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\begin{center}
\vskip 1cm{\LARGE\bf On the Equation $\sigma(n)=n+\phi(n)$
}
\vskip 1cm
\large
Douglas E. Iannucci\\
Towers Condominiums B-16\\
3600 Contant\\
St. Thomas, VI 00802\\
USA\\
\href{mailto:diannuc@gmail.com}{\tt diannuc@gmail.com} 
\end{center}

\vskip .2 in

\begin{abstract}
In this paper we consider the equation $\sigma(n)=n+\phi(n)$, for which $n=2$ is the only known solution. We provide necessary conditions for the existence of any larger solutions.
\end{abstract}

\section{Introduction}
Let $\sigma(n)$ denote the sum of the (positive) divisors of $n$, and let
$\phi(n)$ denote Euler's phi function (the number of positive integers $\leq n$ and
relatively prime to it).
We remark  that~$n=2$ is a solution to the equation 
\begin{equation}\label{titleq}
\sigma(n)=n+\phi(n)
\end{equation}
over natural numbers~$n$. By Hardy and Wright~\cite[Thm.~324]{HW},
the average value of $\sigma(n)$ is $\frac16\pi^2n$. By \cite[Thm.~330]{HW}, the average value of $n+\phi(n)$ is $(\pi^2+6)n/\pi^2$.  Since $\frac16\pi^2\approx1.645$ and $(\pi^2+6)/\pi^2\approx1.608$, it is quite natural to ask if~\eqref{titleq}
has any solutions $n&gt;2$. None are known, and the literature seems devoid of any mention to this specific problem. Other, similar problems have been discussed in publications such as Guy~\cite{guy}; e.g., \S{B-38}, $\phi(m)=\sigma(n)$, and \S{B-42}, $\phi(\sigma(n))=\sigma(\phi(n))$, $\phi(\sigma(n))=n$, $\phi(\sigma(n))=\phi(n)$. Some such problems have been solved and are popular topics of discussion on many internet forums; e.g., $\sigma(n)+\phi(n)=2n$ if and only if~$n$ is a prime or $n=1$. There are three sequences in {\it OEIS\/} (Sloane~\cite{oeis}) related to~\eqref{titleq}; e.g., \seqnum{A070159} (all~$n$ such that $\sigma(n)-n$ divides $\phi(n)$). In fact, all solutions $n&gt;2$ to~\eqref{titleq}, if they exist, would belong to sequence \seqnum{A055940}, which gives composite~$n$ such that $\sigma(n)-n$ divides $\phi(n)$. For, when~$p$ is prime then $\phi(p)/(\sigma(p)-p)=p-1$, hence~2 is the only prime solution to~\eqref{titleq}. Thus we seek those members~$n$ of \seqnum{A055940} where the ratio $\phi(n)/(\sigma(n)-n))=1$; none are known. A similar comment applies to \seqnum{A066679}, giving~$n$ such that $\sigma(n)-n$ is divisible by~$\phi(n)$: all members less than $10^{11}$ are given, the only one of which giving the ratio $(\sigma(n)-n)/\phi(n)=1$ being $n=2$. Are there any more?

We conjecture that~$n=2$ is the only solution to~\eqref{titleq}, and in this paper we obtain necessary conditions for a solution~$n&gt;2$ to exist. 



\section{Preliminaries}
Dividing both sides of~\eqref{titleq} by~$n$ yields




\begin{equation}\label{index}
\frac{\sigma(n)}{n}=1+\frac{\phi(n)}{n}.
\end{equation}
The function~$\sigma(n)/n$ is multiplicative. For all primes~$p$ and natural numbers~$a$,




\begin{equation}\label{geom}
\frac{\sigma(p^a)}{p^a}=1+\frac1{p}+\frac1{p^2}+\cdots+\frac1{p^a}
\end{equation}
is monotonically increasing with~$a$; hence taking the limit of the right hand side of~\eqref{geom} as $a\to\infty$ yields



