\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \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} \vskip .2 in \def\mand{\qquad\mbox{and}\qquad} \def\scr{\scriptstyle} \def\\{\cr} \def\({\left(} \def\){\right)} \def\[{\left[} \def\]{\right]} \def\<{\langle} \def\>{\rangle} \def\fl#1{\left\lfloor#1\right\rfloor} \def\rf#1{\left\lceil#1\right\rceil} \def\cA{{\mathcal A}} \def\cB{{\mathcal B}} \def\cC{{\mathcal C}} \def\cE{{\mathcal E}} \def\cF{{\mathcal F}} \def\cI{{\mathcal I}} \def\cL{{\mathcal L}} \def\cM{{\mathcal M}} \def\cN{{\mathcal N}} \def\cR{{\mathcal R}} \def\cS{{\mathcal S}} \def\cP{{\mathcal P}} \def\cQ{{\mathcal Q}} \def\cT{{\mathcal T}} \def\cX{{\mathcal X}} \def\N{{\mathbb N}} \def\Z{{\mathbb Z}} \def\R{{\mathbb R}} \def\C{{\mathbb C}} \def\Q{{\mathbb Q}} \def\K{{\mathbb K}} \def\L{{\mathbb L}} \def\M{{\mathbb M}} \def\eps{\varepsilon} \def\cA{{\mathcal A}} \def\cB{{\mathcal B}} \def\cC{{\mathcal C}} \def\cD{{\mathcal D}} \def\cE{{\mathcal E}} \def\cF{{\mathcal F}} \def\cG{{\mathcal G}} \def\cH{{\mathcal H}} \def\cI{{\mathcal I}} \def\cJ{{\mathcal J}} \def\cK{{\mathcal K}} \def\cL{{\mathcal L}} \def\cM{{\mathcal M}} \def\cN{{\mathcal N}} \def\cO{{\mathcal O}} \def\cP{{\mathcal P}} \def\cQ{{\mathcal Q}} \def\cR{{\mathcal R}} \def\cS{{\mathcal S}} \def\cT{{\mathcal T}} \def\cU{{\mathcal U}} \def\cV{{\mathcal V}} \def\cW{{\mathcal W}} \def\cX{{\mathcal X}} \def\cY{{\mathcal Y}} \def\o{\omega} \def\cZ{{\mathcal Z}} \def\t{\theta} \def\vec#1{\mathbf{#1}} \def\rem {\mathrm{ \ rem\,}} \def\e{\mathbf{e}} \def\GL{\mathrm{GL}} \def\rank{{\mathrm{rk}\,}} \def\ad{{\mathrm ad}} \def\A{\mathbb{A}} \def\B{\mathbf{B}} \def \C{\mathbb{C}} \def \F{\mathbb{F}} \def \K{\mathbb{K}} \def \Z{\mathbb{Z}} \def \R{\mathbb{R}} \def \Q{\mathbb{Q}} \def \N{\mathbb{N}} \def\Z{\mathbb{Z}} \def \nd{{\, | \hspace{-1.5 mm}/\,}} \def\Zn{\Z_n} \def\z{\zeta} \def\Fp{\F_p} \def\Fq{\F_q} \def \fp{\Fp^*} \def\\{\cr} \def\({\left(} \def\){\right)} \def\fl#1{\left\lfloor#1\right\rfloor} \def\rf#1{\left\lceil#1\right\rceil} \def\SL{\mathrm{SL}} \def\GL{\mathrm{GL}} \def\Sing#1{\mathrm {Sing}\,#1} \def\invp#1{\mbox{\rm {inv}}_p\,#1} \def\Mq{\cM_{m,n}(\F_1)} \def\Mnq{\cM_{n}(\F_q)} \def\Znq{\cZ_{n}(\F_q)} \def\Gnq{\GL_{n}(\F_q)} \def\Snq{\SL_{n}(\F_q)} \def\MZ{\cM_n(\Z)} \def\vt{\vec{t}} \def\MS{\cM_n(\cS)} \def\MA{\cM_n(\cA)} \def\MB{\cM_n(\cB)} \def\SL{\mathrm{SL}} \def\Ln#1{\mbox{\rm {Ln}}\,#1} \def\ord#1{{\mathrm{ord}}_p\,#1} \def\epp{\mbox{\bf{e}}_{p-1}} \def\ep{\mbox{\bf{e}}_p} \def\em{\mbox{\bf{e}}_{m}} \def\ed{\mbox{\bf{e}}_{d}} \def\ii {\iota} \def\wt#1{\mbox{\rm {wt}}\,#1} \def\GR#1{{ \langle #1 \rangle_n }} \def\ab{\{\pm a,\pm b\}} \def\cd{\{\pm c,\pm d\}} \def\Bt {\mbox{\rm {Bt}}} \def\Res#1{\mbox{\rm {Res}}\,#1} \def\Tr#1{\mbox{\rm {Tr}}\,#1} \def\li {\mathrm{li}\,} \def\lam{\lambda} \def\N{\mathbb{N}} \def\s{\sigma} \def\g{\gamma} \def\a{\alpha} \def\b{\beta} \def\e{\epsilon} \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