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\begin{center}
\vskip 1cm{\LARGE\bf 
An Extension of the Euclid-Euler Theorem to \\
\vskip .1in
Certain $\alpha$-Perfect Numbers
}
\vskip 1cm
\large
Paulo J. Almeida\footnote{The first author received support from the Center for Research and Development in Mathematics and Applications (CIDMA) through the Portuguese Foundation for Science and Technology (FCT---Funda\c{c}\~{a}o para a Ci\^{e}ncia e Tecnologia), reference UIDB/04106/2020.} and Gabriel Cardoso\footnote{The second author received support from the Portuguese Foundation for Science and Technology (FCT---Funda\c{c}\~{a}o para a Ci\^{e}ncia e Tecnologia, reference PD/BD/150533/2019).}\\
University of Aveiro\\
Department of Mathematics\\
Campus Universit{\'a}rio de Santiago\\
3810-193 Aveiro\\
Portugal\\
\href{mailto:palmeida@ua.pt}{\tt palmeida@ua.pt}\\
\href{mailto:gabriel.cardoso@ua.pt}{\tt gabriel.cardoso@ua.pt}
\end{center}

\vskip .2 in

\begin{abstract}
In a posthumously published work, Euler proved that all even
perfect numbers are of the form $2^{p-1}(2^p-1)$, where $2^p-1$ is
a prime number. In this article, we extend Euler's method for certain
$\alpha$-perfect numbers for which Euler's result can be generalized. In
particular, we use Euler's method to prove that if $N$ is a $3$-perfect
number divisible by $6$; then either $2\parallel N$ or $3\parallel
N$. As well, we prove that if $N$ is a $\frac{5}{2}$-perfect number
divisible by $5$, then $2^4\parallel N$, $5^2\parallel N$, and $31^2\mid
N$. Finally, for $p\in\{17,257,65537\}$, we prove that there are no
$\frac{2p}{p-1}$-perfect numbers divisible by $p$.
\end{abstract}

\section{Introduction}

A perfect number is equal to the sum of its proper divisors. These numbers have been studied since the ancient Greek times. An old known result is Proposition $36$ of Book IX in the {\it Elements}, where Euclid showed that if $2^p-1$ is a prime number then $2^{p-1}(2^p-1)$ is perfect. Around two thousand years later, L.\ Euler \cite[p.\ 630]{eu} proved that every even perfect number is of Euclid's form. In the same page, there is Euler's formula for an odd perfect number. If $\sigma(N)=2N$ and $N$ is odd, then
\begin{equation}\label{Eulerodd}
	N=p^ek^2,
\end{equation}
where $p$ is a prime number, $(p,k)=1$, and $p\equiv e\equiv \modd{1} {4}$.

Let $\sigma$ to be the sum of divisors function and
\[I(N)=\frac{\sigma(N)}{N}\]
to be the {\it index function} of $N$. Clearly, $I(N)$ is multiplicative. For $p$ a prime number and $a$ a non-negative integer, the index function $I(p^a)$ is monotonically increasing in $a$ but monotonically decreasing in $p$. Also, for any prime number $p$ and any positive integer $a$, we have
\begin{equation}\label{Ip}
	\frac{p+1}{p}\leq I(p^a)&lt;\frac{p}{p-1}.
\end{equation}
Moreover, if $I(N)=\frac{c}{d}$ and $\gcd(c,d)=1$, we have
\begin{equation}\label{pd}
\text{if $p$ is a prime number and $p\mid d$, then $p\mid N$};
\end{equation}
and
\begin{equation}\label{pc}
	\text{ if $p$ is a prime number and $p\mid c$, then $p\mid \sigma(N)$}.
\end{equation}
For any rational number $\alpha$, a positive integer $N$ is an {\it $\alpha$-perfect} number if $I(N)=\alpha$. We say $\alpha$ is an {\it abundancy index} if $\alpha \in \Ima(I)$. The case $\alpha=2$ corresponds to a {\it perfect number} and, when $\alpha$ is an integer, $N$ is a {\it multiperfect number}. More than $5700$ multiperfect numbers have been found (see Achim Flammenkamp's The Multiply Perfect Numbers Page \cite{fla}). Many other rational values for $\alpha$ were studied, some of them with connections with odd perfect numbers \cite{rya,wie}. When a rational $\alpha$ is not in the image of $I(N)$, $\alpha$ is called an {\it abundancy outlaw}. Holdener and Stanton~\cite{hol} gave a list of abundancy outlaws as well as rational numbers for which their abundancy status is unknown. Given a prime $p$, we say that a rational $\alpha$ is a {\it $p$-abundancy outlaw} if there are no $\alpha$-perfect numbers divisible by $p$. 


We will present a generalization of Euler's result on perfect numbers that can be applied to certain $\alpha$-perfect numbers. In particular, it can be applied to the $3$-perfect numbers and $\frac{5}{2}$-perfect numbers. For $p\in\{17,257,65537\}$, we also prove that $\frac{2p}{p-1}$ is a $p$-abundancy outlaw.

The only known $3$-perfect numbers are the following   
\[2^3\cdot 3\cdot 5,\ 2^5\cdot 3\cdot 7,\ 2^9\cdot 3\cdot 11\cdot 31,\ 2^{13}\cdot 3\cdot 11\cdot 43\cdot 127,\]
\[2^8\cdot 5\cdot 7\cdot 19\cdot 37\cdot 73, \mbox{ and }2^{14}\cdot 5\cdot 7\cdot 19\cdot 31\cdot 151.\]
Dickson~\cite[p.\ 33]{dickson} gives a historical account of the $3$-perfect numbers. It is conjectured the numbers above are the only $3$-perfect numbers. Clearly, $N$ is a $3$-perfect number with $2\parallel N$ if and only if $\frac{N}{2}$ is an odd perfect number. Therefore, if the above list of $3$-perfect numbers is complete, then there are no odd perfect numbers. Also, it is easy to see that $\frac{5}{2}$ is not an abundancy outlaw, since $I(24)=\frac{5}{2}$. 