\begin{equation}\label{upssala}
\frac{\sigma(p^a)}{p^a}&lt;\frac{p}{p-1}.
\end{equation}
If $p&lt;q$ are primes then $q/(q-1)\le(p+1)/p$, thus by~\eqref{geom} and~\eqref{upssala},



\begin{equation}\label{prior}
\frac{\sigma(q^b)}{q^b}&lt;\frac{\sigma(p^a)}{p^a}
\end{equation}
for all natural numbers~$a$, $b$. As well,




\begin{equation}\label{prior2}
\frac{q^b}{\phi(q^b)}=\frac{q}{q-1}&lt;\frac{p}{p-1}=\frac{p^a}{\phi(p^a)}
\end{equation}
for all natural numbers~$a$, $b$. By~\eqref{upssala} we have




\begin{equation}\label{mono}
\frac{\sigma(n)}{n}&lt;\prod_{p\mid n}\frac{p}{p-1}=\frac{n}{\phi(n)}.
\end{equation}
As usual we let $\alpha=\frac12(1+\sqrt5)$, the golden ratio. Immediately we obtain the following results.



\begin{theorem}\label{alpha}
If~$n$ is a solution to~\eqref{titleq} then
$$\frac{\sigma(n)}{n}&lt;\alpha&lt;\frac{n}{\phi(n)}.$$
\end{theorem}

\begin{proof}
Setting $x=n/\phi(n)$, by~\eqref{index} and~\eqref{mono} we obtain the quadratic inequality
$$x^2-x-1&gt;0,$$
which has the positive solution $x&gt;\alpha$. Thus the right hand inequality is proved. Otherwise by~\eqref{index}
$$\frac{\sigma(n)}{n}=1+\frac1{x}&lt;1+\frac1{\alpha}=\alpha,$$
thus proving the left hand inequality.
\end{proof}




\begin{theorem}\label{odd}
If~$n&gt;2$ is a solution to~\eqref{titleq} then~$n$ is an odd square.
\end{theorem}

\begin{proof}
If~$a$ and~$b$ are natural numbers, then $\sigma(a)/a\le\sigma(b)/b$ whenever $a\mid b$. Thus if $4\mid n$, 
$$\frac{\sigma(n)}{n}\ge\frac74&gt;\alpha,$$
whence~$n$ is not a solution to~\eqref{titleq} by Theorem~\ref{alpha}. Thus any even solution greater than~2 has the form~$n=2r$, where~$r&gt;1$ is odd. Thus by~\eqref{titleq}, since $\phi(r)&lt;r$,
$$3\sigma(r)=2r+\phi(r)&lt;2r+r=3r,$$
a contradiction. Hence~$n$ is odd. Since $\phi(n)$ is even, the right hand side of~\eqref{titleq} is odd. Hence~$\sigma(n)$ is odd. For odd natural numbers~$n$, $\sigma(n)$ is odd if and only if $n$ is square.
\end{proof}

By Theorem~\ref{odd}, $n=m^2$, where~$m&gt;1$ is odd, for all solutions $n&gt;2$ to~\eqref{titleq}. Since $\phi(m^2)=m\phi(m)$, it follows that by~\eqref{titleq} that




\begin{equation}\label{oddeq}
\sigma(m^2)=m(m+\phi(m)).
\end{equation}
Thus we seek necessary conditions for an odd natural number~$m&gt;1$ to satisfy~\eqref{oddeq}.

