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\begin{center}
\vskip 1cm{\LARGE\bf 
On Integers for Which the Sum of Divisors \\
\vskip  .04in 
is the Square of the Squarefree Core
}
\vskip 1cm
\normalsize
Kevin A. Broughan \\
Department of Mathematics\\
University of Waikato\\
Private Bag 3105 \\
Hamilton, New Zealand\\
\href{mailto:kab@waikato.ac.nz}{\tt kab@waikato.ac.nz} \\
\ \\
Jean-Marie De Koninck \\
D\'epartment de math\'ematiques et de statistique\\
Universit\'e Laval\\
Qu\'ebec G1V 0A6  \\
Canada\\
\href{mailto:jmdk@mat.ulaval.ca}{\tt jmdk@mat.ulaval.ca}\\
\ \\
Imre K\'atai\\
Department of Computer Algebra\\
P\'azm\'any P\'eter s\'et\'any I/C \\
H-1117 Budapest \\
Hungary\\
\href{mailto:katai@compalg.inf.elte.hu}{\tt katai@compalg.inf.elte.hu}\\
\ \\
Florian Luca \\
Centro de Ciencias Matem{\'a}ticas\\
Universidad Nacional Autonoma de M{\'e}xico \\
C. P. 58089 \\
Morelia, Michoac{\'a}n \\
M{\'e}xico \\
\href{mailto:fluca@matmor.unam.mx}{\tt fluca@matmor.unam.mx}\\
\end{center}

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\begin{abstract}
We study integers $n>1$ satisfying the relation $\s(n)=\g(n)^2$, where
$\sigma(n)$ and $\gamma(n)$ are the sum of divisors and the product of
distinct primes dividing $n$, respectively. We show that the only
solution $n$ with at most four distinct prime factors is $n=1782$. We
show that there is no solution which is fourth power free.  We also
show that the number of solutions up to $x>1$ is at most $x^{1/4+\e}$
for any $\e>0$ and all $x>x_{\e}$. Further, call $n$ primitive if no
proper unitary divisor $d$ of $n$ satisfies $\s(d)\mid \g(d)^2$. We
show that the number
of primitive solutions to the equation up to $x$ is less than $x^\e$
for $x>x_{\e}$.
\end{abstract}


\section{Introduction}

 At the Western Number Theory conference in 2000, the second author asked for all positive integer
solutions $n$ to the equation
\begin{equation}
\label{eq:DK}
\s(n)=\g(n)^2
\end{equation}
(denoted ``De Koninck's equation"), where $\s(n)$ is the sum of
all positive divisors of $n$, and $\g(n)$ is the product of the
distinct prime divisors of $n$, the so-called ``core" of $n$. It is easy to
check that $n=1$ and $n=1782$ are solutions, but, as of the time of
writing, no other solutions are known. A computer search for all $n\le 10^{11}$ did not reveal any other solution.
The natural conjecture (coined  the ``De Koninck's conjecture") is that
there are no other solutions. It is included in Richard Guy's compendium \cite[Section B11]{guy}.\par

It is not hard to see, and we prove such facts shortly, that any non-trivial solution $n$ must have at least three
prime factors, must be even, and can never be squarefree. The fourth author \cite{luca}
has a derivation that the number of solutions with a fixed number of
prime factors is finite. Indeed, he did this for the broader class of positive solutions $n$ to the equation $\s(n)=a\g(n)^K$ where $K\ge 2$ and $1\le a\le L$ with $K$ and
$L$ fixed parameters. Other than this, there has been little
progress on De Koninck's conjecture.\par

Here, we show that the above solutions $n=1,~1782$ are the only ones having $\omega(n)\le 4$. As usual, $\omega(n)$ stands for the number of distinct prime factors of $n$. The method relies on elementary upper bounds for the possible exponents of the primes appearing
in the factorization of $n$ and then uses resultants to solve the resulting systems of polynomial equations
whose unknowns are the prime factors of $n$.

We then show that if an integer $n$ is fourth power free
(i.e.  $p^4\nmid n$ for all primes $p$),
then $n$ cannot satisfy De Koninck's equation \eqref{eq:DK}.
We then count the number of potential solutions $n$ up to $x$.  Pollack and Pomerance \cite{PP}, call a positive integer $n$ to be {\it prime--perfect} if $n$ and $\sigma(n)$ share the same set of prime factors. Obviously, any solution $n$ to the De Koninck's equation is also prime--perfect. Pollack and Pomerance show that the set of prime--perfect numbers is infinite and the counting function of
prime--perfects $n\le x$ has cardinality at most
$x^{1/3+o(1)}$ as $x\to\infty$.  By using the results of Pollack and Pomerance, we  show that the number of solutions
$n\le x$ to De Koninck's equation is at most $x^{1/4+\e}$
for any $\e>0$ and all $x>x_{\e}$.
\par


