## 6.7.20

### A formula for $D(x)D(y) - D(xy)$ in terms of the sum-of-aliquot-divisors function, when $\gcd(x,y)=1$

(Note:  This blog post was copied verbatim from this MSE question.)

Hereinafter, we shall let $\sigma(z)$ be the sum of divisors of the positive integer $z$.  Denote the deficiency of $z$ by $D(z) = 2z - \sigma(z)$, and the sum of aliquot divisors of $z$ by $s(z) = \sigma(z) - z$.

We shall compute here a formula for $D(x)D(y) - D(xy)$ in terms of the sum-of-aliquot-divisors function, when $\gcd(x,y)=1$.

Suppose that $\gcd(x,y)=1$.

Then we have
$$D(x)D(y) - D(xy) = (2x - \sigma(x))(2y - \sigma(y)) - (2xy - \sigma(xy))$$
$$= 4xy - 2y\sigma(x) - 2x\sigma(y) + \sigma(x)\sigma(y) - 2xy + \sigma(x)\sigma(y),$$
where we have used the condition $\gcd(x,y)=1$ in the last equation to derive $\sigma(xy)=\sigma(x)\sigma(y)$.

This gives
$$D(x)D(y) - D(xy) = 2xy - 2y\sigma(x) - 2x\sigma(y) + 2\sigma(x)\sigma(y)$$
so that we obtain
$$D(x)D(y) - D(xy) = 2y\bigg(x - \sigma(x)\bigg) - 2\sigma(y)\bigg(x - \sigma(x)\bigg)$$
which simplifies to
$$D(x)D(y) - D(xy) = 2\bigg(x - \sigma(x)\bigg)\bigg(y - \sigma(y)\bigg) = 2\bigg(\sigma(x) - x\bigg)\bigg(\sigma(y) - y\bigg) = 2s(x)s(y).$$

Here are my inquiries:

QUESTIONS

(1) Is it possible to extend the formula
$$D(x)D(y) - D(xy) = 2s(x)s(y)$$
to, say, something that uses three or more arguments (which are pairwise coprime)?

(2) If the answer to Question (1) is YES, what is the closed form for the formula and how can it be proved, in general?

POSTED ATTEMPT

Here is my own attempt for the case of three ($3$) arguments.

Suppose that
$$\gcd(x,y)=\gcd(x,z)=\gcd(y,z)=1.$$

Then we have
$$D(x)D(y)D(z) - D(xyz) = (2x-\sigma(x))(2y-\sigma(y))(2z-\sigma(z))-(2xyz-\sigma(xyz))$$
$$=(4xy-2y\sigma(x)-2x\sigma(y)+\sigma(x)\sigma(y))(2z-\sigma(z))-2xyz+\sigma(x)\sigma(y)\sigma(z)$$
$$=8xyz-4yz\sigma(x)-4xz\sigma(y)+2z\sigma(x)\sigma(y)-4xy\sigma(z)+2y\sigma(x)\sigma(z)+2x\sigma(y)\sigma(z)-\sigma(x)\sigma(y)\sigma(z)-2xyz+\sigma(x)\sigma(y)\sigma(z)$$
$$=2xyz-2yz\sigma(x)-2yz\sigma(x)+2z\sigma(x)\sigma(y)$$
$$+2xyz-2xz\sigma(y)-2xz\sigma(y)+2x\sigma(y)\sigma(z)$$
$$+2xyz-2xy\sigma(z)-2xy\sigma(z)+2y\sigma(x)\sigma(z),$$
from which we obtain
$$=2yz(x-\sigma(x))-2z\sigma(x)(y-\sigma(y))$$
$$+2xz(y-\sigma(y))-2x\sigma(y)(z-\sigma(z))$$
$$+2xy(z-\sigma(z))-2y\sigma(z)(x-\sigma(x))$$
from which we get
$$=2y(x-\sigma(x))(z-\sigma(z))+2z(y-\sigma(y))(x-\sigma(x))+2x(z-\sigma(z))(y-\sigma(y)).$$

This finally gives the formula
$$D(x)D(y)D(z)-D(xyz)=2\bigg(xs(y)s(z)+ys(x)s(z)+zs(x)s(y)\bigg).$$

Checking the formula for $(x,y,z)=(3,5,7)$ gives
$$D(x)D(y)D(z)-D(xyz)=D(3)D(5)D(7)-D(105)=2\cdot{4}\cdot{6}-18=48-18=30$$
$$2\bigg(xs(y)s(z)+ys(x)s(z)+zs(x)s(y)\bigg)=2\bigg(3\cdot s(5)s(7)+5\cdot s(3)s(7)+7\cdot s(3)s(5)\bigg)=2\bigg(3\cdot{1}\cdot{1}+5\cdot{1}\cdot{1}+7\cdot{1}\cdot{1}\bigg)=2\cdot{15}=30.$$

## 4.7.20

### Some modular considerations regarding odd perfect numbers

My first joint paper with Immanuel T. San Diego on odd perfect numbers has been published in Notes on Number Theory and Discrete Mathematics, titled "Some modular considerations regarding odd perfect numbers".

