\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 $q
p^{1/6}$. 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
$$
p^{13} p^{1/6}$, we have that
$$
q<\sigma(q^{\beta})\le 2r^2 \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^5p^{5}$. Let $\gamma$ be such that $r^{\gamma}\| n$. Then
$$
r+1\le \sigma(r^{\gamma})\le 2p^2 q^2