The title says it all.


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


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 chose 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?

In the accepted answer, MSE user Professor Vector writes:

Such a representation of a fraction as the sum of fractions with numerator 1 and different denominators is called Egyption 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: \begin{align} 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 35}+\frac1{ 45}+\frac1{ 231}\\ 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 21}+\frac1{ 231}+\frac1{ 315}\\ 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 33}+\frac1{ 45}+\frac1{ 385}\\ 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 21}+\frac1{ 165}+\frac1{ 693}\\ 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 21}+\frac1{ 135}+\frac1{ 10395} \end{align} There are also solutions of length 11, 13, 15,..., and it can be shown that every odd length $\ge9$ 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?


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 $\gcd(p,m)=1$.

  • $\begingroup$ Reference: Sellers, J. A., Egyptian Fractions and Perfect Numbers, The Mathematics Teacher, 87, no. 1 (January 1994), 60 $\endgroup$ – Arnie Bebita-Dris Mar 25 '20 at 9:39

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 $P\ \implies\ Q$ 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.

  • $\begingroup$ Thank you for making that point clear, @Servaes! $\endgroup$ – Arnie Bebita-Dris Mar 25 '20 at 9:06

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).

  • $\begingroup$ Thank you, @ThomasBloom, for articulating what I had in mind. +1! =) $\endgroup$ – Arnie Bebita-Dris Mar 25 '20 at 10:42
  • 1
    $\begingroup$ I think, @ThomasBloom, that the problem now reduces to exhibiting such an Egyptian fraction decomposition of $1$, where the odd denominators $d \mid n$ are divisors of the odd perfect number $n$. Alas, I think this is wide open. (We do not even know of a single divisor of an odd perfect number, so that further complicates things.) So, in a certain sense, Servaes's answer is still correct. $\endgroup$ – Arnie Bebita-Dris Mar 25 '20 at 11:25
  • $\begingroup$ Lastly, it is known (by work of Ochem and Rao) that an odd perfect number $n$ must satisfy $n > {10}^{1500}$. So essentially, we are having a bad day here! =( $\endgroup$ – Arnie Bebita-Dris Mar 26 '20 at 4:48

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