By restricting to so-called ``primitive" solutions,
using Wirsing's method \cite{wirsing}, we obtain an upper bound of $O(x^\e)$ for all $\e>0$. The notion of primitive that is used is having no proper unitary divisor $d\mid n$ satisfying $\s(d)\mid \g(d)^2$.
In a final section of comments, we make some remarks about the
related problem of identifying those integers $n$ such that  $\g(n)^2 \mid \s(n)$.\par

In summary: the aim of this paper is to present items of evidence
for the truth of De Koninck's conjecture, and to indicate the
necessary structure of a possible counter example. Any non-trivial
solution other than $1782$ must be even, have one prime divisor to
power 1 and possibly one prime divisor to a power congruent to $1$
modulo $4$, with other odd prime divisors being to even powers. At least one prime divisor must appear with an exponent
$4$ or more. Finally, any counter example must be greater than
$10^{11}$.\par

We use the following notations, most of which have been recorded already: $\s(n)$ is the sum of divisors, $\g(n)$ is the product of the distinct primes dividing $n$, if $p$ is prime $v_p(n)$ is the highest power of $p$ which divides $n$, $\omega(n)$ is the number of distinct prime divisors of $n$, and $\cK$ is the set of all solutions to $\s(n)=\g(n)^2$. The symbols $p,q,p_i$ and $q_i$ with $i=1,2,\ldots$ are reserved for odd primes.\par

\section{Structure of solutions}

First we derive the shape of the members of  $\cK$.

\begin{lemma}
\label{lem:form}
If $n>1$ is in $\cK$, then
$$
n=2^e p_1\prod_{i=2}^s p_i^{a_i},
$$
where $e\ge 1$ and $a_i$ is even for all $i=3,\ldots,s$. Furthermore, either
$a_2$ is even in which case $p_1\equiv 3\pmod 8$, or
$a_2\equiv 1\pmod 4$ and $p_1\equiv p_2 \equiv 1\pmod 4$.
\end{lemma}
\medskip

\begin{proof}
Firstly, we note that $n$ must be even: indeed, if $n>1$ satisfies
$\s(n)=\g(n)^2$ and $n$ is odd, then $\s(n)$ must be odd so that the exponent
of each prime dividing $n$ must be even, making $n$ a perfect
square. But then $n<\s(n) = \g(n)^2 \le n$, a contradiction.\par

Secondly, since $n$ is even, it follows that $2^2\| \gamma(n)^2$.
Write
$$n=2^e\prod_{i=1}^s p_i^{a_i}
$$ with distinct odd primes $p_1,\ldots,p_s$ and positive integer exponents $a_1,\ldots,a_s$,
where the primes are arranged in such a way that the odd exponents
appear at the beginning and the even ones at the end. Using
the fact that $\sigma(2^e)=2^{e+1}-1$ is odd, we get that $2^2\|
\prod_{i=1}^s \sigma(p_i^{a_i})$. Thus, there are at most two
indices $i$ such that $\sigma(p_i^{a_i})$ is even, with all the other
indices being odd.  But if $p$ is odd and
$\sigma(p^a)$ is also odd, then $a$ is even. Thus, either only $a_1$
is odd, or only $a_1$ and $a_2$ are odd. Now let us show that there
is at least one exponent which is $1$. Assuming that this is not so,
the above argument shows that $a_1\ge 3$ and that $a_i\ge 2$ for
$i=2,\ldots,s$. Thus,
$$
4p_1^2\prod_{i=2}^s p_i^2=\gamma(n)^2=\sigma(n)\ge
\sigma(2)\sigma(p_1^3)\prod_{i=2}^s
\sigma(p_i^2)>3p_1^3\prod_{i=2}^s p_i^2,
$$
leading to $p_1<4/3$, which is impossible. Hence, $a_1=1$. Finally, if $a_2$ is even, then $2^2\| \sigma(p_1)$ showing that
$p_1\equiv 3\pmod 8$, while if $a_2$ is odd, then $2\| \sigma(p_1)$ and $2\| \sigma(p_2^{a_2})$, conditions which easily lead to the conclusion that
$p_1\equiv p_2\equiv 1\pmod 4$ and $a_2\equiv 1\pmod 4$.
\end{proof}

\section{Solutions with $\omega(n)\le 4$}

\begin{theorem}
\label{thm:main}
Let $n\in \cK$ with
$\omega(n)\le 4$. Then $n=1$ or $n=1782$.
\end{theorem}

\begin{proof}

Using Lemma \ref{lem:form}, we write $n=2^{\alpha} pm$, where $\alpha>0$ and $m$ is coprime to $2p$.

