-m-(p^a-1)p^b)\\
      &=\sum_{k=0}^{p^a}(-1)^ke_k(0,p^b,2p^b,\ldots,(p^a-1)p^b)(x-m)^{p^a-k}\\
      &=\sum_{k=0}^{p^a}(-1)^kp^{bk}e_k(0,1,2,\ldots,p^a-1)(x-m)^{p^a-k}.
\end{split}
\]
We claim that $p^{bk}e_k(0,1,2,\ldots,p^a-1)\equiv0\pmod{p^{a+1}}$ for every $1\leqslant k\leqslant p^a$.

If $k=1$, then indeed
\[
p^be_1(0,1,2,\ldots,p^a-1)=p^b\cdot\frac{(p^a-1)p^a}{2}=\frac{p^a-1}{2}\cdot p^{a+b}\equiv0\pmod{p^{a+1}},
\]
as required.

Next, suppose that $k\geqslant 2$. If either $p^a-p^{a-1}<k\leqslant p^a$ or $p-1\nmid k$, then by Proposition~\ref{prop_3_2} $e_k(0,1,2,\ldots,p^a-1)\equiv0\pmod{p^a}$, so $p^{bk}e_k(0,1,2,\ldots,p^a-1)\equiv0\pmod{p^{a+1}}$, as required.

It remains only to deal with $k$ such that $2\leqslant k\leqslant p^a-p^{a-1}$ and $p-1\mid k$. Let $t$ be the non-negative integer such that $p^t\|k$. Then $p^t\|\frac{k}{p-1}$ and by Proposition~\ref{prop_2_1},
\[
p^{a-1-t}\mid \binom{p^{a-1}}{\frac{k}{p-1}}.
\]
Since by Proposition~\ref{prop_3_2}
\[
e_k(0,1,2,\ldots,p^a-1)\equiv
(-1)^k(-1)^{k/(p-1)}\binom{p^{a-1}}{\frac{k}{p-1}}\pmod{p^a},
\]
it follows that
\[
p^{bk}e_k(0,1,2,\ldots,p^a-1)\equiv
0\pmod{p^{bk+a-1-t}}.
\]
Thus, in order to prove our claim it suffices to prove that $bk+a-1-t\geqslant a+1$, that is, $bk\geqslant t+2$. Indeed, if $t=0$, then $bk\geqslant2=t+2$ since $k\geqslant2$, and if $t\geqslant1$, then $bk\geqslant bp^t\geqslant 3^t\geqslant t+2$, so our claim is proved.

