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

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 $$i(q)$$?

• 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! =) Commented Feb 21, 2022 at 4:41
• Anyway, please do flesh out your comment as an actual answer, so that I can upvote. Thanks again! =) Commented Feb 21, 2022 at 4:44
• $G \times J^2 = H$, @mathlove. (Remark 2.1, page 3 of the preprint.) Commented Feb 21, 2022 at 5:47
• In this question, $H = i(q)$. Commented Feb 21, 2022 at 5:47
• Yes, I agree. $\$ Commented Feb 21, 2022 at 12:02

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.