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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}\\
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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 $qp^{1/6}$. 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
$$
p^{13}p^{5}$. Let $\gamma$ be such that $r^{\gamma}\| n$. Then
$$
r+1\le \sigma(r^{\gamma})\le 2p^2 q^2p^{1/6}$, we have that
$$
q<\sigma(q^{\beta})\le 2r^2q^{1/3}>p^{1/18}$. Now $\gamma\le 89$, for if not, then
$$
p^50$. Then
\begin{equation}
\label{eq:2}
q^w r^{\delta}\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\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^51$ 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 $B/\gamma(B)^20$ 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