We first consider the case $p=3$. If additionally $m=1$, we then get that
$\sigma(n)=6^2$, and we get no solution. On the other hand, if $m>1$, then $\sigma(m)$ is a divisor of $\gamma(n)^2/4$ and must therefore be odd. This means that every prime
factor of $m$ appears with an even exponent. Say $q^{\beta}\| m$. Then
$$
\sigma(q^{\beta})=q^{\beta}+\cdots+q+1
$$
is coprime to $2q$ and is larger than $3^2+3+1>9$. Thus, there exists a prime factor of $m$ other than 3 or $q$, call it $r$, which divides
$q^{\beta}+\cdots+q+1$, implying that it also divides $m$ and that it appears in the factorization of $m$ with an even exponent. Since $\omega(n)\le 4$, we have $m=q^{\beta} r^{\gamma}$. Now
$$
q^{\beta}+\cdots+q+1=3^i r^j\qquad {\text{\rm and}}\qquad r^{\gamma}+\cdots+r+1=3^{k} q^{\ell},
$$
where $i+k\le 2$ and $j,\ell\in \{1,2\}$. Thus,
$$
(q^{\beta}+\cdots+q+1)(r^{\gamma}+\cdots+r+1)=3^{i+k} q^\ell r^j.
$$
The left--hand side of this equality is greater than or equal to $3q^{\beta} r^{\gamma}$. In the case where $\beta> 2$, we have $\beta\ge 4$, so that $q^4 r^2\le q^{\beta} r^{\alpha}\le 9 q^2r^2$, giving $q\le 3$, which is a contradiction.
The same contradiction is obtained if $\gamma>2$.

Thus, $\beta=\gamma=2$. If $l=j=2$, we then get that
$$
(q^2+q+1)(r^2+r+1)=3^{i+k}q^2r^2,
$$
leading to $\sigma(2^{\alpha})\mid 3^{2-i-j}$. The only possibility is $\alpha=1$ and $i+j=1$, showing that $i=0$ or $j=0$. Since the problem is symmetric,
we treat only the case $i=0$. In that case, we get
$q^2+q+1=r^2$, which is equivalent to $(2q+1)^2+3=(2r)^2$, which has no convenient solution $(q,r)$.


If $j=\ell=1$, we then get that
$$
q^2 r^2<(q^2+q+1)(r^2+r+1)<9qr,
$$
implying that $qr<9$, which is false.

Hence,  it remains to consider the case $j=2$ and $\ell=1$, and viceversa. Since the problem is symmetric in $q$ and $r$, we  only look at $j=2$ and $\ell=1$. In that case, we have
$$
q^2r^2<(q^2+q+1)(r^2+r+1)= 3^{i+k} r^2 q,
$$
so that $q<3^{i+k}$. Since $q>3$, this shows that $i=k=1$ and $q\in \{5,7\}$. Therefore, $r^2+r+1=75,~147$, and neither gives a convenient solution $n$.

From now on, we can assume that $p>3$, so that $p+1=2^u m_1$, where $u\in \{1,2\}$ and $m_1>1$ is odd. Let $q$ be the largest prime factor of $m_1$. Clearly, $p+1\ge 2q$, so that $q<p$. Moreover, since $\omega(n)\le 4$ we have
$$
p<4q^4<q^6,
$$
so that
$q>p^{1/6}$. Let again $\beta$ be such that $q^{\beta}\|n$. We can show that $\beta\le 77$. Indeed, assuming that $\beta\ge 78$, we first observe that
$$
p^{13}<q^{78}\le q^{\beta}<\sigma(q^{\beta}),
$$
and write
$$
\sigma(q^{\beta})=2^v m_2,
$$
where $v\in \{0,1\}$ and $m_2$ is coprime to $2q$. If $m_2$ divides $p^2$, we get that
$$
p^{13}<\sigma(q^{\beta})\le 2p^2,
$$
which is a contradiction. Thus, there exists another prime factor $r$ of $n$, and $m_2\le p^2r^2$. Hence,
$$
p^{13}<\sigma(q^{\beta})<2p^2r^2<p^3 r^2,
$$
implying that $r>p^{5}$. Let $\gamma$ be such that $r^{\gamma}\| n$. Then
$$
r+1\le \sigma(r^{\gamma})\le 2p^2 q^2<p^5,
$$
which is a contradiction. Thus, $\beta\le 77$.

Say $r$ doesn't appear in the factorization of $(p+1)\sigma(q^{\beta})$. Then we need to solve the system of equations
$$
p+1=2^u q^w\qquad {\text{\rm and}}\qquad q^{\beta}+\cdots+q+1=2^v p^z,
$$
where $\beta\in \{1,\ldots,77\}, ~u\in \{1,2\},~0\le v\le 2-u,~\{w,z\}\subseteq \{1,2\}$, which we can solve with resultants. This gives us a certain number of possibilities for the pair $(p,q)$. If $\omega(n)=3$, we have $\sigma(n)=4p^2q^2$, and we find $n$. If $\omega(n)= 4$, then $\sigma(r^{\gamma})$ is a divisor of $2p^2q^2$
and we find certain possibilities for the pair $(r,\gamma)$. Then we extract $n$ from the relation $\sigma(n)=4p^2q^2r^2$.

