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\begin{document}
\title{Odd perfect numbers of a special form\footnote{2000 Mathematics 
Subject Classification:
11A05, 11A25.}
\footnote{Key words and phrases: Odd perfect numbers.}}
\author{Tomohiro Yamada}
\date{}
\maketitle

\begin{abstract}
 We shall show that there is a effectively computable
upper bound of odd perfect numbers whose Euler factors
are powers of fixed exponent.
\end{abstract}

\section{Introduction}\label{intro}
 We denote by $\sigma(N)$ the sum of divisors of $N$.
 $N$ is said to be perfect if $\sigma(N)=2N$. Though
it is not known whether or not an odd perfect number
exists, many conditions which must be satisfied by
such a number are known.
 Suppose $N$ is an odd perfect number. Euler has
shown that $N=p^{\alpha} q_1^{2\beta_1}\cdots q_t^{2\beta_t}$
for distinct odd primes $p, q_1, \cdots, q_t$ with
$p\equiv \alpha\equiv 1\pmod{4}$.
 Steuerwald\cite{St} proved that we cannot have
$\beta_1=\cdots=\beta_t=1$. McDaniel\cite{Mc} proved
that we cannot have $\beta_1\equiv\cdots\equiv\beta_t\equiv1\pmod{3}$.
 If $\beta_1=\cdots=\beta_t=\beta$, then it is known
that $\beta\ne 2$(Kanold\cite{Ka1}), $\beta\ne 3$(Hagis
and McDaniel\cite{HMD}), $\beta\ne 5, 12, 17, 24, 62$
(McDaniel and Hagis\cite{MDH}), and $\beta\ne 6, 8, 11, 14, 18$
(Cohen and Williams\cite{CW}).
 In their paper \cite{HMD}, Hagis and McDaniel conjecture
that $\beta_1=\cdots=\beta_t=\beta$ does not occur.
 We cannot prove this conjecture. But we can prove that
for any {\it fixed} $\beta$, all of the odd perfect numbers
$N$ can be effectively determined.
 Our result is as follows.
\begin{thm}\label{thm1}
 Let $\beta\ge 1$. If $N=p^{\alpha} q_1^{2\beta}\cdots q_t^{2\beta}$ is an odd perfect number, then
\begin{equation}\label{eq0}
\omega(N)\le 4\beta^2+2\beta+3.
\end{equation}
and
\begin{equation}\label{eq1}
N\le 2^{4^{4\beta^2+2\beta+3}}.
\end{equation}
\end{thm}

\section{Lemmas}\label{lemmas}
 Let us denote by $v_p(n)$ the solution $e$ of $p^e||n$.
 For distinct primes $p$ and $q$, we denote by $o_q(p)$
the exponent of $p \mod q$ and we define $a_q(p)=v_q(p^d-1)$,
where $d=o_q(p)$.
 Clearly $o_q(p)$ divides $q-1$  and $a_q(p)$ is a positive
integer. Now we quote some elementary properties of $v_q(\sigma(p^x))$.
Lemmas \ref{lm1} and \ref{lm2} are well-known. Lemma \ref{lm1}
follows from Theorems 94 and 95 in Nagell\cite{Nag}. Lemma \ref{lm2}
has been proved by Zsigmondy\cite{Zsi} and rediscovered by many authors
such as Dickson\cite{Dic} and Kanold\cite{Ka2}.

\begin{lem}\label{lm1}
 Let $p, q$ be distinct primes with $q\ne2$ and $c$ be a positive integer.
 If $p\equiv 1\pmod{q}$, then $v_q(\sigma(p^c))=v_q(c+1)$.
 Moreover, if $p\nequiv 1\pmod{q}$, then $v_q(\sigma(p^c))=a_q(p)+v_q(c+1)$
if $o_q(p)\mid (c+1)$ and $v_q(\sigma(p^c))=0$ otherwise.
\end{lem}

\begin{lem}\label{lm2}
 If $a>b\ge 1$ are coprime integers, then $a^n-b^n$ has
a prime factor which does not divide $a^m-b^m$ for any $m<n$,
unless $(a, b, n)=(2, 1, 6)$ or $a-b=n=1$, or $n=2$ and $a+b$ is a power of $2$.
\end{lem}

