Reference request regarding odd 4-perfect numbers

Question

I am reading a paper by Broughan and Zhou (2006) that is dealing with odd 4-perfect numbers. The title of the paper is "Odd multiperfect numbers of abundancy four." In this paper they mention casually that if N is odd and 4 perfect then N has at least 22 distinct prime factors. Unfortunately they do not do a good job of referencing their work. The references they have put there have led me nowhere. I have scoured almost the whole internet without finding the source of their claim. Can someone who is familiar with this result provide the necessary reference, so that I can incorporate that result in my work. Thank you in advance.

https://researchcommons.waikato.ac.nz/handle/10289/1796

Answer 1

It is indeed hard to track down a source, but this is a basic fact that has probably been proven independently many times. If you want to use it in your work, just give a quick proof and mention that you're not the first to come up with it. (Or even cite this StackExchange answer, but don't claim that I'm the first to come up with the idea, because I'm not.)

The idea is that if $n = p_1^{k_1} \cdots p_m^{k_m}$, then $$ \sigma(n) = (1 + p_1 + p_1^2 + \dots + p_1^{k_1}) \cdots (1 + p_m + p_m^2 + \dots + p_m^{k_m}) $$ with a factor of $(1 + p_i + p_i^2 + \dots + p_i^{k_i})$ for every prime power $p_i^{k_i}$ in $n$. We can see this by expanding the product: each divisor of $N$ is obtained by choosing a term from every factor.

Since $1 + p_i + p_i^2 + \dots + p_i^{k_i} = \frac{p_i^{k_i+1}-1}{p_i-1} < \frac{p_i}{p_i-1} \cdot p_i^{k_i}$, we have $\sigma(n) < \frac{p_1}{p_1-1} \cdot \frac{p_2}{p_2-1} \cdots \frac{p_m}{p_m-1} \cdot n$. So if we want $\sigma(n)=4n$, then we must have $\frac{p_1}{p_1-1} \cdot \frac{p_2}{p_2-1} \cdots \frac{p_m}{p_m-1} > 4$.

For a fixed number $m$ of prime factors, the product $\frac{p_1}{p_1-1} \cdot \frac{p_2}{p_2-1} \cdots \frac{p_m}{p_m-1}$ is maximized when the primes are taken to be as small as possible. So we get the following results:

• In general, $m$ must be at least $4$, because $\frac21 \cdot \frac32 \cdot \frac54 = \frac{15}{4} < 4$, but $\frac21 \cdot \frac32 \cdot \frac54 \cdot \frac76 = \frac{35}{8} > 4$.
• If $n$ is odd, then the primes diving $n$ must start from $3$ rather than $2$. In this case, the product $\frac32 \cdot \frac54 \cdot \frac76 \cdots \frac{73}{72}$ (with $20$ factors) is still too small: it is about $3.96677$. Including the next factor $\frac{79}{78}$ brings the product over $4$, to about $4.01763$. Therefore $m \ge 21$.
• If $n$ is odd and not divisible by $3$, then the primes dividing $n$ must start from $5$. In this case, going to the $141^{\text{st}}$ prime, $811$, is still not enough; we must go to the $142^{\text{nd}}$ prime, $821$. Therefore $m \ge 140$.

The paper you cite actually claims that when $n$ is odd we have $m \ge 22$, and when $n$ is odd and not divisible by $3$ we have $m \ge 142$. It's possible that they're referring to a result that did some clever case checking on top of the arithmetic above. But it's more likely that someone along the line was sloppy, and confused the index of the last prime in the product for the total number of primes.