In order to apply our generalization of the Euclid-Euler theorem we will need to find all solutions of certain two and three-variable exponential diophantine equations. We will use the general method developed by Styer \cite{sty} which formalizes and extends a method used by Guy, Lacampagne, and Selfridge \cite{GLS} (also see the paper~\cite{sco} for similar results).

\section{Generalizing Euler's method}\label{GenMet}
Let $\alpha$ be a rational number and $N&gt;1$ be an $\alpha$-perfect number, then $I(N)=\alpha$. Using \eqref{Ip} and the multiplicativity of $I(N)$, there exist positive integers $r$ and $m$, prime numbers $p_i$, and positive integers $a_i$, with $1\leq i\leq r$; such that
\begin{equation}\label{Nr}
	N=m\prod_{i=1}^{r}p_i^{a_i},
\end{equation}
\begin{equation}\label{alpha}
	\alpha\prod_{i=1}^{r}\frac{p_i-1}{p_i}\leq 1,
\end{equation}	
and
\[\gcd\left (m,\prod_{i=1}^rp_i^{a_i}\right )=1.\]
Therefore,
\[\alpha m\prod_{i=1}^rp_i^{a_i}=\sigma(N)=\sigma(m)\prod_{i=1}^r\frac{p_i^{a_i+1}-1}{p_i-1}.\]
Hence, for some integers $k$ and $d$, we have
\begin{equation}\label{sigm}
\sigma(m)=\frac{\beta k}{d}\prod_{i=1}^rp_i^{a_i+1}
\end{equation}
and
\begin{equation}\label{m}
m=\frac{k}{d}\prod_{i=1}^r
\left (p_i^{a_i+1}-1\right ),
\end{equation}
where
\[\beta=\alpha\prod_{i=1}^r\frac{p_i-1}{p_i}\leq 1\]
and
\[d=\gcd\left (\beta\prod_{i=1}^rp_i^{a_i+1},\prod_{i=1}^r\left (p_i^{a_i+1}-1\right )\right ).\]

We will try to find a lower bound for $\sigma(m)$, by summing divisors of $m$ that are explicitly indicated in Eq.~\eqref{m}.  A comparison of the lower bound with Eq.~\eqref{sigm}, will give us contradictions or conditions on the form of $N$. From now on, we will always consider $\beta=1$.

\section{The Euclid-Euler theorem: $r=1$ and $\alpha=\frac{p}{p-1}$}
Suppose $N=p^a m$ is an $\alpha$-perfect number, where $p$ is a prime number and $\alpha=\frac{p}{p-1}$ satisfies Condition~\eqref{alpha} with equality. Then $I(N)=\alpha$.

By \eqref{pd}, if $p\neq 2$ then $2\mid N$. Therefore,
\[I(N)\geq\frac{3}{2}\frac{p+1}{p}&gt;\frac{p}{p-1}=\alpha.\]
Hence, $p=2$ and we have the case studied by Euler~\cite{eu}. The Euler theorem is a well known result, but we write its proof here to give context for our results.

\begin{theorem}[Euler] 
If $N$ is an even perfect number, then $N=2^{p-1}(2^p-1)$, where $p$ and $2^p-1$ are prime numbers.
\end{theorem}
\begin{proof}
Suppose  $N=2^am$ and $\sigma(N)=2N$. Then
$$ 2^{a+1}m  = \sigma(N) = (2^{a+1}-1)\sigma(m).$$
Since $\gcd(2^{a+1}-1,2^{a+1})=1$, then there exists a positive integer $k$ such that $m=k(2^{a+1}-1)$ and $\sigma(m)=k2^{a+1}$. Since $k$ and $m$ divide $m$, and $\sigma(m)=k+m$, then $2^{a+1}-1$ is a prime number and $k=1$. Now, $2^{a+1}-1$ can only be a prime number if $a+1$ is a prime number $p$. Therefore, $N=2^{p-1}(2^p-1)$.
\end{proof}

\section{$3$-perfect numbers divisible by $6$: $r=2$ and $\alpha=3$}
In this section, we will consider $r=2$, $p_1=2$, and $p_2=3$ in Eq.~\eqref{Nr}. We will prove that if a $3$-perfect number $N$ is divisible by $6$, then either $2\parallel N$ or $3\parallel N$. Although Steuerwald~\cite{ste} prove this result in 1954, we will use the method introduced in Section \ref{GenMet}. Before we proceed, we need results about the difference between powers of $2$ and $3$. These results are special cases of Pillai's conjecture and Mih\v{a}ilescu's theorem \cite{Mih}, which proves the Catalan conjecture. It is worth noting that the solutions of equation $|2^a-3^b|=1$ were first determined by Levi ben Gershon, also known as Gersonides (1288--1344), in his treatise {\it De Numeris Harmonicis}.

\begin{lemma}\label{pillai} The only solutions of the diophantine equation
\begin{equation}\label{2a3b1}
	2^a-3^b=-1
\end{equation}
are  $(1,1)$ and $(3,2)$. Also, the only solutions of the diophantine equation
\begin{equation}\label{2a3b2c}
2^a-3^b=2^c-1,
\end{equation}
 are  $(2,1,1)$, $(4,2,3)$, and $(a,0,a)$, for all $a\in\mathbb{Z}$.
\end{lemma}
\begin{proof}
We start by solving the equation $|2^a-3^b| =1$.
Since $2$ is a primitive root of $3^b$, for any $b\geq 1$; then the solutions of \(2^x\equiv \pm \modd{1} {3^b}\) are \({\displaystyle x=3^{b-1}k}\), for $k\geq 0$. But, for $b\geq 3$ we have
\[2^{3^{b-1}}&gt;2\cdot 3^b.\]
Therefore, $b\in\{0, 1, 2\}$ and so we obtain the solutions $(a,0,a)$ and $(2,1,1)$ of Eq.~\eqref{2a3b2c}. Also, we obtain the solutions $(1,1)$ and $(3,2)$ of Eq.~\eqref{2a3b1}.