It follows by the above discussion that
\[
f(x)\equiv (x-m)^{p^a}\pmod{p^{a+1}}.
\]
On the one hand, the coefficient of $x$ in $f(x)$ is
\[
N=e_{p^a-1}(m,m+p^b,m+2p^b,\cdots m+(p^a-1)p^b).
\]
Now, the coefficient of $x$ in $(x-m)^{p^a}$ is $p^a(-1)^{p^a-1}m^{p^a-1}$.  Therefore, $N\!\equiv\! p^a(-1)^{p^a-1}$\linebreak $\cdot m^{p^a-1}\pmod{p^{a+1}}$. Since $p\nmid m$, it follows that $p^a\|N$, as required.
\end{proof}
%
The next three theorems handle the case where $p=2$. The first theorem deals with the case when $b\geqslant2$, while the second and the third theorem treat the case when $b=1$.
%
\begin{theorem}\label{tm_4_2}
If $a,b,m$ are positive integers such that $2\nmid m$ and $b\geqslant2$, then the maximal prime power $2^e$ which divides the numerator of the fraction
\[
\frac{1}{m}+\frac{1}{m+2^b}+\frac{1}{m+2\cdot2^b}+\frac{1}{m+3\cdot2^b}
+\ldots+\frac{1}{m+(2^a-1)2^b},
\]
when written in reduced form, is $2^a$.
\end{theorem}
%
\begin{proof}
Set
\[
  N=e_{2^a-1}(m,m+2^b,m+2\cdot2^b,\cdots,m+(2^a-1)2^b),
\]
where $e_{2^a-1}$ denotes the $(2^a-1)$th elementary symmetric function. As in the proof of Theorem~\ref{tm_4_1}, it suffices to prove that $2^a\|N$. To do so, consider the polynomial
\[
\begin{split}
  f(x)&=(x-m)(x-(m+2^b))(x-(m+2\cdot2^b))\cdots (x-(m+(2^a-1)2^b))\\
      &=(x-m)(x-m-2^b)(x-m-2\cdot2^b)\cdots (x-m-(2^a-1)2^b)\\
      &=\sum_{k=0}^{2^a}(-1)^ke_k(0,2^b,2\cdot2^b,\ldots,(2^a-1)2^b)(x-m)^{2^a-k}\\
      &=\sum_{k=0}^{2^a}(-1)^k2^{bk}e_k(0,1,2,\ldots,2^a-1)(x-m)^{2^a-k}.\\
\end{split}
\]
We claim that if $b\geqslant2$, then $2^{bk}e_k(0,1,2,\ldots,2^a-1)\equiv0\pmod{2^{a+1}}$ for every $1\leqslant k\leqslant 2^a$. For simplicity, let us denote $e_k(0,1,2,\ldots,2^a-1)$ by $e_k$. If $k=1$, then $2^be_1\equiv0\pmod{2^{b+a-1}}$ by Proposition~\ref{prop_3_3}, and since $b\geqslant2$, it follows that $a+1\leqslant b+a-1$, so $2^be_1\equiv 0\pmod{2^{a+1}}$, as claimed. If $k=2$, then $2^{2b}e_2\equiv 0\pmod{2^{2b+a-2}}$ by Proposition~\ref{prop_3_3}, and since $b\geqslant2$, it follows that $a+1\leqslant 2b+a-2$, so $2^{2b}e_2\equiv 0\pmod{2^{a+1}}$, as claimed. Next, suppose that $2<k\leqslant2^{a-1}$. If $2\nmid k$, then $2^{bk}e_k\equiv0\pmod{2^{bk+a}}$ by Proposition~\ref{prop_3_3}, and since $a+1<bk+a$, it follows that $2^{kb}e_k\equiv 0\pmod{2^{a+1}}$, as claimed. If $2\mid k$, let $t$ be the positive integer such that $2^t\|k$. By propositions~\ref{prop_2_1} and \ref{prop_3_3}, we deduce that  $2^{kb}e_k\equiv0\pmod{2^{bk+a-1-t}}$. Since $t\geqslant1$ and $b\geqslant2$, it follows that $2+t<2\cdot2^{t}\leqslant bk$, that is, $a+1< bk+a-1-t$, so $2^{bk}e_k\equiv 0\pmod{2^{a+1}}$, as claimed. If $2^{a-1}<k\leqslant 2^{a-1}+2$, then $2^{bk}e_k\equiv0\pmod{2^{bk+a-1}}$
by Proposition~\ref{prop_3_3}, and since $a+1\leqslant bk+a-1$, it follows that $2^{bk}e_k\equiv 0\pmod{2^{a+1}}$, as claimed. Finally, if $2^{a-1}+2<k\leqslant 2^a$, then $2^{bk}e_k\equiv0\pmod{2^{bk+a}}$ by Proposition~\ref{prop_3_3}, so $2^{bk}e_k\equiv 0\pmod{2^{a+1}}$, as claimed.

To conclude, we proved that if $b\geqslant2$, then $2^{bk}e_k\equiv0\pmod{2^{a+1}}$ for every $1\leqslant k\leqslant 2^a$. Consequently, $f(x)\equiv(x-m)^{2^a}\pmod{2^{a+1}}$. As in the proof of Theorem~\ref{tm_4_1}, it follows that $N\equiv 2^a(-1)^{2^a-1}m^{2^a-1}\equiv 2^am^{2^a-1}\pmod{2^{a+1}}$, and since $2\nmid m$, we deduce that $2^a\|N$, as required.
\end{proof}