## 16.4.20

### When does $\gcd(m,\sigma(m^2))$ equal $\gcd(m^2,\sigma(m^2))$? What are the exceptions?

This post is taken verbatim from this MSE question.

(This question is related to this earlier one.)

Let $\sigma(x)$ be the sum of divisors of the positive integer $x$.  The greatest common divisor of the integers $a$ and $b$ is denoted by $\gcd(a,b)$.

Here are my questions:

When does $\gcd(m,\sigma(m^2))$ equal $\gcd(m^2,\sigma(m^2))$?  What are the exceptions?

I tried searching for examples and counterexamples via Sage Cell Server, it gave me these outputs for the following GP scripts:

for(x=1, 100, if(gcd(x,sigma(x^2))==gcd(x^2,sigma(x^2)),print(x)))

All positive integers from $1$ to $100$ (except for the integer $99$) satisfy $\gcd(m,\sigma(m^2))=\gcd(m^2,\sigma(m^2))$.

for(x=1, 1000, if(gcd(x,sigma(x^2))<>gcd(x^2,sigma(x^2)),print(x)))

The following integers in the range $1 \leq m \leq 1000$ DO NOT satisfy $\gcd(m,\sigma(m^2))=\gcd(m^2,\sigma(m^2))$:
$$99 = {3^2}\cdot{11}$$
$$154 = 2\cdot 7\cdot 11$$
$$198 = 2\cdot{3^2}\cdot{11}$$
$$273 = 3\cdot 7\cdot 13$$
$$322 = 2\cdot 7\cdot 23$$
$$396 = {2^2}\cdot{3^2}\cdot{11}$$
$$399 = 3\cdot 7\cdot 19$$
$$462 = 2\cdot 3\cdot 7\cdot 11$$
$$469 = 7\cdot 67$$
$$495 = {3^2}\cdot 5\cdot 11$$
$$518 = 2\cdot 7\cdot 37$$
$$546 = 2\cdot 3\cdot 7\cdot 13$$
$$553 = 7\cdot 79$$
$$620 = {2^2}\cdot 5\cdot 31$$
$$651 = 3\cdot 7\cdot 31$$
$$693 = {3^2}\cdot 7\cdot 11$$
$$741 = 3\cdot 13\cdot 19$$
$$742 = 2\cdot 7\cdot 53$$
$$770 = 2\cdot 5\cdot 7\cdot 11$$
$$777 = 3\cdot 7\cdot 37$$
$$792 = {2^3}\cdot{3^2}\cdot 11$$
$$798 = 2\cdot 3\cdot 7\cdot 19$$
$$903 = 3\cdot 7\cdot 43$$
$$938 = 2\cdot 7\cdot 67$$
$$966 = 2\cdot 3\cdot 7\cdot 23$$
$$990 = 2\cdot{3^2}\cdot 5\cdot 11$$

MY ATTEMPT

I know that primes $m_1 := p$ and prime powers $m_2 := q^k$ satisfy the equation, since then we have
$$\gcd(m_1, \sigma({m_1}^2)) = \gcd(p, \sigma(p^2)) = 1 = \gcd(p^2, \sigma(p^2)) = \gcd({m_1}^2, \sigma({m_1}^2)),$$
and
$$\gcd(m_2, \sigma({m_2}^2)) = \gcd(q^k, \sigma(q^{2k})) = 1 = \gcd(q^{2k}, \sigma(q^{2k})) = \gcd({m_2}^2, \sigma({m_2}^2)).$$

This shows that there are infinitely many solutions to the equation
$$\gcd(m, \sigma(m^2)) = \gcd(m^2, \sigma(m^2)).$$

Follow-Up Questions

What can be said about solutions to $\gcd(m, \sigma(m^2)) = \gcd(m^2, \sigma(m^2))$ for which the number of distinct prime factors $\omega(m)$ satisfies

(a) $\omega(m)=2?$

(b) $\omega(m)=3?$

## 15.4.20

### If $q^k n^2$ is an odd perfect number with special prime $q$, then $\gcd(\sigma(q^k),\sigma(n^2))=i(q)=\gcd(n^2, \sigma(n^2))$ (subject to a certain condition).

Let $N = q^k n^2$ be an odd perfect number with special prime $q$.

From this paper in NNTDM, we have the equation
$$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right).$$

In particular, we know that the index $i(q)$ is an integer greater than $5$ by a result of Dris and Luca.

Here is a conditional proof (copied from MathOverflow) that
$$\gcd(\sigma(q^k),\sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)).$$

First, since we have
$$\sigma(q^k)\sigma(n^2) = \sigma({q^k}{n^2}) = \sigma(N) = 2N = 2{q^k}{n^2}$$
we obtain
$$\sigma(q^k) = \frac{2 q^k n^2}{\sigma(n^2)} = \frac{2n^2}{\sigma(n^2)/q^k} = \frac{2n^2}{i(q)}$$
and
$$\sigma(n^2) = \frac{2 q^k n^2}{\sigma(q^k)} = {q^k}\cdot\bigg(\frac{2n^2}{\sigma(q^k)}\bigg) = {q^k}{i(q)},$$
so that we get
$$\gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg).$$

Now, since $\gcd(q, n) = \gcd(q^k, 2n^2) = 1$ and $i(q)$ is odd, we get
$$\gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

Hence, we conclude that $G:=\gcd(\sigma(q^k),\sigma(n^2))=\gcd\bigg({n^2}/{i(q)}, i(q)\bigg)$.