Now, suppose $c\geq 2$. If $c$ is even, then $3\mid 2^c-1$ and we obtain a contradiction.
If $c$ is odd, we have $2^{a} \equiv\modd{1} {3}$ so $a$ is even. Therefore, there exists a positive integer $a'$ such that $a = 2a'$. As $c=2c'+1$, for some positive integer $c'$, we have $2^{2a'} - 3^{b} = 2^{2c'+1}-1$. So $(-1)^{b+1} \equiv \modd{-1} {4}$. Hence, $b$ is even and we have $b = 2b'$, for some non-negative integer $b'$. Then we have
\[2^{2a'} - 3^{2b'} = 2^{2c'+1}-1.\]
Therefore,
\[(-1)^{a'} - (-1)^{b'} \equiv  \modd{2 \cdot (-1)^{c'}-1} {5}.\]
If $a'$ and $b'$ have the same parity we have a contradiction. If $a'$ is odd and $b'$ is even we have a contradiction as well. Therefore, $a'$ is even and $b'$ is odd. So we have $4 \mid a$ and $b \equiv\modd{2} {4}$, which implies $1 - (-1) \equiv  \modd{2 \cdot (-1)^{c'}-1} {5}$. Hence, $c'$ is odd and $c \equiv \modd{3} {4}$.
Since $c &gt; 1$, then $a &gt; c$. Therefore, we have \({\displaystyle 2^c(2^{a-c}-1)=3^b-1}\). Hence,
\({\displaystyle 2^c\parallel 3^b-1}\).

Since $b \equiv \modd{2} {4}$, then $2^3 \parallel 3^b-1$. Hence, $c=3$ and we can rewrite Eq.~\eqref{2a3b2c} as
\begin{center}
\({\displaystyle \left(2^{\frac{a}{2}}-3^{\frac{b}{2}}\right) \left(2^{\frac{a}{2}}+3^{\frac{b}{2}}\right) = 7}.\)
\end{center}

Therefore, $a=4$ and $b=2$. Hence, if  $c\geq 2$, the only solution of Eq.~\eqref{2a3b2c} is $(4,2,3)$. Therefore, we obtain the stated result,
\end{proof}


We can now state and give a new proof of Steuerwald's theorem on $3$-perfect numbers.

\begin{theorem}\label{even3}
Suppose $N$ is a $3$-perfect number of the form $N=2^a\,3^bm$, where $a,b\geq 1$ and $\gcd(6,m)=1$. Then $a=1$ and $b\neq 1$, or $a\neq 1$ and $b=1$.
\end{theorem}


\begin{proof}
Let $N=2^a\,3^bm$ such that $a,b\geq 1$, $\gcd(6,m)=1$, and $\sigma(N)=3N$. Then
\[3N=2^a\,3^{b+1}m=\sigma(N)=(2^{a+1}-1) \, \frac{3^{b+1}-1}{2} \, \sigma(m).\]
Therefore,
\[\frac{\sigma(m)}{m}=\frac{2^{a+1}\,3^{b+1}}{(2^{a+1}-1)(3^{b+1}-1)}.\]
Let $d = \gcd\left(2^{a+1}\, 3^{b+1}, (2^{a+1}-1)(3^{b+1}-1)\right)$. It is easy to see that $d = 2^s \, 3^t$, where $1 \leq s \leq a+1$ and $0 \leq t \leq b+1$.
Since
\[\gcd\left (\frac{2^{a+1}3^{b+1}}{2^s\,3^t},\frac{(2^{a+1}-1)(3^{b+1}-1)}{2^s\,3^t}\right )=1,\]
then
\[\sigma(m)=\frac{2^{a+1}\,3^{b+1}}{2^s\,3^t}k\mbox{ and }m=\frac{2^{a+1}-1}{3^t} \, \frac{3^{b+1}-1}
{2^s}k,\]
for some positive integer $k$.

Let us consider the following three cases, which will establish the claim.

\bigskip

\noindent \textit{Case A:} Suppose that $t \neq 0$ and let
\[M=\max\left (\frac{2^{a+1}-1}
{3^t},\frac{3^{b+1}-1}{2^s}\right ).\]
Then we have
\begin{align*}
\frac{\sigma(m)}{k} &amp; =\frac{2^{a+1}3^{b+1}}{2^s\,3^t}\\
&amp; = \frac{\left (\left (2^{a+1}-1\right )+1\right )\left (\left (3^{b+1}-1\right )+1\right )}{2^s\,3^t}\\
&amp; =\frac{2^{a+1}-1}{3^t}\,\frac{3^{b+1}-1}{2^s}+\frac{2^{a+1}-1}{3^t} \, \frac{1}{2^s}+\frac{3^{b+1}-1}{2^s} \, \frac{1}{3^t}+\frac{1}{2^s\,3^t}\\
&amp; &lt;\frac{m}{k}+\frac{M}{2}+\frac{M}{3}+1\\
&amp; &lt;\frac{m}{k}+M+1.
\end{align*}
Therefore, 
\begin{equation}\label{smaller}
\sigma(m)&lt;m+Mk+k.
\end{equation}
By definition of $M$, we have $m=uMk$, with $u\in \left \{\frac{2^{a+1}-1}{3^t},\frac{3^{b+1}-1}{2^s}\right \}$. By definition of $s$ and $t$, we have $u\in\mathbb{Z}$. Then $Mk\,\mid\,m$. If $Mk \neq m$ and $M \neq 1$, then $m$, $Mk$, and $k$ are different divisors of $m$. Thus, 
\begin{equation}\label{larger}
	\sigma(m)\geq m+Mk+k.
\end{equation} 
The combination of inequalities~\eqref{smaller} and \eqref{larger}, give us a contradiction.