 By Lemma \ref{lm2}, we obtain the following lemmas.

\begin{lem}\label{lm3}
 Suppose $p$ is a prime and $n$ is a positive integer.
 If $d\mid (n+1)$, $d>1$ and $(p, d)$ satisfies neither
$(p, d)=(2, 6)$ nor $(p, d)=(2^m-1, 2)$ for some integer $m$,
then there exists a prime $q$ with $o_q(p)=d$ and $q\mid\sigma(p^n)$.
\end{lem}

\begin{proof}
 We can apply Lemma \ref{lm2} with $(a, b)=(p, 1)$ and
we see that there exists a prime $p$ such that $o_p(q)=d$.
 $q$ does not divide $p-1$ since $o_p(q)=d>1$.
 On the other hand, $q$ divides $p^{n+1}-1$ since $1<d\mid (n+1)$.
 Hence $q\mid\sigma(p^n)$.
\end{proof}

\begin{lem}\label{lm4}
 If $p$ is a prime and $n$ is a positive integer,
then $\omega(\sigma(p^n))\ge \tau(n+1)-1$ unless $p=2$
and $n\equiv 5\pmod{6}$, or $p$ is a Mersenne prime and $n$ is odd.
\end{lem}

\begin{proof}
 If $d\mid (n+1)$ and $d\ne 1$, then $(p, n, d)$
satisfies the condition of Lemma \ref{lm3}.
 Hence $\sigma(p^n)$ has a prime factor $q_d$ satisfying $o_p(q_d)=d$.
 Thus we have $\omega(\sigma(p^n))\ge \#\{d : d\mid (n+1), d>1\}=d(n+1)-1$.
\end{proof}

 The following lemma is Lemma 2 of Danpat, Hunsucker and Pomerance \cite{DHP}.

\begin{lem}\label{lm5}
 If $p, q$ are distinct primes with $q\ne 2$
satisfying $\sigma(q^x)=p^y$ for some positive integers $x, y$,
then $p\equiv 1\pmod{q}$ or $a_q(p)=1$.
\end{lem}

 By this lemma and Lemma \ref{lm1}, we immediately deduce the following result.

\begin{lem}\label{lm6}
 Let $p, q$ be distinct primes satisfying the condition
of Lemma \ref{lm5} and $q^f\mid\sigma(p^e)$. Then $q^{f-1} o_q(p)$ divides $e+1$.
\end{lem}
\begin{proof}
 By Lemma \ref{lm5}, $p\equiv 1\pmod{q}$ or $a_q(p)=1$.
 If $p\equiv 1\pmod{q}$, then $q^f\mid\sigma(p^e)$
implies $q^f=q^f o_q(p)\mid (e+1)$ by Lemma \ref{lm1}.
 If $a_q(p)=1$, then $o_q(p)\mid (e+1)$ and $v_q(e+1)\ge f-a_q(p)=f-1$ by Lemma \ref{lm1}.
 In both cases, $q^{f-1} o_q(p)$ divides $e+1$.
\end{proof}

 Moreover, we quote a result in Kanold\cite{Ka1}.

\begin{lem}\label{lm7}
 Let $N=p^{\alpha}q_1^{2\beta_1}\cdots q_r^{2\beta_r}$ be
an odd perfect number and $l$ be a common divisor of
$2\beta_1+1, \cdots, 2\beta_r+1$. Then $l^4\mid N$.
 Moreover, if $l$ is a power of a prime $q$, then $p\ne q$.
\end{lem}

\section{Proof of the first part of Theorem \ref{thm1}}
 Let $N=p^{\alpha} q_1^{2\beta}\cdots q_r^{2\beta}$ be an odd perfect number.