For the prime $p=2$, we are left with the case $b=1$.
This case will be handled according to two sub-cases: $m=1$ and $m>1$.
%
\begin{theorem}\label{tm_4_3}
If $a$ is a positive integer, then the maximal prime power $2^e$ which divides the numerator of the fraction
\[
1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}
+\ldots+\frac{1}{1+2(2^a-1)},
\]
when written in reduced form, is $2^{2a}$.
\end{theorem}
%
\begin{proof}
Note that
\[
\begin{split}
1+&\frac{1}{3}+\frac{1}{5}
+\ldots+\frac{1}{1+2(2^a-1)}\\
&=\sum_{1\leqslant s<2^{a+1}\atop\hbox{\tiny $s$ is odd}}\frac{1}{s}
=\sum_{1\leqslant s<2^a\atop\hbox{\tiny $s$ is odd}}\left(\frac{1}{s}+\frac{1}{2^{a+1}-s}\right)
=2^{a+1}\sum_{1\leqslant s<2^a\atop\hbox{\tiny $s$ is odd}}\frac{1}{s(2^{a+1}-s)}.
\end{split}
\]
Hence, in order to prove the claim it suffices to prove that the numerator of the fraction
\[
\sum_{1\leqslant s<2^a\atop\hbox{\tiny $s$ is odd}}\frac{1}{s(2^{a+1}-s)},
\]
when written in reduced form, is divisible by $2^{a-1}$ but not by $2^a$. For simplicity, for every $0\leqslant k\leqslant 2^{a-1}$, let us denote the following $k$th elementary symmetric expression:
\[
e_k(1(2^{a+1}-1),3(2^{a+1}-3),\ldots,(2^a-1)(2^a+1)),
\]
by $e_k$. As in the proof of Theorem~\ref{tm_4_1}, it suffices to prove that $2^{a-1}\|e_{2^{a-1}-1}$. To do so, consider the polynomial
\[
f(x)=\prod_{1\leqslant s<2^{a+1}\atop\hbox{\tiny $s$ is odd}}\hspace*{-0.2cm}(x-s).
\]
On the one hand, by Bauer's Theorem~\cite[p. 127]{Hardy&Wright},
\[
  f(x)\equiv(x^2-1)^{2^{a-1}}
  =\sum_{k=0}^{2^{a-1}}\binom{2^{a-1}}{k}(-1)^kx^{2^a-2k}\pmod{2^{a+1}}.
\]
On the other hand,
\[
\begin{split}
  f(x)&=
  \prod_{1\leqslant s<2^{a+1}\atop\hbox{\tiny $s$ is odd}}\hspace*{-0.2cm}(x-s)=
  \prod_{1\leqslant s<2^a\atop\hbox{\tiny $s$ is odd}}\hspace*{-0.2cm}(x-s)(x-(2^{a+1}-s))\\
  &\equiv\prod_{1\leqslant s<2^a\atop\hbox{\tiny $s$ is odd}}(x^2+s(2^{a+1}-s))
  =\sum_{k=0}^{2^{a-1}}e_kx^{2^a-2k}\pmod{2^{a+1}}.
\end{split}
\]
By comparing the coefficient of $x^{2^a-2k}$ we obtain that
\[
e_k\equiv\binom{2^{a-1}}{k}(-1)^k\pmod{2^{a+1}},
\]
for every $0\leqslant k\leqslant 2^{a-1}$. In particular, $2^{a-1}\|e_{2^{a-1}-1}$, as required.
\end{proof}