If $Mk=m$ or $M=1$, then
$$\frac{2^{a+1}-1}{3^t}=1\mbox{ or }\frac{3^{b+1}-1}{2^s} = 1.$$
Therefore, $2^{a+1}-3^t=1$ or $3^{b+1}-2^{s} = 1$. By Lemma~\ref{pillai}, we have $a=1$ or $b=1$.

\bigskip

\noindent \textit{Case B:} Suppose that $t = 0$ and
\[2^{a+1}-1 \neq\frac{3^{b+1}-1}{2^s}.\]
Let \[M'=\min\left(2^{a+1}-1,\frac{3^{b+1}-1}{2^s}\right).\]
If $M'=1$ then
$$2^{a+1}-1=1\mbox { or }\frac{3^{b+1}-1}{2^s}=1.$$
Therefore, $2^{a+1}=2$ or $3^{b+1}-2^s=1$.
As $a,b \geq 1$, by Lemma~\ref{pillai}, we conclude that $b=1$.

If $M'\neq 1$ then 
\[m, \left (2^{a+1}-1\right )k, \frac{3^{b+1}-1}{2^s}k,\mbox{ and }k,
\]
are different divisors $m$. Therefore,
\begin{align*}
\sigma(m) &amp; \geq m+(2^{a+1}-1)k+\frac{3^{b+1}-1}{2^s}k+k\\
&amp; &gt;(2^{a+1}-1)\, \frac{3^{b+1}-1}{2^s}k+\frac{2^{a+1}-1}{2^s}k+\frac{3^{b+1}-1}{2^s}k+\frac{k}
{2^s} \\
&amp; =\sigma(m),
\end{align*}
where the strict inequality results from $s\geq 1$. So, we obtain a contradiction.

\bigskip

\noindent \textit{Case C:} Suppose that $t = 0$ and
\[2^{a+1}-1 = \frac{3^{b+1}-1}{2^s}.\]
Then $2^{a+1+s}-3^{b+1} = 2^s - 1$. By Lemma~\ref{pillai}, we have $(a,b)\in\{(0,0), (0,1)\}$. Since $a,b \geq 1$ we obtain a contradiction. 

Hence, we must have $a=1$ or $b=1$.

Next, we prove that we cannot have $a=1$ and $b=1$, simultaneously.

If $a=1$, let $N=2N'$, where $N'$ is odd. Then 
\[6N'=3N=\sigma(N)=3\sigma(N')\]
Therefore, $\sigma(N')=2N'$. So $N'$ is an odd perfect number. By Euler's formula~\eqref{Eulerodd}, if $3\mid N'$ then $3^2\mid N'$. Hence, we cannot have $b=1$. Thus, we have $a=1$ and $b\neq 1$, or $a\neq 1$ and $b=1$.
\end{proof}

\section{Fermat prime numbers: $r=2$ and $\alpha=\frac{2p}{p-1}$}
In Eq.~\eqref{Nr}, if $r=2$, $p_1=2$, and $p_2\neq 3$, then $\alpha$ is not an integer. Moreover, if $p_2$ is not a Fermat number, then we must have a prime $p\notin\{p_1,p_2\}$ such that $p\mid\frac{p_2-1}{2}$. Using \eqref{pd}, we obtain $p\mid N$. For this reason, and in order to keep $r=2$ without any other known prime divisors of $N$, we will only consider the cases where $p_2\neq 3$ and $p_2$ is a Fermat prime number. Let $F_n=2^{2^n}+1$ denote the $n$-th Fermat number, for any non-negative integer $n$. The next result states some properties of Fermat numbers that will be used later.

\begin{lemma}\label{Fnlem}
Let $n$ be a non-negative integer and $F_n$ be the $n$-th Fermat number. Then
\begin{enumerate}
	\item $F_{n-1}\mid F_n^{2^n}-1$, for $n\geq 1$;
	\item $\ord_{2^{2^n+s}}(F_n)=2^s$, for $n\geq 1$ and $s$ a non-negative integer. 
\end{enumerate}
\end{lemma}
\begin{proof}
Let $n\geq 1$. Clearly, $F_n=\left (F_{n-1}-1\right )^2+1$. So there exists an integer $t$, such that
\begin{align*}
F_n^{2^n}-1 &amp; =  \left (\left (F_{n-1}-1\right )^2+1\right )^{2^n}-1\\
&amp; = -1+ \left (\left (F_{n-1}^2-2F_{n-1}\right )+2\right )^{2^n}\\
&amp; =  -1+\sum_{j=0}^{2^n} {2^n\choose j}\left (F_{n-1}^2-2F_{n-1}\right )^j2^{2^n-j}\\
&amp; =  -1+tF_{n-1}+2^{2^n}\\
&amp; =  tF_{n-1}+F_0F_1\cdots F_{n-1}. 
\end{align*}
Hence,  $F_{n-1}\mid F_n^{2^n}-1$.

The statement {\it 2.} is clearly true for $s=0$. We prove by induction that 
\begin{equation}\label{Fn}
F_n^{2^{s-1}}\equiv\modd{1+2^{2^n+s-1}} {2^{2^n+s}},
\end{equation}
for any $s\geq 1$.