The cyclotomic polynomials~$\Phi_n(x)$  may be defined recursively by $\Phi_1(x)=x-1$ and
$$x^n-1=\prod_{d\mid n}\Phi_d(x)$$
for all~$n&gt;1$. Thus 




\begin{equation}\label{cycsig}
\sigma(p^a)=\prod_{\substack{d\mid a+1\\d&gt;1}}\Phi_d(p)
\end{equation}


for primes~$p$ and natural numbers~$a$. Nagell~\cite[Thm.~94]{Nag}, has shown



\begin{lemma}\label{nagell}
Let~$p$ and~$q$ be distinct odd primes and let $h=e_q(p)$, where~$e_q(p)$ denotes the exponent to which~$p$ belongs modulo~$q$. Then $q\mid\Phi_n(p)$ if and only if $n=hp^{\gamma}$ for some~$\gamma\ge0$. If~$\gamma&gt;0$ then $q\|\Phi_{hp^{\gamma}}(p)$.
\end{lemma}
For prime~$p$ and natural number~$n$ we denote the $p$-valuation of~$n$ by~$v_p(n)$. Thus $v_p(n)=k$ if and only if $p^k\|n$. By~\eqref{cycsig} and Lemma~\ref{nagell},




\begin{equation}\label{vsig}
v_q(\sigma(p^a))=\begin{cases}0,&amp;\text{if $h\nmid a+1$;}\\
v_q(a+1),&amp;\text{if $h=1$;}\\
v_q(p^h-1)+v_q(a+1),&amp;\text{if $h&gt;1$, $h\mid a+1$.}\end{cases}
\end{equation}



\section{Lower Bounds on $\boldsymbol{\omega(m)}$}
The arithmetic function~$\omega$ counts the number of prime power components of a natural number~$n$;  i.e., if~$n$ has the unique prime factorization $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$, then $\omega(n)=k$. 



\begin{theorem}\label{om2}
If~$m$ is a solution to~\eqref{oddeq} then~$\omega(m)&gt;2$.
\end{theorem}
\begin{proof}
That $\omega(m)&gt;1$ follows from~\eqref{prior2} and Theorem~\ref{alpha}, since otherwise
$$\frac{m}{\phi(m)}\le\frac32&lt;\alpha,$$
contradicting Theorem~\ref{alpha}.

Similarly if~$\omega(m)=2$ then~$3\mid m$, otherwise
$$\frac{m}{\phi(m)}\le\frac54\cdot\frac76&lt;\alpha.$$
Thus~$m=3^ap^b$ for some odd prime~$p&gt;3$ and natural numbers~$a$ and~$b$. Then~$p\ge11$ (by~\eqref{prior} and Theorem~\ref{alpha}), otherwise 
$$\frac{\sigma(m^2)}{m^2}\ge\frac{\sigma(3^2)}{3^2}\cdot\frac{\sigma(7^2)}{7^2}=\frac{13}9\cdot\frac{57}{49}&gt;\alpha.$$
Then $p\le13$ (by~\eqref{prior2} and Theorem~\ref{alpha}), otherwise 
$$\frac{m}{\phi(m)}\le\frac32\cdot\frac{17}{16}&lt;\alpha.$$
Suppose $p=11$. Thus $m=3^{a}11^{b}$. Thus by~\eqref{oddeq},
$$\sigma(m^2)=\sigma(3^{2a}11^{2b})=3^{2a-1}11^{2b-1}(3\cdot11+\phi(3\cdot11)).$$
Since $33+\phi(33)=53$, we have $53\mid\sigma(3^{2a})\sigma(11^{2b})$. By~\eqref{vsig}, this is impossible since $e_{53}(3)=52$ and $e_{53}(11)=26$.

Likewise $p=13$ implies $63\mid\sigma(3^{2a})\sigma(11^{2b})$ for some natural numbers~$a$ and~$b$; thus $3\mid\sigma(11^{2b})$, which is impossible since $e_3(11)=2$.
\end{proof}



\begin{theorem}\label{om3}
If~$m$ is a solution to~\eqref{oddeq} then~$\omega(m)&gt;3$.
\end{theorem}
\begin{proof}
By Theorem~\ref{om2} it suffices to show that the assumption $\omega(m)=3$ is untenable. By~\eqref{prior}, we have $3\mid m$, since otherwise
$$\frac{m}{\phi(m)}\le\frac54\cdot\frac76\cdot\frac{11}{10}&lt;\alpha,$$
contradicting Theorem~\ref{alpha}.
 
