Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. (Note that the divisor sum $\sigma$ is a multiplicative function.)

A number $P$ is said to be perfect if $\sigma(P)=2P$. If a perfect number $N$ is odd, then $N$ is called an odd perfect number. Euler proved that a hypothetical odd perfect number $N$ must have the form $$N = q^k n^2$$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

It is known that $$i(q)=\gcd(n^2,\sigma(n^2))=\frac{n^2}{\sigma(q^k)/2}=\frac{\sigma(n^2)}{q^k},$$ where $i(q)=\sigma(N/{q^k})/{q^k}$ is the index of $N$ at the (special) prime $q$, as initially defined by Broughan, Delbourgo, and Zhou, and whose results were eventually improved upon by Chen and Chen.

In a recent preprint, Dris proves that the following implication holds: $$i(q) \text{ is squarefree } \implies \frac{\sigma(q^k)}{2} \text{ is not squarefree.}$$ We likewise obtain the biconditional $$i(q) \text{ is a square } \iff \frac{\sigma(q^k)}{2} \text{ is a square.}$$ This implies that we have the chain of implications $$i(q) \text{ is a square } \implies \frac{\sigma(q^k)}{2} \text{ is a square } \implies \frac{\sigma(q^k)}{2} \text{ is not squarefree.}$$

These findings highly suggest that $\sigma(q^k)/2$ is not squarefree.

My question is as follows:

What are the remaining cases to consider for this problem, specifically all the possible premises for $i(q)$?

  • $\begingroup$ Thank you very much for your time and attention! Therefore, the remaining case is when $G = \gcd\bigg(\sigma(q^k)/2, \sigma(n^2)/q^k\bigg) = \gcd(F, H)$ is squarefree. Do you concur, @mathlove! =) $\endgroup$ Commented Feb 21, 2022 at 4:41
  • $\begingroup$ Anyway, please do flesh out your comment as an actual answer, so that I can upvote. Thanks again! =) $\endgroup$ Commented Feb 21, 2022 at 4:44
  • $\begingroup$ $G \times J^2 = H$, @mathlove. (Remark 2.1, page 3 of the preprint.) $\endgroup$ Commented Feb 21, 2022 at 5:47
  • $\begingroup$ In this question, $H = i(q)$. $\endgroup$ Commented Feb 21, 2022 at 5:47
  • 1
    $\begingroup$ Yes, I agree. $\ $ $\endgroup$
    – mathlove
    Commented Feb 21, 2022 at 12:02

2 Answers 2


FYI, here says that "Any arbitrary positive integer $n$ can be represented in a unique way as the product of a square and a square-free integer: $n=m^{2}k$. In this factorization, $m$ is the largest divisor of $n$ such that $m^2$ is a divisor of $n$". Using this, one can say that $m=1$ if and only if $n$ is squarefree, and that $k=1$ if and only if $n$ is a square. Therefore, the remaining case is when $m\gt 1$ and $k\gt 1$.


I am posting this self-answer, mainly for my own benefit.

I searched for

"A number is either squarefree, a square, or"

in Google, and one of the results that it returned was this Math 446 - Homework #3 PDF file.

I quote the relevant part of that file (8 (c)) here:

Let $m \geqslant 2$ be an integer. Then either $m$ is squarefree, or $m$ is a square, or $$m = a\cdot{b},$$ where $a$ is squarefree, and $b$ is a square.


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