\begin{theorem}\label{tm_4_4}
Suppose that $a$ is a positive integer and $m>1$ is an odd integer. In addition, let $c$ be the positive integer such that $2^c\|m-1$. Then the maximal prime power $2^e$ which divides the numerator of the fraction
\[
\frac{1}{m}+\frac{1}{m+2}+\frac{1}{m+4}+\frac{1}{m+6}
+\ldots+\frac{1}{m+2(2^a-1)},
\]
when written in reduced form, is either $2^{a+\min\{a,c\}}$ if $a\neq c$, or at least $2^{2a+1}$ if $a=c$.
\end{theorem}
%
\begin{proof}
Consider the polynomial
\[
f(x)=(x-m)(x-(m+2))(x-(m+4))\cdots (x-(m+2(2^a-1))),
\]
and let $u$ be the odd positive integer such that $m=1+2^cu$. As in the proof of Theorem~\ref{tm_4_1},
\[
f(x)=\sum_{k=0}^{2^a}(-1)^k2^ke_k(x-m)^{2^a-k},
\]
where we denote $e_k=e_k(0,1,2,\ldots,2^a-1)$ for simplicity. Note that in order to prove our claim, it suffices to find the maximal power of $2$ which divides the coefficient of $x$ in $f(x)$. Since
\[
\begin{split}
  f(x)
  &=\sum_{k=0}^{2^a}\left((-1)^k2^ke_k\sum_{j=0}^{2^a-k}\binom{2^a-k}{j}x^j(-m)^{2^a-k-j}\right)\\
  &=\sum_{k=0}^{2^a}\sum_{j=0}^{2^a-k}(-1)^{2^a-j}m^{2^a-k-j}2^ke_k\binom{2^a-k}{j}x^j,
\end{split}
\]
it follows that the coefficient $A$ of $x$ in $f(x)$ is
\[
\begin{split}
A=\sum_{k=0}^{2^a}&(-1)^{2^a-1}m^{2^a-k-1}2^ke_k\binom{2^a-k}{1}
%&=-\sum_{k=0}^{2^a}m^{2^a-k-1}2^ke_k(2^a-k)\\
=-\sum_{k=0}^{2^a}(1+2^cu)^{2^a-k-1}2^ke_k(2^a-k)\\
&=-\sum_{k=0}^{2^a}\sum_{i=0}^{2^a-k-1}\binom{2^a-k-1}{i}(2^cu)^i 2^ke_k(2^a-k)\\
&=-\underbrace{\sum_{k=0}^{2^a}2^ke_k(2^a-k)}_{B}-\sum_{k=0}^{2^a}\sum_{i=1}^{2^a-k-1}
\underbrace{\binom{2^a-k-1}{i}2^{ci+k}u^ie_k(2^a-k)}_{C_{ki}}.
\end{split}
\]
First, we shall prove that $C_{ki}\equiv0\pmod{2^{a+c+1}}$ for all $0\leqslant k\leqslant 2^a$ and $1\leqslant i\leqslant2^a-k-1$, except when $(k,i)=(0,1)$. Let $t$ be the non-negative integer such that $2^t\|k$.

If either $3\leqslant k\leqslant2^{a-1}$ is odd or $2^{a-1}<k\leqslant 2^a$, then by Proposition~\ref{prop_3_3}, $e_k\equiv0\pmod{2^{a-1}}$, so $2^{ci+k}e_k\equiv0\pmod{2^{ci+k+a-1}}$, and since $ci+k+a-1>a+c+1$, it follows that $C_{ki}\equiv0\pmod{2^{a+c+1}}$, as claimed. If $3\leqslant k\leqslant2^{a-1}$ is even, then by Proposition~\ref{prop_3_3}, $e_k\equiv\binom{2^{a-1}}{k}\pmod{2^a}$, so $e_k\equiv 0\pmod{2^{a-1-t}}$ by Proposition~\ref{prop_2_1}. In addition, since $2^a-k\equiv0\pmod{2^t}$, it follows that $e_k(2^a-k)\equiv0\pmod{2^{a-1}}$, so $2^{ci+k}e_k(2^a-k)\equiv0\pmod{2^{ci+k+a-1}}$. Since $ci+k+a-1>a+c+1$, it follows that $C_{ki}\equiv0\pmod{2^{a+c+1}}$, as claimed.