Thus $m=3^ap^bq^c$ for odd primes $3&lt;p&lt;q$ and natural numbers~$a$, $b$, and~$c$. Similarly, $5\nmid m$ and $7\nmid m$ since otherwise
$$\frac{\sigma(m^2)}{m^2}\ge\frac{13}9\cdot\frac{57}{49}&gt;\alpha.$$ 
Furthermore $p\le23$ since otherwise
$$\frac{m}{\phi(m)}\le\frac32\cdot\frac{29}{28}\cdot\frac{31}{30}&lt;\alpha.$$ 
Thus $p\in\{11,\,13,\,17,\,19,\,23\}$. We shall prove that all the five cases are impossible. Note that by~\eqref{oddeq},




\begin{equation}\label{aux3}
\sigma(3^{2a})\sigma(p^{2b})\sigma(q^{2c})=3^{2a-1}p^{2b-1}q^{2c-1}(3pq+2(p-1)(q-1)).
\end{equation}

Suppose $p=23$. Then $q\le31$ since otherwise
$$\frac{m}{\phi(m)}\le\frac32\cdot\frac{23}{22}\cdot\frac{37}{36}&lt;\alpha.$$ 
Note that $3\mid\sigma(23^{2b})\sigma(q^{2c})$ by~\eqref{aux3}, thus by~\eqref{vsig} we have $q\equiv1\pmod3$, hence~$q=31$. Furthermore $3\mid2c+1$, hence $\Phi_3(31)\mid3^{2a}\cdot23^{2b-1}\cdot31^{2c-1}\cdot1153$ by~\eqref{aux3}. But this is impossible since $331\mid\Phi_3(31)$. Thus $p\ne23$.

Suppose $p=19$. Then $q\le43$ since otherwise
$$\frac{m}{\phi(m)}\le\frac32\cdot\frac{19}{18}\cdot\frac{47}{46}&lt;\alpha.$$ 
Thus $q\in S$, where $S=\{23,\,29,\,31,\,37,\,41,\,43\}$. By~\eqref{aux3} we have

$$\sigma(3^{2a})\sigma(19^{2b})\sigma(q^{2c})=3^{2a}19^{2b-1}q^{2c-1}(31q-12).$$

We remark that $3\nmid\sigma(19^{2b})$, since otherwise by~\eqref{cycsig} and~\eqref{vsig} we have $127\mid31q-12$ (as $\Phi_3(19)=3\cdot127$), but this is false for all $q\in S$. Hence $3\mid\sigma(q^{2c})$, thus by~\eqref{cycsig} and~\eqref{vsig}, $q\in\{31,\,37,\,43\}$ and $\Phi_3(q)\mid3^{2a}19^{2b-1}(31q-12)$; this is false. Thus $p\ne19$.


Suppose $p=17$. Thus by~\eqref{aux3} 
$$\sigma(3^{2a})\sigma(17^{2b})\sigma(q^{2c})=3^{2a-1}17^{2b-1}q^{2c-1}(83q-32).$$
By~\eqref{cycsig} and~\eqref{vsig} we have $17\mid\sigma(q^{2c})$, thus $q\equiv1\pmod{17}$, hence $q\ge103$. Then
$$\frac{m}{\phi(m)}\le\frac32\cdot\frac{17}{16}\cdot\frac{103}{102}&lt;\alpha,$$
contradicting Theorem~\ref{alpha}. Thus $p\ne17$.

Suppose $p=13$. Then $q\ge37$ since otherwise
$$\frac{\sigma(m^2)}{m^2}\ge\frac{13}{9}\cdot\frac{183}{169}\cdot\frac{993}{961}&gt;\alpha.$$
Furthermore $v_3(m)\le2$ since otherwise
$$\frac{\sigma(m^2)}{m^2}\ge\frac{1093}{729}\cdot\frac{183}{169}&gt;\alpha.$$
Hence we consider the two cases, $v_3(m)=2$ and $v_3(m)=1$. 