Since $F_n\equiv \modd{F_n} {2^{2^n+1}}$, the congruence~\eqref{Fn} is valid for $s=1$. Now, suppose \eqref{Fn} is valid for $s$. Then there exists an integer $t$ such that
\begin{align*}
F_n^{2^s} &amp; =  \left (F_n^{2^{s-1}}\right )^2\\
&amp; =  \left (1+\left (2^{2^n+s-1}+2^{2^n+s}t\right )\right )^2\\
&amp; =  1+ 2^{2^n+s}+2^{2^n+s+1}t+2^{2^{n+1}+2s-2}+2^{2^{n+1}+2s}t^2+2^{2^{n+1}+2s}t\\
&amp; \equiv \modd{1+ 2^{2^n+s}} {2^{2^n+s+1}}.
\end{align*}
Therefore, Eq.~\eqref{Fn} is true for any $s\geq 1$. So we have 
\[\ord_{2^{2^n+s}}(F_n)\neq 2^{s-1}.\]
Since $\ord_{2^{2^n+s}}(F_n)$ is a power of $2$ and 
\[F_n^{2^{s}}\equiv \modd{1} {2^{2^n+s}},\]
then statement {\it 2.} is obtained.
\end{proof}

Before applying Euler's method we still need two technical results about some exponential diophantine equations. Our proof of these results is inspired by Styer's work \cite{sty}. 

\begin{lemma}\label{largec}
Let $n\geq 1$ and  $F_n$ be the $n$-th Fermat number. Suppose there exists a non-negative integer $s$ and a prime number $q_n$, such that
\begin{equation}\label{qn1}
	q_n \mid F_n^{2^{s}}-1
\end{equation}
and
\begin{equation}\label{qn2}
\ord_{F_n}(2) \mid \ord_{q_n}(2). 
\end{equation}
Then the exponential diophantine equation 
\begin{equation}\label{2apb2clarge}
	2^a-F_n^b = 2^c - 1,
\end{equation}
has no solutions $(a,b,c)$ with $b\geq 1$ and $c\geq 2^n+s$.
\end{lemma}
\begin{proof}
Suppose there exists a non-negative integer $s$ and a prime number $q_n$ satisfying Cond.~\eqref{qn1} and \eqref{qn2}. Also, suppose that exists a solution $(a,b,c)$ of Eq.~\eqref{2apb2clarge} with $b\geq 1$ and $c\geq 2^n+s$. Since $b\geq 1$, we have $a&gt;c$. Therefore, we have 
\[F_n^b\equiv \modd{1} {2^{2^n+s}}.\]
By Lemma \ref{Fnlem}, we have \(\ord_{2^{2^n+s}}(F_n)=2^{s}\). Hence, $2^{s}\mid b$. Now, Cond.~\eqref{qn1} and Eq.~\eqref{2apb2clarge} imply $2^a-1\equiv \modd{2^c-1} {q_n}$. Let $t=\ord_{q_n}(2)$. Then $a\equiv \modd{c} {t}$. Hence, exists $a'$ such that $a=c+a't$. But, by Eq.~\eqref{qn2}, $\ord_{F_n}(2)\,\mid\,t$, which implies
\[2^c-1\equiv 2^a\equiv 2^c\,\left (2^{t}\right )^{a'}\equiv\modd{ 2^c} {F_{n}}.\]
Since the previous equation has no solutions, we obtain a contradiction. Thus, we obtain the stated result. 
\end{proof}

\begin{lemma}\label{LemaCondensado}
Let $F_n$ be the $n$-th Fermat number and consider the following exponential diophantine equations 
\begin{equation}\label{2apb1}
	2^a-F_n^b = -1
\end{equation}	
and
\begin{equation}\label{2apb2c}
	2^a-F_n^b = 2^c - 1.
\end{equation}
Then 
\begin{itemize}
	\item[(a)] when $n\in\{1,2,3,4\}$, Equation~\eqref{2apb1} only holds for $(a,b)=(2^n,1)$;
	\item[(b)] when $n\in\{2,3,4\}$, Equation~\eqref{2apb2c} only holds for 
\[(a,b,c) \in \{(a,0,a)\,\mid\, a\in\mathbb{Z}\}\cup\{(2^n+1,1,2^n)\};\]
	\item[(c)] when $n=1$, Equation~\eqref{2apb2c} only holds for 
\[(a,b,c) \in \{(a,0,a)\,\mid\, a\in\mathbb{Z}\}\cup\{(3,1,2),(7,3,2),(5,2,3)\}.\]
\end{itemize}  
\end{lemma}

\begin{proof}
Suppose  $n\in\{1,2,3,4\}$, $2^a-F_n^b=-1$, and $a \geq 2^n+1$. Therefore, $2^{2^n+1} \mid F_n^{b}-1$. So $b$ is even and we have $b=2b'$, for some positive integer $b'$. But, since $3\mid 5^2-1$, $9 \mid 17^2-1$, $43 \mid 257^2-1$, and $11 \mid 65537^2-1$, we always obtain a contradiction. Therefore, $(2^n,1)$ is the only solution of Eq.~\eqref{2apb1}, and we obtain statement {\it (a)}.

Now, we consider Eq.~\eqref{2apb2c} with $n\in\{1,2,3,4\}$. Notice that we always have $a\geq c$. If $b=0$ then $a=c$ and so we have $(a,b,c)=(a,0,a)$. Since $2^c(2^{a-c}-1) = F_n^b-1$ and $2^{2^n} \mid F_n^b-1$, for any positive integer $b$, we have $c\geq 2^n$. Therefore, if $b=1$ then $c=2^n$ and $a=2^n+1$. 