Now say $r$ appears in the factorization of $(p+1)\sigma(q^{\beta})$. We then write
\begin{equation}
\label{eq:1}
p+1=2^u q^w r^{\delta}\qquad {\text{\rm and}}\qquad \sigma(q^{\beta})=2^v p^z r^{\eta},
\end{equation}
where $u\in \{1,2\},~w\in \{1,2\},~0\le v\le 2-u,~z\in \{0,1,2\},~\delta+\eta\in \{1,2\}$. If $z=0$, then since $q>p^{1/6}$, we have that
$$
q<\sigma(q^{\beta})\le 2r^2<r^3,
$$
so that $r>q^{1/3}>p^{1/18}$. Now $\gamma\le 89$, for if not, then
$$
p^5<r^{90}\le r^{\gamma}<\sigma(r^{\gamma})<2p^2q^2<p^5,
$$
which is false.

Suppose now that $z>0$. Then
\begin{equation}
\label{eq:2}
q^w r^{\delta}<p<4q^{w} r^{\delta}
\end{equation}
from the first relation of \eqref{eq:1}, while
\begin{equation}
\label{eq:3}
\frac{q^{\beta}}{2r^{\eta}}<p^z<\frac{2q^{\beta}}{r^{\eta}}
\end{equation}
from the second relation of \eqref{eq:1}. If $z=1$, we get from \eqref{eq:2} and \eqref{eq:3} that
$$
r^{\delta+\eta}<2q^{\beta-w}\qquad {\text{\rm and}}\qquad r^{\delta+\eta}>\frac{q^{\beta-w}}{8}.
$$
From the above left inequality  and the fact that $\delta+\eta\ge 1$, we read that $\beta-w\ge 1$, and then from the right one that $9r^2>8 r^{\delta+\eta}>q^{\beta-w}\ge q$, and thus $r^2\ge 3r>q^{1/2}$,  so that
$r>q^{1/4}>p^{1/24}$. It now follows easily that $\gamma\le 119$, for if not, then $\gamma\ge 120$ would give
$$
p^5<r^{120}\le r^{\gamma}<\sigma(r^{\gamma})\le 2p^2q^2<p^5,
$$
which is a contradiction. Finally, if $z=2$, we get from \eqref{eq:3} that
$$
\frac{q^{\beta/2}}{{\sqrt{2}}r^{\eta/2}}<p<\frac{{\sqrt{2}}q^{\beta/2}}{r^{\eta/2}},
$$
which combined with \eqref{eq:2} yields
$$
r^{\delta+\eta/2}<{\sqrt{2}}q^{\beta/2-w}\qquad {\text{\rm and}}\qquad r^{\delta+\eta/2}>\frac{q^{\beta/2-w}}{4{\sqrt{2}}}.
$$
From the above left inequality and because $\delta+\eta/2\ge 1/2$, we read that $\beta/2>w$, implying that $\beta/2-w\ge 1/2$. Thus,
$$
4{\sqrt{2}} r^{2}\ge 4{\sqrt{2}} r^{\delta+\eta/2}>q^{\beta/2-w}\ge q^{1/2}
$$
and therefore
$$
r^8> 32 r^4\ge  (4{\sqrt{2}} r^{\delta+\eta/2})^2>q>p^{1/6},
$$
showing that $r>p^{1/48}$. This shows that $\gamma\le 239$, for if $\gamma\ge 240$, then
$$
p^5<r^{240}\le r^{\gamma}<\sigma(r^{\gamma})<2p^2q^2<p^5,
$$
which is a contradiction. Thus, we need to solve
\begin{eqnarray*}
p+1 & = & 2^u q^w r^{\delta};\\
\sigma(q^{\beta}) & = & 2^v p^z r^{\eta};\\
\sigma(r^{\gamma}) & = & 2^{\lambda} p^s q^t,\\
\end{eqnarray*}
where $1\le \beta\le 77$, $1\le \gamma\le 239$, $u\in \{1,2\}$, $u+v+\lambda\le 2$, $1\le w\le 2$, $w+t\le 2$, $\delta+\eta\in \{1,2\}$, $z\in\{0,1,2\}$ and $s\in \{0,1,2\}$.
This can be solved with resultants and it gives us a certain number of possibilities for the triplet $(p,q,r)$.  From $\sigma(n)=4p^2 q^2 r^2$, we extract $n$ by solving the equation for $\a$, given $p,q$ and $r$. Failure to detect an integer value for $\a$ means the candidate solution fails. A computer program went through all these steps and confirmed the conclusion of Theorem \ref{thm:main}.