In the former case we have $v_{13}(m)\le2$ since otherwise
$$\frac{\sigma(m^2)}{m^2}\ge\frac{121}{81}\cdot\frac{5229043}{4826809}&gt;\alpha,$$
leaving two subcases, $v_{13}(m)=2$ or $v_{13}(m)=1$. The first subcase yields by~\eqref{aux3},
$$11^2\cdot30941\sigma(q^{2c})=3^4\cdot13^3q^{2c-1}(21q-8),$$
implying both that $q=30941$ and $11\mid21q-8$, a contradiction. Similarly, the second subcase  yields
$$11^2\cdot61\sigma(q^{2c})=3^3\cdot13^3q^{2c-1}(21q-8),$$
implying both that $q=61$ and $11\mid21q-8$, a contradiction. 

In the latter case, viz., $v_3(m)=1$, we have by~\eqref{aux3}




\begin{equation}\label{aux313}
\sigma(13^{2b})\sigma(q^{2c})=3^2\cdot13^{2b-2}q^{2c-1}(21q-8).
\end{equation}
Note that 




\begin{equation}\label{insert313b}
b&gt;1, 
\end{equation}
otherwise~\eqref{aux313} yields $61\sigma(q^{2c})=3q^{2c-1}(21q-8)$ whence $q=61$. Thus $\sigma(61^{2c})=3\cdot61^{2c-2}\cdot19\cdot67$, thus $c=1$, thus the contradiction $13\cdot97=19\cdot67$. Since $b&gt;1$, it follows from~\eqref{aux313} that $13\mid\sigma(q^{2c})$.

We show here $q\not\equiv1\pmod{13}$. Otherwise by~\eqref{aux313} and~\eqref{vsig} $2b-2=v_{13}(\sigma(q^{2c}))=v_{13}(2c+1)$, hence $2c+1\ge13^{2b-2}$, hence $2c-1&gt;10^b$. Thus (since~\eqref{aux313} implies $q^{2c-1}\mid\sigma(13^{2b})$) we have $\sigma(13^{2b})\ge q^{2c-1}&gt;q^{10^b}&gt;13^{10^b}&gt;13^{2b+1}&gt;\sigma(13^{2b})$, a contradiction. Hence $q\not\equiv1\pmod{13}$; therefore by~\eqref{vsig}, 




\begin{equation}\label{insert313c}
c\equiv1\pmod3; 
\end{equation}
furthermore




\begin{equation}\label{aux313t0}
2b-2=v_{13}(\sigma(q^{2c})=v_{13}(q^2+q+1)+v_{13}(2c+1).
\end{equation}
Let $t=v_{13}(2c+1)$. Suppose $t&gt;0$. Since $v_{13}(q+2+q+1)&lt;3\log_{13}q$, from~\eqref{aux313t0} it follows that




\begin{equation}\label{aux313t1}
2b-2&lt;3\log_{13}q+t.
\end{equation}
Since $13^t\mid2c+1$, we have $2c-1\ge13^{t}-2$, hence by~\eqref{aux313}
$$13^{2b+1}&gt;\sigma(13^{2b})\ge q^{2c-1}\ge q^{13^{t}-2},$$
hence




\begin{equation}\label{aux313t2}
2b-2&gt;(13^{t}-2)\log_{13}q-3.
\end{equation}
By~\eqref{aux313t1} and~\eqref{aux313t2} it follows (since $\log_{13}(q)&gt;1$) that
$$t&gt;(13^t-5)\log_{13}q-3&gt;13^t-8,$$
a contradiction if $t\ge1$. Therefore $t=0$. Thus by~\eqref{vsig}, \eqref{aux313},



\begin{equation}\label{aux31331}
2b-2=v_{13}(\sigma(q^{2c}))=v_{13}(q^2+q+1).
\end{equation}