From now on, we assume $b\geq 2$. Since $F_n$ is a Fermat number, we have $a\geq 2^{n+1}+1$. As $2^n+n&lt;2^{n+1}$, then Eq.~\eqref{2apb2c} implies
\begin{equation}\label{bs}
F_n^b\equiv\modd{ 1-2^c} {2^{2^n+n}}.
\end{equation}
By Lemma \ref{Fnlem}, we have \(\ord_{2^{2^n+n}}(F_n)=2^n\). Then, for each $c$, exist integers $b_c$ and $b_c'$ such that $b=b_c+2^nb_c'$. Notice that $b_c$ and $b_c'$ also depend on which $n$ we are considering. By Lemma \ref{Fnlem}, we have $F_{n-1}\mid F_n^{2^n}-1$. Hence, 
\begin{equation}\label{as}
2^a-F_n^{b_c}\equiv \modd{2^c-1} {F_{n-1}}.
\end{equation}
As $\ord_{F_{n-1}}(2)=2^n$, then exist integers $a_c$ and $a_c'$ such that $a=a_c+2^na_c'$.
But then 
\begin{equation}\label{m1Fn}
2^{a_c}(-1)^{a_c'}\equiv \modd{2^c-1} {F_{n}}.
\end{equation}
We now analyze Eq.~\eqref{2apb2c} for each $n\in\{1,2,3,4\}$.

Suppose $n=1$. 

Since $s=2$ and $q_1=13$ satisfy the conditions of Lemma \ref{largec}, then Eq.~\eqref{2apb2c} has no solutions for $c\geq 4$. Therefore, $c=2$ or $c=3$.

Clearly, $(7,3,2)$ is the only solution of Eq.~\eqref{2apb2c}, with $c=2$ and $a\leq 7$. If $a \geq 8$ and $c=2$, then $5^b \equiv  \modd{-3} {256}$. Therefore, $b \equiv  \modd{35} {64}$, i.e., there exists a non-negative integer $b'$ such that $b = 35 + 64b'$. Then
$$2^a - 5^{35} \cdot 5^{64b'} = 3.$$
As $641 \mid 5^{64}-1$ and $5^{35} \equiv  \modd{516} {641}$, then
$$2^a  \equiv  \modd{519} {641}.$$
But this congruence has no solutions.

If $c=3$ and $a\leq 5$, then $(5,2,3)$ is the only solution of Eq.~\eqref{2apb2c}. Suppose $c=3$ and $a\geq 6$, then $5^b \equiv \modd{-7} {64}$. Therefore, we have $b \equiv  \modd{10} {16}$; i.e., there exists a non-negative integer $b'$ such that $b = 10 + 16b'$. Then
$$2^a - 5^{10} \cdot 5^{16b'} = 7.$$
As $13 \mid 5^{16}-1$ and $5^{10} \equiv  \modd{-1} {13}$, then
$$2^a \equiv  \modd{6} {13}.$$
Therefore, $a \equiv \modd{5} {12}$. On the other hand, since  $5^{10} \equiv \modd{9} {17}$ and $17 \mid 5^{16}-1$, then
$$2^a \equiv \modd{-1} {17}.$$
Hence, we have $a \equiv\modd{4} {8}$, which gives us a contradiction. Thus, we obtain statement {\it (c)}.

Let us now consider $n=2$. 

Since $s=4$ and $q_1=18913$ satisfy the conditions of Lemma \ref{largec}, then Eq.~\eqref{2apb2c} has no solutions for $c\geq 8$.

For each $c \in \{4,5,6,7\}$, consider $b=b_c+2^nb_c'$ to be the solutions of Eq.~\eqref{bs} and $a=a_c+2^na_c'$ to be the solutions of Eq.~\eqref{as} (when these equations have solution). The values of $b_c$ and $a_c$ are given in Table~\ref{table1}.

\begin{table}[ht]
	\centering 
\begin{tabular}{ccc}
	\hline\hline 
	$c$ &amp; $b_c$ &amp; $a_c$ \\\hline
	$4$ &amp; $3$ &amp; $3$ \\
	$5$ &amp; $2$ &amp; No solutions \\
	$6$ &amp; $0$ &amp; $2$ \\	
	$7$ &amp; $0$ &amp; $2$\\\hline
\end{tabular}
\caption{Solutions of Eq.~\eqref{bs} and Eq.~\eqref{as} for $n=2$.}\label{table1}
\end{table}
Hence, when $c=5$, Eq.~\eqref{as} has no solutions. Also, when $c\in\{4,6,7\}$, Eq.~\eqref{m1Fn} has no solutions. 	 

Next, we consider $n=3$.  

Since $s=4$ and $q_3=193$ satisfy the conditions of Lemma \ref{largec}, then Eq.~\eqref{2apb2c} has no solutions for $c\geq 12$.

For each $c \in \{8,9,10,11\}$, consider $b=b_c+2^nb_c'$ to be the solutions of Eq.~\eqref{bs} and $a=a_c+2^na_c'$ to be the solutions of Eq.~\eqref{as} (when these equations have solution). The values $b_c$ and $a_c$ are given in table~\ref{table2}.

\begin{table}[ht]
	\centering 
	\begin{tabular}{ccc}
		\hline\hline 
	$c$ &amp; $b_c$ &amp; $a_c$ \\\hline
	$8$ &amp; $7$ &amp; $7$ \\
	$9$ &amp; $6$ &amp; No solutions \\
	$10$ &amp; $4$ &amp; $1$ \\	
	$11$ &amp; $0$ &amp; $3$\\\hline
\end{tabular}
	\caption{Solutions of Eq.~\eqref{bs} and Eq.~\eqref{as} for $n=3$.}\label{table2}
\end{table}
Hence, when $c=9$, Eq.~\eqref{as} has no solutions. Also, when $c\in\{8,10,11\}$, Eq.~\eqref{m1Fn} has no solutions. 	

Finally, consider $n=4$.

Since $s=5$ and $q_4=38899171806337$ satisfy the conditions of Lemma \ref{largec}, then Eq.~\eqref{2apb2c} has no solutions for $c\geq 21$.