\end{proof}



\section{The case of fourth power free $n$}


\begin{theorem}
\label{thm:4free}
If $n>1$ is in $\cK$, then $n$ is not fourth power free.
\end{theorem}

\begin{proof}
Let us assume that the result is false, that is, that there exists some $n\in \cK$
which is  fourth power free. By Lemma \ref{lem:form} we can write
$$
n=2^e p_1 p_2^{a_2} \prod_{i=1}^k q_i^2,
$$
where $a_2\in \{0,1\}$. Let ${\mathcal Q}=\{q_1,\ldots,q_k\}$. The idea is to exploit the fact that there exist at most two elements $q\in {\mathcal Q}$ such that $q\equiv 1\pmod 3$. If there were three or more such elements, then $3^3$ would divide
$\prod_{q\in {\mathcal Q}} \sigma(q^2)$ and therefore a divisor of $\gamma(n)^2$, which is a contradiction.

We begin  by showing that $k\le 8$. To see this, let
$$
{\mathcal R}=\left\{r\in {\mathcal Q}: \gcd\left(\sigma(r^2),\prod_{q\in {\mathcal Q}} q\right)=1\right\}.
$$
Then $\prod_{r\in {\mathcal R}} \sigma(r^2)$ divides $p_1^2$ (if $a_2=0$) and $p_1^2p_2^2$ if $a_2>0$. It follows that $\sigma(r^2)$ is either a multiple of $p_1$ or of $p_2$ for each $r\in {\mathcal R}$. Since there can be at most two $r$'s for which $\sigma(r^2)$ is a multiple of $p_1$, and at most two $r$'s for which $\sigma(r^2)$ is a multiple of $p_2$, we get that $\#{\mathcal R}\le 4$. When $r\in {\mathcal Q} \setminus {\mathcal R}$, we have, since $\sigma(r^2)>9$, that $\sigma(r^2)=r^2+r+1$ is a multiple of some prime $q_{i_r}>3$
for some $q_{i_r}\in {\mathcal Q}$. Now, since $q_{i_r}$ is a prime divisor of $r^2+r+1$ larger than 3, it must satisfy $q_{i_r}\equiv 1\pmod 3$.  Since $i_r$ can take the same value for  at most two  distinct primes $r$, and there are at most two distinct values of the index $i_r$, we get that $k-\#{\mathcal R}\le 4$, which implies that $k\le 8$, as claimed.

Next rewrite the equation $\s(n)=\g(n)^2$ as
\begin{equation}
\label{eq:*}
\left(\frac{2^{e+1}-1}{4}\right) \prod_{i=1}^k \left( \frac{q_i^2+q_i+1}{q_i^2}\right)=\left(\frac{p_1^2}{p_1+1}\right) \left(\frac{p_2^{2\delta_2}}{\sigma(p_2^{a_2})}\right),
\end{equation}
where $\delta_2=0$ if $a_2=0$ and $\delta_2=1$ if $a_2>0$. The left--hand side  of \eqref{eq:*} is at most
\begin{equation}
\label{eq:2toe}
\left(\frac{2^{e+1}-1}{4}\right) \left(\prod_{q\le 23} \frac{q^2+q+1}{q^2}\right)<0.73 (2^{e+1}-1).
\end{equation}
First assume that $a_2=0$. Then the right--hand side of \eqref{eq:*} is
\begin{equation}
\label{eq:***}
\frac{p_1^2}{p_1+1}\ge \frac{9}{4}=2.25.
\end{equation}
If $e=1$, then the left--hand side of inequality \eqref{eq:*} is, in light of \eqref{eq:2toe}, smaller than $0.73(2^2-1)<2.22$, which
contradicts the lower bound provided in \eqref{eq:***}. Thus, $e\in \{2,3\}$, and
$$
\frac{p_1^2}{p_1+1}\le 0.73(2^4-1)=10.95,
$$
so that $p_1\le 11$. Since $p_1\equiv 3\pmod 8$, we get that $p_1\in \{3,11\}$. If $p_1=11$, then $3\in {\mathcal Q}$. If $p_1=3$, then since $e\in \{2,3\}$, we get that either $5$ or $7$ is in ${\mathcal Q}$.

If $3\in {\mathcal Q}$, then $13\mid 3^2+3+1$,
$ 61\mid 13^2+13+1$ and $97\mid 61^2+61+1$ are all three in ${\mathcal Q}$ and are congruent to $1$ modulo $3$, a contradiction.

If $5\in {\mathcal Q}$, then $31\mid 5^2+5+1$, $331\mid 31^2+31+1$ and $7\mid 331^2+331+1$ are all in ${\mathcal Q}$, a contradiction.

If $7\in {\mathcal Q}$, then $7$, $19\mid 7^2+7+1$ and $127\mid 19^2+19+1$ are all in ${\mathcal Q}$, a contradiction.

Assume next that $a_2>0$. Then, by Lemma 1, $p_1\equiv p_2\equiv 1\pmod 4$. Since $e\in \{1,2,3\}$, it follows that one of $3,5,7$ divides $n$.