 First assume that $2\beta+1=l^{\gamma}$,
where $l$ is a prime and $\gamma$ is a positive integer.
 Various results referred in the Introduction of
this paper allows us to assume that $\beta\ge 8$
without loss of generality.

 By Lemma \ref{lm7}, $p\ne l$ and $l^{4\gamma}$ divides $N$.
 Hence $l=q_{i_0}$ for some $i_0$ and $v_l(N)=2\beta$.
 We divide $q_1,\cdots, q_r$ into four disjoint sets. Let
\begin{equation*}
 S=\{i:q_i\equiv 1\pmod{l}\},
\end{equation*}
\begin{equation*}
 T=\{i:q_i\nequiv 1\pmod{l}, i\ne i_0, q_j\mid\sigma(q_i^{2\beta})\text{ for some }1\le j\le r\},
\end{equation*}
and
\begin{equation*}
 U=\{i:q_i\nequiv 1\pmod{l}, i\ne i_0, q_j\nmid\sigma(q_i^{2\beta})\text{ for any }1\le j\le r\}.
\end{equation*}

 Then $i\in S\cup T\cup U\cup \{i_0\}$ and thus we have
\begin{equation}\label{eqa1}
r\le \#S+\#T+\#U+1.
\end{equation}

\begin{lem}\label{lma1}
\begin{equation*}
\#S\le 2\beta.
\end{equation*}
\end{lem}
\begin{proof}
 For $i\in S$, we have $l\mid\sigma(q_i^{2\beta})$ by Lemma \ref{lm1}. Hence
\begin{equation*}
\#S\le v_l(\prod_{i\in S}\sigma(q_i^{2\beta})) \le v_l(2N)=v_l(N)=2\beta.
\end{equation*}
\end{proof}

\begin{lem}\label{lma2}
\begin{equation*}
\#T\le (2\beta)^2.
\end{equation*}
\end{lem}
\begin{proof}
 If $i\in T$, then $q_j\mid\sigma(q_i^{2\beta})$ for some $j\in S$. Hence we have \\
$\sum_{j\in S} v_{q_j}(\sigma(q_i^{2\beta}))\ge 1$ for $i \in T$.
 By Lemma \ref{lma1} we have

\begin{equation*}
\begin{split}
\#T&\le \sum_{i \in T} \sum_{j\in S} v_{q_j}(\sigma(q_i^{2\beta}))=\sum_{j\in S} v_{q_j}(\prod_{i \in T}\sigma(q_i^{2\beta}))\\
&\le \sum_{j \in S} v_{q_j}(2N)=\sum_{j \in S} v_{q_j}(N)\le (2\beta)^2.
\end{split}
\end{equation*}

\end{proof}

\begin{lem}\label{lma3}
\begin{equation*}
\#U\le 1.
\end{equation*}
\end{lem}
\begin{proof}
 If $i\in U$ and $q$ is a prime dividing $\sigma(q_i^{2\beta})$, then $q=p$
since $q\mid 2N$, $\sigma(q_i^{2\beta})$ is odd, and $q\ne q_j$ for any $j$.
 Thus $\sigma(q_i^{2\beta})=p^{\zeta_i}$ for some positive integer $\zeta_i$.

 We shall show $q_i^{2\beta}\mid\sigma(p^{\alpha})$.
 If $q_i$ divides $\sigma(q_j^{2\beta})$ for some $j$,
then $q_i$ divides $2\beta+1=l^{\gamma}$ or
$1<o_{q_i}(q_j)\mid (2\beta+1)=l^{\gamma}$ by Lemma \ref{lm1}.
 The former case cannot occur since $q_i\ne q_{i_0}=l$.
 If $1<o_{q_i}(q_j)\mid (2\beta+1)=l^{\gamma}$, then $o_{q_i}(p)=l^t$
for some integer $t>0$ and therefore $q_i\equiv 1\pmod{l}$,
which is inconsistent with the assumption $q_i\in U$.
 Since $q_i^{2\beta}\mid N\mid\sigma(N)$,
we conclude that $q_i^{2\beta}\mid\sigma(p^{\alpha})$.