This is equivalent to
$$G = \frac{1}{i(q)}\cdot\gcd\bigg(n^2, (i(q))^2\bigg) = \frac{1}{i(q)}\cdot\bigg(\gcd(n, i(q))\bigg)^2.$$

But we also have
$$\gcd(n, i(q)) = \gcd\bigg(n, \gcd(n^2, \sigma(n^2))\bigg)$$
$$= \gcd\bigg(\sigma(n^2), \gcd(n, n^2)\bigg) = \gcd(n, \sigma(n^2)).$$

Consequently, we obtain
$$G = \frac{1}{i(q)}\cdot\bigg(\gcd(n, \sigma(n^2))\bigg)^2 = \frac{\bigg(\gcd(n, \sigma(n^2))\bigg)^2}{\gcd(n^2, \sigma(n^2))}.$$

In particular, we get
$$\gcd(\sigma(q^k), \sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)),$$
if and only if $\gcd(n^2, \sigma(n^2)) = \gcd(n, \sigma(n^2))$.

## 31.3.20

### K. A. P. Dagal's "A New Operator for Egyptian Fractions"

My friend, Keneth Adrian P. Dagal, has a new preprint out there (uploaded to the math.HO subject area in arXiv), titled "A New Operator for Egyptian Fractions".

Here is the abstract:

This paper introduces a new equation for rewriting two unit fractions to another two unit fractions. This equation is useful for optimizing the elements of an Egyptian Fraction. Parity of the elements of the Egyptian Fractions are also considered. And lastly, the statement that all rational numbers can be represented as Egyptian Fraction is re-established.

## 25.3.20

### Summing Odd Fractions to One, and Odd Perfect Numbers

(This blog post is lifted verbatim from this MSE question, and the answers and a comment contained therein.)

The title says it all.

Question

What exactly is the relationship between Egyptian/unit fractions with odd denominators, and odd perfect numbers?

Motivation

In a comment underneath the question Summing Odd Fractions to One:

From the list $\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},\frac{1}{11}$..... is it possible to choose a limited number of terms that sum to one?
This can be done with even fractions: $\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{12},\frac{1}{24}$

it is stated that:

This would be true if an odd perfect number existed :) MSE user idok

Is this claim true/valid?

Such a representation of a fraction as the sum of fractions with numerator 1 and different denominators is called Egyptian fraction, because that was the way fractions were written in ancient Egypt. It's clear that for 1, we must have an odd number of summands, because otherwise the numerator of the sum would be even and the denominator odd. As it turns out, the minimal number is 9, and there are the following 5 solutions:
$1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{35}+\frac{1}{45}+\frac{1}{231}$

$1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{21}+\frac{1}{231}+\frac{1}{315}$

$1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{33}+\frac{1}{45}+\frac{1}{385}$

$1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{21}+\frac{1}{165}+\frac{1}{693}$

$1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{21}+\frac{1}{135}+\frac{1}{10395}$
There are also solutions of length 11, 13, 15,..., and it can be shown that every odd length $\ge 9$ is possible. This information (and further references) can be found in this article.

Does this answer make the existence of an odd perfect number more likely?

Background

The topic of odd perfect numbers likely needs no introduction, but I include this section here for completion.

A positive integer $n$ is said to be perfect if $\sigma(n)=2n$, where $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$.  If $N$ is odd and $\sigma(N)=2N$, then $N$ is called an odd perfect number.  It is currently unknown whether there is an odd perfect number, despite extensive computer searches.

Euler proved that an odd perfect number, if one exists, must have the form $N=p^k m^2$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Comment underneath the question: Reference: Sellers, J. A., Egyptian Fractions and Perfect Numbers, The Mathematics Teacher, 87, no. 1 (January 1994), 60

The claim is true because $1$ is the sum of finitely many fractions with odd denominator and unit numerator. More generally, for any statement $P$ the implication $P \Longrightarrow Q$ is true if $Q$ is true. This says nothing about the truth value of $P$, however. In this particular case, this doesn't make the existence of odd perfect numbers any more or less likely. In this sense the quoted comment is a bit misleading.

I don't think the answer by Servaes is right, because there is a direct (non-trivial) link. Suppose $n$ is an odd perfect number. Then

$$\sum_{d\mid n} d = 2n.$$

Divide both sides by $n$ and we get

$$\sum_{d\mid n} \frac{1}{d} = 2.$$

Subtracting $1$ from both sides we have written $1$ as the sum of $1/d$ where $d$ are all odd numbers (since all are divisors of $n$, which is odd).

Further Research:

The following MSE questions are also tangentially related to this blog post:

On the decomposition of $1$ as the sum of Egyptian fractions with odd denominators

and

On the decomposition of $1$ as the sum of Egyptian fractions with odd denominators - Part II