Since $q^2+q+1&lt;q^3$, from~\eqref{aux313} and~\eqref{aux31331} it follows that
\begin{gather*}
13^{2b-1}&gt;\sigma(13^{2b})\ge q^{2c-1}=(q^3)^{\frac{2c-1}3}&gt;(q^2+q+1)^{\frac{2c-1}3}\ge(13^{2b-2})^{\frac{2c-1}3}\\
=13^{\frac23(b-1)(c-2)}.\end{gather*}
Thus $2b-1&gt;\frac23(b-1)(2c-1)$; equivalently, $(b-1)(c-2)&lt;\frac94$. Therefore $(b-1)(c-2)\le2$. By~\eqref{insert313b} it follows that




\begin{equation}\label{313cb}
c\le\frac{2b}{b-1}.
\end{equation}
By~\eqref{insert313c} we have $c=1$ if $b\ge3$. Suppose $b=2$.  By~\eqref{insert313c}, \eqref{313cb},
$c=4$ or $c=1$. But $c=4$ implies, by~\eqref{aux313},
$$\sigma(13^4)\sigma(q^8)=3^2\cdot13^2q^7(21q-8),$$
hence by~\eqref{cycsig} $q^7\mid\Phi_5(13)=30941$, a contradiction. Therefore $c=1$ by~\eqref{insert313b}. Hence by~\eqref{aux313},




\begin{equation}\label{aux313b}
\sigma(13^{2b})(q^2+q+1)=3^2\cdot13^{2b-2}q(21q-8).
\end{equation}

By~\eqref{insert313b}, \eqref{aux313b}, we have $13^2\mid q^2+q+1$, hence $q\ge191$.

We remark here that~\eqref{aux313b} is equivalent to
$$q=\frac{13^{2b+1}-1}{13^{2b-2}(71q-3061)+(q+1)}.$$
Therefore
$$q&lt;\frac{13^{2b+1}}{13^{2b-2}(71q-3061)}=\frac{2197}{71q-3061},$$
a contradiction since $q\ge191$. Therefore $p\ne13$.

Finally, suppose $p=11$. By~\eqref{geom} and~\eqref{prior}, we have $3\|m$, otherwise
$$\frac{\sigma(m^2)}{m^2}\ge\frac{121}{81}\cdot\frac{133}{121}&gt;\alpha,$$
contradicting Theorem~\ref{alpha}. Thus by~\eqref{aux3},




\begin{equation}\label{aux311}
13\sigma(11^{2b})\sigma(q^{2c})=3\cdot11^{2b-1}q^{2c-1}(53q-20). 
\end{equation}

We show here $q\not\equiv1\pmod{11}$. Otherwise by~\eqref{vsig}, \eqref{aux311}, $2b-1=v_{11}(\sigma(q^{2c}))=v_{11}(2c+1)$, hence $2c+1\ge11^{2b-1}$. Thus by~\eqref{aux311},
$$11^{2b+1}&gt;\sigma(11^{2b})\ge q^{2c-1}\ge q^{11^{2b-1}-2},$$
thus $2b+1&gt;(11^{2b-1}-2)\log_{11}q&gt;11^{2b-1}-2$, a contradiction.

Therefore $q\not\equiv1\pmod{11}$. By~\eqref{vsig}, $e_{11}(q)=5$ (whence $q&gt;13$) and thus




\begin{equation}\label{aux311a}
5\mid2c+1;\end{equation}
furthermore by~\eqref{aux311}



\begin{equation}\label{aux311b}
2b-1=v_{11}(\sigma(q^{2c}))=v_{11}\left(\Phi_5(q)\right)+v_{11}(2c+1). 
\end{equation}
Let $t=v_{11}(2c+1)$. Suppose $t&gt;0$. Since $\Phi_5(q)&lt;q^5$, we have by~\eqref{aux311b} 




\begin{equation}\label{aux311c}
2b-1&lt;5\log_{11}q+t. 
\end{equation}
On the other hand, since $2c+1\ge11^t$, \eqref{aux311} yields $11^{2b+1}&gt;\sigma(11^{2b})\ge q^{2c-1}\ge q^{11^t-2}$, hence