If $3\mid n$, then $3\in {\mathcal Q}$.

If $5\mid n$, and $5$ is one of $p_1$ or $p_2$, then $3\mid \sigma(p_1p_2^{a_2})\mid n$, while if $5\in {\mathcal Q}$, then $31=5^2+5+1$
is not congruent to $1$ modulo $4$ and divides $n$, implying that it belongs to ${\mathcal Q}$, and thus $3\mid 31^2+31+1\mid n$.

Finally, if $7\mid n$, then $7$ cannot be $p_1$ or $p_2$, meaning that $7$ is in ${\mathcal Q}$ and therefore that $3\mid 7^2+7+1$, which implies that $3\mid n$.

To sum up, it is always the case that when $a_2>0$, necessarily $3$ divides $n$.

Hence, $13=3^2+3+1$ divides $n$, so that either $13\in {\mathcal Q}$, or not. If $13\not\in {\mathcal Q}$, then $7\mid 13+1$ is in ${\mathcal Q}$, in which case
$19\mid 7^2+7+1$ divides $n$ and it is not congruent to $1$ modulo $4$, implying that $19\in {\mathcal Q}$ and thus that $127\mid 19^2+19+1$
divides $n$ and is not congruent to $1$ modulo $4$, so that $127\in {\mathcal Q}$. Hence, all three numbers $7,19,127$ are in ${\mathcal Q}$, which again is a contradiction.

If $13\in {\mathcal Q}$, then $61\mid 13^2+13+1$ divides $n$.

If $61$ is one of $p_1$ or $p_2$, then $31\mid \sigma(p_1p_2^{a_2})$
and $31\equiv 3\pmod 4$, so that $31\in {\mathcal Q}$.  Next $331\mid 31^2+31+1$ is a divisor of $n$ and it is not congruent to $1$ modulo $4$, implying that it belongs to ${\mathcal Q}$ and therefore that $13,31,331$ are all in ${\mathcal Q}$, a contradiction.

Finally, if $61\in {\mathcal Q}$, then  $97\mid 61^2+61+1$ is a divisor of $n$.
If $97\in {\mathcal Q}$ we get a contradiction since $13$ and $61$ are already  in ${\mathcal Q}$, while if $97$ is one of $p_1$ or $p_2$, then
$7\mid \sigma(p_1p_2^{a_2})$ is a divisor of $n$ and therefore necessarily in ${\mathcal Q}$, again a contradiction.
\end{proof}


\section{Counting the elements in $\cK\cap [1,x]$}

Let $\cK(x)=\cK\cap [1,x]$.

\begin{theorem}
\label{thm:count}
The estimate
$$
\#\cK(x)\le x^{1/4+o(1)}
$$
holds as $x\to\infty$.
\end{theorem}

\begin{proof}

By Theorem 1.2 in \cite{PP}, we have $\#{\mathcal K}(x)=x^{1/3+o(1)}$ as $x\to\infty$. It remains to improve the exponent $1/3$ to $1/4$.
We recall the following result from \cite{PP}.


\begin{lemma}
\label{lem:PP}
If $\sigma(n)/n=N/D$ with $(N,D)=1$, then given $x\ge 1$ and $d\ge 1$
$$
\#\{n\le x: D=d\}= x^{o(1)}
$$
as $x\rightarrow\infty$.
\end{lemma}

Now let $n\in \mathcal{K}(x)$, assume that $n>1$ and write it in the form $n=A\cdot B$ with $A$ squarefree, $B$ squarefull and $(A,B)=1$.
By Lemma \ref{lem:form}, we have $A\in \{1,p_1,2p_1,p_1p_2,2p_1p_2\}$. Then
\begin{equation}
\label{eq:ND}
\frac{N}{D}=\frac{\s(n)}{n}= \frac{\g(n)^2}{n}=\frac{\g(A)^2}{A}\cdot\frac{\g(B)^2}{B}= \frac{A}{B/\g(B)^2},
\end{equation}
and $(A,B/\g(B)^2)=1$.
Since $\sigma(n)> n$, it follows that $B/\gamma(B)^2<A$. Thus,
$$
B/\gamma(B)^2<{\sqrt{A B/\gamma(B)^2}}\le {\sqrt{n}}\le \sqrt{x}.
$$
By Lemma \ref{lem:form} again, we can write $B=\delta C^2D$, where $C$ is squarefree, $D$ is $4$-full, $\delta\in \{1,2^3\}$, and
where $\delta$, $C$ and $D$ are pairwise coprime. Then $B/\gamma(B)^2=\delta/\gamma(\delta)^2 \times D/\gamma(D)^2$, so therefore $D/\gamma(D)^2\le B/\gamma(B)^2<x^{1/2}$. Because $D$ is $4$-full it follows that
$D/\gamma(D)^2$ is squarefull and so the number of choices for $D/\gamma(D)^2$ is $O(x^{1/4})$.  Hence, the number of choices for $B/\gamma(B)^2\in \{D/\gamma(D)^2, 2D/\gamma(D)^2\}$ is also $O(x^{1/4})$, which together with Lemma \ref{lem:PP} and formula \eqref{eq:ND} implies the desired conclusion.
\end{proof}