For each $c \in \{16,17,18,19,20\}$, consider $b=b_c+2^nb_c'$ to be the solutions of Eq.~\eqref{bs} and $a=a_c+2^na_c'$ to be the solutions of Eq.~\eqref{as} (when these equations have solution). The values $b_c$ and $a_c$ are given in table~\ref{table3}.
\begin{table}[h!]
\centering 
\begin{tabular}{ccc}
	\hline\hline 
	$c$ &amp; $b_c$ &amp; $a_c$ \\\hline
	$16$ &amp; $15$ &amp; $15$ \\
	$17$ &amp; $14$ &amp; No solutions \\
	$18$ &amp; $12$ &amp; No solutions \\	
	$19$ &amp; $8$ &amp; No solutions\\
	$20$ &amp; $0$ &amp; $4$\\\hline
\end{tabular}
\caption{Solutions of Eq.~\eqref{bs} and Eq.~\eqref{as} for $n=4$.}\label{table3}
\end{table}

Hence, when $c\in\{17,18,19\}$, Eq.~\eqref{as} has no solutions. When $c\in\{16,20\}$, Eq.~\eqref{m1Fn} has no solutions. 

Thus, statement {\it (b)} is obtained.
\end{proof}

We are now in conditions to prove a generalization of the Euclid-Euler theorem for $\frac{2F_n}{F_n-1}$-perfect numbers, when $n\in\{1,2,3,4\}$. The proof of the next theorem follows the same steps as the proof of Theorem \ref{even3}.
\begin{theorem}\label{TeoremaCondensado}
Let $F_n$ be the $n$-th Fermat number. Then
\begin{enumerate}	
	\item if exists $N$ such that $\frac{\sigma(N)}{N}=\frac{F_1}{2}$ and $F_1 \mid N$, then $2^4 \parallel N, F_1^2 \parallel N$, and $31^2 \mid N$.
\item if $n\in\{2,3,4\}$ then $\frac{2F_n}{F_n-1}$ is a $F_n$-abundancy outlaw.
\end{enumerate}
\end{theorem}

\begin{proof}
Let $n \in \{1,2,3,4\}$ and $F_n=2^{2^n}+1$. We can write $N=2^a F_n^bm$ such that $a,b\geq 1$, $\gcd(2F_n,m)=1$, and
$$\sigma(N)=\frac{2F_n}{F_n-1}N.$$ 
Then
	$$\frac{2F_n}{F_n-1} = \frac{\sigma(N)}{N} = \frac{2^{a+1}-1}{2^a} \, \frac{F_n^{b+1}-1}{F_n^b(F_n-1)} \, \frac{\sigma(m)}{m}.$$
	Therefore,
	$$\frac{\sigma(m)}{m} = \frac{2^{a+1} \, F_n^{b+1}}{(2^{a+1}-1)(F_n^{b+1}-1)}.$$
	Let 
\[d = \gcd\left(2^{a+1} \, F_n^{b+1}, (2^{a+1}-1)(F_n^{b+1}-1)\right).\]
As $F_n-1 \mid F_n^{b+1}-1$, then $d = 2^s \, F_n^t$, where $2^n \leq s \leq a+1$ and $0 \leq t \leq b+1$.
	Since
	\[\gcd\left  (\frac{2^{a+1}F_n^{b+1}}{2^s\, F_n^t},\frac{(2^{a+1}-1)(F_n^{b+1}-1)}{2^s\, F_n^t}\right )=1,\]
	then
	\[\sigma(m)=\frac{2^{a+1}F_n^{b+1}}{2^s\, F_n^t}k\mbox{ and }m=\frac{2^{a+1}-1}{F_n^t} \, \frac{F_n^{b+1}-1}
	{2^s}k,\]
	for some positive integer $k$.

Let us consider the following three cases, which will establish the claim.

\bigskip

\noindent \textit{Case A:} Suppose that $t \neq 0$ and let
	\[M=\max\left(\frac{2^{a+1}-1}
	{F_n^t},\frac{F_n^{b+1}-1}{2^s}\right ).\]
	Then we have that
	\begin{align*}
		\frac{\sigma(m)}{k} &amp; =\frac{2^{a+1}F_n^{b+1}}{2^s\, F_n^t}\\
		&amp; =\frac{2^{a+1}-1}{F_n^t}\,\frac{F_n^{b+1}-1}{2^s}+\frac{2^{a+1}-1}{F_n^t} \, \frac{1}{2^s}+\frac{F_n^{b+1}-1}{2^s} \, \frac{1}{F_n^t}+\frac{1}{2^s\, F_n^t}\\
		&amp; &lt;\frac{m}{k}+\frac{M}{F_n-1}+\frac{M}{F_n}+1\\
		&amp; &lt;\frac{m}{k}+M+1.
	\end{align*}
Therefore,
\begin{equation}\label{smallerF}
	\sigma(m)&lt;m+Mk+k.
\end{equation}
If $Mk \neq m$ and $M \neq 1$, then $m$ has at least the divisors $m$, $Mk$, and $k$. Thus, 
\begin{equation}\label{largerF}
\sigma(m)\geq m+Mk+k.
\end{equation}
The combination of inequalities~\eqref{smallerF} and \eqref{largerF}, give us a contradiction.
	
If $Mk=m$ or $M=1$, then
	$$\frac{2^{a+1}-1}{F_n^t}=1\mbox{ or }\frac{F_n^{b+1}-1}{2^s} = 1.$$
	Thus $2^{a+1}-F_n^t=1$ or $F_n^{b+1}-2^{s} = 1$.
By Lemma \ref{LemaCondensado}, the only solutions of these equations are $(a,t)=(0,0)$ and $(b,s)=(0,2^n)$. Hence, we obtain a contradiction because $a,b \geq 1$.

\bigskip
	
\noindent \textit{Case B:} Suppose $t = 0$ and
	\[2^{a+1}-1 \neq\frac{F_n^{b+1}-1}{2^s}.\]
	Let \[M'=\min\left(2^{a+1}-1,\frac{F_n^{b+1}-1}{2^s}\right).\]
	If $M'=1$ then
	$$2^{a+1}-1=1\mbox { or }\frac{F_n^{b+1}-1}{2^s}=1.$$
	Thus $2^{a+1}=2$ or $F_n^{b+1}-2^s=1$.
	As $a,b\geq 1$, by Lemma \ref{LemaCondensado}, we have a contradiction.
	