\begin{equation}\label{aux311d}
2b+1&gt;(11^t-2)\log_{11}q.
\end{equation}
Then~\eqref{aux311c}, \eqref{aux311d}, together imply
$$t&gt;(11^t-7)\log_{11}q-2&gt;11^t-9,$$
which is impossible for $t\ge1$. Therefore $t=0$. Hence by~\eqref{aux311b}




\begin{equation}\label{aux311e}
2b-1=v_{11}\left(\Phi_5(q)\right).
\end{equation}
Since $\Phi_5(q)&lt;q^5$, we have by~\eqref{aux311}, \eqref{aux311e},
$$11^{2b+1}&gt;\sigma(11^{2b})\ge q^{2c-1}=(q^5)^{\frac{2c-1}5}&gt;\left(\Phi_5(q)\right)^{\frac{2c-1}5}\ge(11^{2b-1})^{\frac{2c-1}5}.$$
Thus $2b+1&gt;\frac15(2b-1)(2c+1)$; equivalently, 




\begin{equation}\label{aux311f}
c&lt;\frac{6b+2}{2b-1}.
\end{equation}
By~\eqref{vsig}, $3\nmid\sigma(11^{2b})$ since $e_3(11)=2$. Thus by~\eqref{aux311}, we have $3\mid\sigma(q^{2c})$, hence by~\eqref{vsig} $3\mid2c+1$. Thus by~\eqref{aux311a}, $15\mid2c+1$, hence $c\ge7$. Thus by~\eqref{aux311f} we have $b=1$, $c=7$. Thus by~\eqref{aux311}
$$7\cdot13\cdot19\sigma(q^{14})=3\cdot11q^{13}(53q-20),$$
implying $q^{13}\mid1729$, which is impossible. Therefore $p\ne11$, thus completing the proof.\end{proof}




\section{Results on the shape of $\boldsymbol{m}$}
By the {\it shape\/} of a natural number, we mean the nature of the
exponents of the prime factors in the unique prime factorization of
that number. We have already seen in Theorem~\ref{odd} that a
solution~$n&gt;2$ to~\eqref{titleq} must be an odd square, $n=m^2$, as
in~\eqref{oddeq}. We seek necessary conditions on the shape of~$m$.




\begin{theorem}\label{squff}
If~$m$ is a solution to~\eqref{oddeq} then~$m$ is neither squarefree nor the square of a squarefree natural number.
\end{theorem}
\begin{proof}
First suppose that~$m$ is squarefree. By~\eqref{cycsig}, $\sigma(m^2)$ is the product of cyclotomic polynomials $\Phi_3(p)$ evaluated at the prime divisors~$p$ of~$m$. Thus by Lemma~\ref{nagell}, if~$q$ is a prime dividing $\sigma(m^2)$, then either $q=3$ or $q\equiv1\pmod3$. Thus by~\eqref{oddeq}, since $m\mid\sigma(m^2)$,
$$m=3^{\epsilon}p_1p_2\cdots p_k,$$
where $\epsilon\in\{0,\,1\}$ and $p_1&lt;p_2&lt;\cdots&lt;p_k$ are primes such that $p_j\equiv1\pmod3$, $1\le j\le k$. By Lemma~\ref{nagell} we have $3\|\Phi_3(p_j)$, $1\le j\le k$, hence $v_3(\sigma(m^2))=k$. Since $k\ge3$ by Theorem~\ref{om3}, we have $3^2\mid\phi(m)$, hence $3^{\epsilon}\|m+\phi(m)$, hence
$$v_3(m(m+\phi(m))=2\epsilon.$$
Thus by~\eqref{oddeq}, $k=2\epsilon$, a contradiction since $k\ge3$ and $\epsilon\le1$. Hence~$m$ cannot be squarefree.