 We can apply Lemma \ref{lm6} with $(q, f, e)=(q_i, 2\beta, \alpha)$
and deduce that $q_i^{2\beta-1}$ divides $(\alpha+1)/2$
since $q_i$ is odd and $\alpha\equiv 1\pmod{4}$.
 Hence $(\alpha+1)/2$ must be divisible by $\prod_{i\in U}q_i^{2\beta-1}$
and therefore $d(\alpha+1)\ge 2(2\beta)^{\#U}$. By Lemma \ref{lm4},
we have $\omega(\sigma(p^{\alpha}))\ge 2(2\beta)^{\#U}-1$.

 On the other hand, we have $\omega(\sigma(p^{\alpha}))\le \omega(N)+1$
since $\sigma(p^{\alpha})\mid 2N$. Thus, from Lemmas \ref{lma1} and \ref{lma2} we obtain
\begin{equation*}
\begin{split}
2(2\beta)^{\#U}-1\le \omega(N)+1\le \#S+\#T+\#U+3\le \#U+(2\beta)^2+2\beta+3.
\end{split}
\end{equation*}

 Since $\#U\le (2\beta)^{\#U-1}\le (2\beta)^{\#U}/16$
by the assumption that $\beta\ge 8$, we have
\begin{equation}
\frac{31}{16}(2\beta)^{\#U}\le 2(2\beta)^{\#U}-\#U \le (2\beta)^2+2\beta+4\le \frac{21}{16}(2\beta)^2,
\end{equation}
and therefore $\#U\le 1$.
\end{proof}

 By Lemmas \ref{lma1}, \ref{lma2}, \ref{lma3} and (\ref{eqa1}), we have $\omega(N)\le r+1\le 4\beta^2+2\beta+3$,
which is the desired result.


 Next we assume that $2\beta+1=l_1^{\gamma_1}l_2^{\gamma_2}\cdots l_s^{\gamma_s}$
with $s\ge 2$, where $l_1, \cdots, l_s$ are distinct primes.
 By Lemma \ref{lm7}, $l_i^{4\gamma_i}$ divides $N$ for each $i$.
 This clearly implies that there are at least $s-1$ primes in $l_i$'s
each of which is equal to $q_j$ for some $j$. Hence
we may assume that $l_i=q_i$ for $i=1, \cdots, s-1$.

 Let $S=\{i:q_i\equiv 1\pmod{l_1}\}$. As in the prime-power case,
we derive that $\#S\le 2\beta$.
 By Lemma \ref{lm4}, each $\sigma(q_j^{2\beta})$ has at least
one prime factor $q$ with $o_q(q_j)=d$ for any $d>1$ dividing $2\beta+1$.
 If we denote by $w$ the number of divisors of $2\beta+1$ divisible by $l_1$,
then $w=\gamma_1(\gamma_2+1)\cdots(\gamma_s+1)\ge 2^{s-1}$.
 Thus each $\sigma(q_j^{2\beta})$ has at least $2^{s-1}-1$ prime factors
$\equiv 1\pmod{l_1}$ and different from $p$, namely, belonging to $S$.

 Hence we conclude that $r\le 2\beta\#S/(2^{s-1}-1)\le (2\beta)^2$
and therefore $\omega(N)\le 4\beta^2+1$, which is more than we desired.
 This completes the proof of the first part of Theorem \ref{thm1}.

 To obtain the second part of our theorem it remains to apply the result of Nielsen\cite{Nie},
who has shown that $M\le (d+1)^{4^l}$ for any positive integer $n, d, M, l$ satisfying
$\sigma(M)/M=n/d$ and $\omega(M)=l$.
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\end{thebibliography}{}
\vskip 12pt

{\small Tomohiro Yamada}\\
{\small Department of Mathematics\\Faculty of Science\\Kyoto University\\Kyoto, 606-8502\\Japan}\\
{\small e-mail: \protect\normalfont\ttfamily{tyamada@math.kyoto-u.ac.jp}}
\end{document}