A positive integer $d$ is said to be a {\it unitary divisor} of $n$ if $d\mid n$ and $(d,n/d)=1$; it is said to be a {\it proper unitary divisor} of $n$ if it also satisfies $1<d<n$. We will say that an integer $n\in \cK(x)$ is {\it primitive} if no proper unitary divisor $d$ of $n$ satisfies $\s(d)\mid \g(d)^2$. Let us denote this subset of $\cK(x)$ by $\cH(x)$. Elements of $\cH(x)$ can be considered as the {\it primitive solutions} of $\s(n)=\g(n)^2$. For example, the number $n=1782\in \cH(x)$ since, although the proper divisor $d=6$ of $n$ satisfies  $\s(d)\mid \g(d)^2$, it fails to be unitary. Also, it is interesting to observe that the condition $\s(d)\mid \g(d)^2$ seems to be very restrictive: for instance, the only positive integers $d<10^8$ satisfying this condition are 6 and 1782; this is already an indication that the set  $\cH(x)$ is very thin. As a matter of fact, we now prove the following result.


\begin{theorem}
\label{thm:primitive}
Let $\e>0$ be given. Then, given any $\e>0$,
$$
\# \cH(x) = O(x^\e).
$$
\end{theorem}

\begin{proof}
Let $n\in\cH(x)$ and assume that $x>0$ is large. Let $a$ be the largest divisor of $n$ such that all prime factors $p\mid a$ satisfy $p\le \log x$. Write $n=a\cdot b$ and write down the standard factorization of $b$ into primes as
$$
b=p_1^{\b_1}\cdots p_k^{\b_k},\qquad {\text{\rm where}}\qquad p_1<\cdots<p_k.
$$
Set $M:= \lceil \log x/\log\log x\rceil$. Then, since $b\le n\le x$ and since for each $i$, we have $\log x<p_i$, we get
$$
(\log x)^{\b_1+\cdots +\b_k} < p_1^{\b_1}\cdots p_k^{\b_k}=b\le x,
$$
implying that
$$
\b_1+\cdots +\b_k<\frac{\log x}{\log\log x}, \quad {\text{\rm so that}}\quad k\le M.
$$

Now assume that the positive integer $a$ is given and that there is some positive integer $b$ such that $n=a\cdot b$ is a primitive element of $\cK$. We will show how to find $b$ from $a$ using the knowledge of the exponents $\beta_1,\ldots,\beta_k$.

Firstly, if $a$ is already primitive, we then have $b=1$. So, suppose that $a$ is not primitive.

Since $\s(a)\s(b)=\g(a)^2\g(b)^2$, and the two factors on the right hand side are coprime, we must have
$$
d:=\frac{\s(a)}{(\s(a),\g(a)^2)}\mid \g(b)^2.
$$
 Hence, let $p_1$ be the least prime dividing the left--hand side of the above relation. Note that the left--hand side is not $1$, since otherwise we would have $\s(a)\mid \g(a)^2$, which is not possible since $n$ is primitive.

Now replace $a$ by $ap_1^{\b_1}$ and proceed. If at step $i<k$, we have $d=1$, then the choice of the $\b_i$'s for $i=1,\ldots,k$ fails to generate an element of $\cK$. We can then move on to the next choice. With success at every step, we generate $b$ from $a$ by finding primes $p_1,\ldots,p_k$ such that $a\cdot p_1^{\b_1}\cdots p_k^{\b_k}\in\cK$.

To complete the proof, we only need to find an upper bound for
$$
\#\{\text{choices for $a$}\}\cdot \#\{\text{choices for $(\b_1,\ldots,\b_k)$}\},
$$
and this is the same as in Wirsing's proof \cite{wirsing} or \cite[Theorem 7.8, pp. 1008-1010]{pomerance} for the case of multiperfect numbers:
\begin{eqnarray*}
\#\{\text{choices for $(\b_1,\ldots,\b_k)$}\} & \le & \#\{(\b_1,\ldots,\b_k): \b_1+\cdots + \b_k\le M\}\le 2^M,\\
\#\{\text{choices for $a$}\} & \le  & \#\{n\le x: p\mid n\Rightarrow p\le \log x\}\\
&\le& \#\{n\le x: p\mid n\Rightarrow \log^\frac{3}{4} x<p\le \log x\}\\&~~&\times
\#\{n\le x: p\mid n\Rightarrow p\le \log^\frac{3}{4} x\}\\
&\le& 2^{4M}\times 2^M=2^{5M},
\end{eqnarray*}
in which case we obtain the upper bound
$$
2^{6M}=x^\frac{6\log 2}{\log\log x}=x^{o(1)}\qquad {\text{\rm as}}\qquad x\to\infty,
$$
for the number of primitive $n\in\cK(x)$, which completes the proof of this theorem.
\end{proof}