If $M'\neq 1$ then $m$ has at least the divisors
	\[m,\left (2^{a+1}-1\right )k,\frac{F_n^{b+1}-1}{2^s}k\mbox{ and }k.
	\]
	Therefore,
	\begin{align*}
		\sigma(m) &amp; \geq m+(2^{a+1}-1)k+\frac{F_n^{b+1}-1}{2^s}k+k\\
		&amp; &gt;(2^{a+1}-1)\, \frac{F_n^{b+1}-1}{2^s}k+\frac{2^{a+1}-1}{2^s}k+\frac{F_n^{b+1}-1}{2^s}k+\frac{k}
		{2^s} \\
		&amp; =\sigma(m),
	\end{align*}
	where the strict inequality results from $s\geq 2^n$. Hence, we obtain a contradiction.

\bigskip
	
\noindent 	\textit{Case C:} Suppose $t = 0$ and
	\[2^{a+1}-1 = \frac{F_n^{b+1}-1}{2^s}.\]
	Then $2^{a+1+s}-F_n^{b+1} = 2^s - 1$. By Lemma \ref{LemaCondensado}, we have
	\begin{equation*}
		\begin{split}
	(F_n,a,b,s)\in\{(5,-1,-1,1),\ (5,0,0,2),\ (5,1,&amp;1,3),\ (5,4,2,2),\ (17,0,0,4),\\
	&amp; \quad (257,0,0,8),\ (65537,0,0,16)\}.
\end{split}
\end{equation*}
	Since $a,b\geq 1$, we only have solutions for $F_n=5$ and then $(F_n,a,b) = (5,1,1)$ or $(F_n,a,b)=(5,4,2)$.
	
	If $(F_n,a,b)=(5,1,1)$ then
	$$\frac{\sigma(N)}{N}=\frac{5}{2} = \frac{3}{2} \cdot \frac{6}{5} \cdot \frac{\sigma(m)}{m}.$$
	Therefore, $9 \mid m$. But then
	$$\frac{\sigma(N)}{N}=\frac{5}{2} \geq \frac{3}{2} \cdot \frac{6}{5} \cdot \frac{13}{9}&gt;\frac{5}{2}.$$
	Hence, $(F_n,a,b)=(5,4,2)$ and so $31^2\mid N$.

\end{proof}

\section{Acknowledgment}

This work was supported by The Center for Research and Development in Mathematics and Applications (CIDMA) through the Portuguese Foundation for Science and Technology (FCT---Funda\c{c}\~{a}o para a Ci\^{e}ncia e Tecnologia), reference UIDB/04106/2020 and reference PD/BD/150533/2019. 

\begin{thebibliography}{10}

\bibitem{dickson}
L.~E. Dickson, {\em History of the Theory of Numbers, Volume I: Divisibility
  and Primality}, Chelsea Publishing Company, 1919.

\bibitem{eu}
L.~Euler, {\em Commentationes Arithmeticae Collectae}, Vol.~2, Imperial Academy
  of Sciences, St.\ Petersburg, 1849.

\bibitem{fla}
A.~Flammenkamp, The multiply perfect numbers page,
  \url{http://wwwhomes.uni-bielefeld.de/achim/mpn.html}, 2022.

\bibitem{GLS}
R.~K. Guy, C.~B. Lacampagne, and J.~L. Selfridge, Primes at a glance, {\em
  Math. Comp.} {\bf 48} (1987), 183--202.

\bibitem{hol}
J.~A. Holdener and W.~G. Stanton, Abundancy ``outlaws" of the form
  $(\sigma({N})+t)/{N}$, {\em J. Integer Sequences} {\bf 10} (2007),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL10/Holdener/holdener7.html}{\tt Article 07.9.6}.

\bibitem{Mih}
P.~Mih\v{a}ilescu, Primary cyclotomic units and a proof of catalans conjecture,
  {\em J. Reine Angew. Math.} {\bf 2004} (2004), 167--195.

\bibitem{rya}
R.~Ryan, A simpler dense proof regarding the abundancy index, {\em Math. Mag.}
  {\bf 76} (2003), 299--301.

\bibitem{sco}
R.~Scott, On the equations $p^x-b^y=c$ and $a^x+b^y=c^z$, {\em J. Number
  Theory} {\bf 44} (1993), 153--165.

\bibitem{ste}
R.~Steuerwald, Ein {S}atz {\"u}ber nat{\"u}rliche {Z}ahlen ${N}$ mit
  $\sigma({N})=3{N}$, {\em Arch. Math. (Basel)} {\bf 5} (1954), 449--451.

\bibitem{sty}
R.~Styer, Small two-variable exponential diophantine equations, {\em Math.
  Comp.} {\bf 60} (1993), 811--816.

\bibitem{wie}
P.~Weiner, The abundancy ratio, a measure of perfection, {\em Math. Mag.} {\bf
  73} (2000), 307--310.

\end{thebibliography}

\newpage

\bigskip
\hrule
\bigskip

\noindent{\it 2020 Mathematics Subject Classification}: Primary 11A25; Secondary 11N25, 11B83.

\noindent {\it Keywords:} Euclid-Euler theorem, $\alpha$-perfect number, Fermat number, abundancy index, abundancy outlaw.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences \seqnum{A005820}, \seqnum{A019434}, \seqnum{A214409}, and \seqnum{A214413}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  July 2 2022; 
revised version received   July 3 2022; October 5 2022; October 6 2022; 
October 12 2022.
Published in {\it Journal of Integer Sequences}, October 14 2022.

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