Next, if $k=0$, then $C_{0i}=\binom{2^a-1}{i}2^{ci}u^i2^a$. Now, for $i\geqslant2$, we get that $ic+a\geqslant c+a+1$, so $C_{0i}\equiv0\pmod{2^{a+c+1}}$. If $k=1$, then $C_{1i}=\binom{2^a-2}{i}2^{ci+1}e_1u^i(2^a-1)$. Suppose that $i\geqslant 2$. Since $e_1\equiv0\pmod{2^{a-1}}$ by Proposition~\ref{prop_3_3}, and since $(ci+1)+(a-1)\geqslant c+a+1$, it follows that $C_{1i}\equiv0\pmod{2^{a+c+1}}$. If $i=1$, then $C_{11}=(2^a-2)2^{c+1}e_1u(2^a-1)$, so $C_{11}\equiv0\pmod{2^{a+c+1}}$ since $2\mid2^a-2$.

If $k=2$, then $C_{2i}=\binom{2^a-3}{i}2^{ci+2}e_2u^i(2^a-2)$. By Proposition~\ref{prop_3_3},
$e_2\equiv0\pmod{2^{a-2}}$. In addition, since $2\mid 2^a-2$ and $(ci+2)+(a-2)+1\geqslant c+a+1$, it follows that $C_{2i}\equiv0\pmod{2^{a+c+1}}$, as required. To conclude, it follows that
\[
  A\equiv-B-C_{01}=-B-(2^a-1)2^{a+c}u
\pmod{2^{a+c+1}}.
\]
Next, note that once we prove that $2^{2a}\|B$, then in the case $a>c$, we obtain that $2a\geqslant a+c+1$, so $A\equiv-(2^a-1)2^{a+c}u\pmod{2^{a+c+1}}$, which implies that  $2^{a+c}\|A$, as required. In the case $a<c$, we obtain that $2a+1<a+c+1$, so $A\equiv-B\pmod{2^{2a+1}}$, which implies that $2^{2a}\|A$, as required. Finally, in the case $a=c$, we obtain
\[
A\equiv-2^{2a}\left(\frac{B}{2^{2a}}+(2^a-1)u\right)\pmod{2^{2a+1}},
\]
and since $\frac{B}{2^{2a}}+(2^a-1)u$ is a sum of two odd numbers, it follows that $2^{2a+1}\mid A$, as required.

So, it suffices to prove that $2^{2a}\|B$. To do so, consider the polynomial
\[
\begin{split}
  f(x)
  &=\prod_{k=0}^{2^a-1}(x-(1+2k))=\prod_{k=0}^{2^a-1}((x-1)+2k)\\
  &=\sum_{k=0}^{2^a}(-1)^ke_k(0,2,4,6,\ldots,2(2^a-1))(x-1)^{2^a-k}\\
  &=\sum_{k=0}^{2^a}\sum_{i=0}^{2^a-k}(-1)^{k+i}2^ke_k(0,1,2,3,\ldots,2^a-1)
  \binom{2^a-k}{i}x^{2^a-k-i}\\
\end{split}
\]
On the one hand, by recalling that $e_k=e_k(0,1,2,3,\ldots,2^a-1)$, we deduce that the coefficient of $x$ in $f(x)$ is
\[
\sum_{k=0}^{2^a}(-1)^{2^a-1}2^ke_k
  \binom{2^a-k}{2^a-k-1}=-\sum_{k=0}^{2^a}2^ke_k(2^a-k)=-B
\]
On the other hand, by Theorem~\ref{tm_4_3}, it follows that $2^{2a}\|B$, as required.
\end{proof}
%
We stress that if $a=c$ in Theorem~\ref{tm_4_4}, then the numerator of the corresponding fraction may be divisible by larger power of $2$ than $2^{2a+1}$. As an illustrative example, for $a=2$ and $m=1021$, we obtain that
\[
\frac{1}{1021}+\frac{1}{1023}+\frac{1}{1025}+\frac{1}{1027}=
\frac{4294946816}{1099501142025}.
\]
Since here $2^2\|m-1$, it follows that $a=c$, so by Theorem~\ref{tm_4_4}, we deduce that the numerator $4294946816$ is divisible by $2^5$, although it is actually divisible by $2^{12}$.


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% \end{linenumbers}
\end{document} 