Now suppose that~$m$ is the square of a squarefree natural number. As above, $\sigma(m^2)$ is the product of cyclotomic polynomials $\Phi_5(p)$ evaluated at the prime divisors~$p$ of~$m$, hence if~$q$ is a prime dividing $\sigma(m^2)$, then either $q=5$ or $q\equiv1\pmod5$; consequently
$$m=5^{\epsilon}p_1^2p_2^2\cdots p_k^2,$$
where $\epsilon\in\{0,\,2\}$ and $p_1&lt;p_2&lt;\cdots&lt;p_k$ are primes such that $p_j\equiv1\pmod5$, $1\le j\le k$. Again, as above, $v_5(\sigma(m^2))=k$. Since $k\ge3$ by Theorem~\ref{om3}, we have $5^3\mid\phi(m)$, hence $5^{\epsilon}\|m+\phi(m)$, hence
$$v_5(m(m+\phi(m))=2\epsilon.$$
Thus $k=2\epsilon$, hence~$k=4$. Thus by~\eqref{prior2}
$$\frac{m}{\phi(m)}\le\frac54\cdot\frac{11}{10}\cdot\frac{31}{30}\cdot\frac{41}{40}\cdot\frac{61}{60}&lt;\alpha,$$
contradicting Theorem~\ref{alpha}. Hence~$m$ is not the square of a squarefree natural number. 
\end{proof}

\section{Concluding Remarks}
A natural number~$n$ is said to be {\it perfect\/} if $\sigma(n)=2n$.
No odd perfect numbers are known, and the question regarding their
existence remains what is widely believed to be the oldest unsolved
problem in all of mathematics. It seems the question of
whether~\eqref{titleq} has any solutions $n&gt;2$ parallels that regarding
the existence of odd perfect numbers. For example, the only even
solution to~\eqref{titleq} is~2, while the even solutions to
$\sigma(n)=2n$ are in one-to-one correspondence with the set of
Mersenne primes, and are thus characterized completely. On the other
hand, no odd solutions to~\eqref{titleq}, or to $\sigma(n)=2n$, are
known. In either case, the only results known are those stating
necessary conditions for such solutions to exist. Considering the
difficulty involved in proving Theorem~\ref{om3}, it is clear that the
problem regarding solutions $n&gt;2$ to~\eqref{titleq} is no less
difficult than that of the existence of odd perfect numbers.

All calculations for this paper were done by hand. This was in fact
quite easy, since the bulk of the work comprised comparing rational
numbers to the golden ratio, $\alpha$. This entailed computing the
first few convergents of the rational number in question until a
partial quotient exceeded~1, a task easily done by hand. Otherwise,
some of the factorizations of cyclotomic polynomials given were found
in Brillhart et al.~\cite{Brill}.


\begin{thebibliography}{9}

\bibitem{Brill}
J. Brillhart, D. H. Lehmer, J. L. Selfridge, B. Tuckerman, and S.
Wagstaff, {\it Factorizations of\/} $b^n\pm1$, $b=2$, 3, 5, 6, 7, 10,
11, 12 {\it up to high powers,\/} Amer. Math. Soc., 3rd edition,
2002.

\bibitem{guy}
R. Guy, {\it Unsolved Problems in Number Theory,\/}, 3rd edition,
Springer-Verlag, 2004.

\bibitem{HW}
G. Hardy and E. Wright, {\it An Introduction to the Theory of
Numbers,\/} Oxford at the Clarendon Press, 4th edition, 1960.

\bibitem{Nag}
T. Nagell,
\emph{Introduction to Number Theory,\/}
Wiley, 1951.

\bibitem{oeis}
N. J. A. Sloane, {\it On-Line Encyclopedia of Integer Sequences\/},
\url{https://oeis.org}.

\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25.

\noindent \emph{Keywords: }
sum of divisors, Euler totient.

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\noindent (Concerned with sequences
\seqnum{A055940},
\seqnum{A066679}, and
\seqnum{A070159}.)

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\vspace*{+.1in}
\noindent
Received  January 9 2017;
revised version received May 22 2017. 
Published in {\it Journal of Integer Sequences}, June 25 2017.

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