%====================================================================================================
\section{Final remarks}
Here, we briefly consider another question related to the problem
of De Koninck,
namely the one which consists in identifying those integers $n$ satisfying $\g(n)^2\mid \s(n)$. There is an infinite set of solutions $n=2^i
3^j$ with $i\equiv 5~(\bmod ~6),~ j\equiv 1~(\bmod ~2)$. If $n=2^i 3^j$
satisfies $\g(n)^2\mid \s(n)$, then these two congruence conditions are also
satisfied.

Indeed, first let $i=5+6k,~j=1+2m$ and $n=2^i 3^j$. Then
$\s(2^i)=2^{6(k+1)}-1\equiv 0~(\bmod ~9)$ and $3^{2(m+1)}-1\equiv 0~(\bmod
~8)$ so that $3^2\mid \s(2^i)$ and $2^2\mid \s(3^j)$.\par

Now assume that $n=2^i 3^j$ and that the integers $r$ and $s$ are such that $2^2\mid \s(3^s)$ and
$3^2\mid \s(2^r)$.  It is well--known and quite easy to prove by elementary arguments that
\begin{eqnarray*}
v_2(\s(3^s))+1&=&v_2((3+1)(s+1))\ge 3,\text{ and}\\
v_3(\s(2^r))&=&v_3((2+1)(r+1))\ge 2,
\end{eqnarray*}
so by the first of these equations $s$ is odd. By the second equation, we see that
$r\equiv 2 \pmod 3$, so that $r\equiv 2\pmod 6$ or $r\equiv 5\pmod 6$. If
the first of these was true, then $r$ would be even, so that $3\mid
\s(2^r)$ would not be possible. Thus, we must have $r\equiv 5\pmod 6$.

Observe that this infinite set $2^i 3^j$ does not exhaust all of the non-trivial
solutions, even those with only two distinct prime factors. For example,
$n=p^{q-2}q^{p-2}$ with
$p=2,~q=1093$ or
$p=83,~q=4871$ are both solutions, since in either case we have
\begin{eqnarray*}
p^2\mid q^{p-1}-1 \quad \mbox{ and } \qquad ~~q^2\mid p^{q-1}-1,
\end{eqnarray*}
and such divisibilities yield $p^2 q^2 \mid \s(p^{q-2} q^{p-2})$.
Note also that there are many non-trivial solutions with $3$ prime factors, for example $17$
solutions up to $10^6$ and $25$ up to $4\times 10^6$. Typical
solutions have the form
$$
\{2^3 3^3 5^5,~2^5 3^5 7^1,~2^9 3^4 11^1\}.
$$

As a final note, let us mention that, given any arbitrary integer $k\ge 2$, one can easily check that the more general property $\g(n)^k|\s(n)$ is indeed satisfied by infinitely many positive integers $n$, namely those of the form
$$n=2^{2 i  3^{k-1}-1}  3^{j 2^{k-1} -1} \qquad (i\ge 1,\ j\ge 1).$$

\section{Acknowledgements}
Research of F.~L. was supported in part by Grant SEP-CONACyT 79685, of I.~K. by a grant from the European Union and the European Social Fund and of J-M.D.K. by a grant from NSERC.
\bigskip
\begin{thebibliography}{9}


\bibitem{guy}
R. K. Guy, {\it Unsolved Problems in Number Theory, Third Edition}, Springer, 2004.

\bibitem{luca} F. Luca,
On numbers $n$ for which the prime factors of $\s(n)$ are among the
prime factors of $n$, {\it Result. Math.\/}, {\bf 45} (2004),
79--87.

\bibitem{pomerance} C. Pomerance and A. S\'ark\"ozy, Combinatorial Number Theory, in {\it Handbook of Combinatorics vol I}, R. L. Graham, M. Gr\"otschel and L. Lov\`asz, eds., Elsevier Science, 1995.

\bibitem{PP}
 P. Pollack and C. Pomerance, Prime--perfect numbers, {\it INTEGERS}, to appear.

\bibitem{wirsing}
E. Wirsing, Bemerkung zu der Arbeit \"uber vollkommene Zahlen, {\it Math. Ann.} {\bf 137} (1959), 316--318.


 \end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: 
Primary 11A25; Secondary 11A41.

\noindent {\it Keywords: } 
sum of divisors, squarefree core of an integer, De Koninck's
conjecture.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received February 23 2012;
revised versions received  July 26 2012; August 14 2012; September 3 2012.
Published in {\it Journal of Integer Sequences}, September 8 2012.

\bigskip
\hrule
\bigskip

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