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

Question

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! =) — Jose Arnaldo Bebita Dris, Feb 21 '22 at 04:41
Anyway, please do flesh out your comment as an actual answer, so that I can upvote. Thanks again! =) — Jose Arnaldo Bebita Dris, Feb 21 '22 at 04:44
$G \times J^2 = H$, @mathlove. (Remark 2.1, page 3 of the preprint.) — Jose Arnaldo Bebita Dris, Feb 21 '22 at 05:47
So, applying the logic in your answer below, since $H$ is an integer, then: (1) $J = 1$ if and only if $H$is squarefree. (2) $G = 1$ if and only if $H$ is a square. (3) The remaining case is when $G > 1$ and $J > 1$. This case occurs exactly when $G$ is squarefree. — Jose Arnaldo Bebita Dris, Feb 21 '22 at 05:57
@mathlove: Hmmm, from naive logic, under the assumption that $G$ is squarefree, I am only getting (from the equation $H = G \times {J^2}$) that: (1) $J=1$ if and only if $H$ is squarefree. (2) $G=1$ if and only if $H$ is a square. (3) The remaining case is indeed when $G > 1$ and $J > 1$. Do you have an unconditional proof for your claim in this previous comment? — Jose Arnaldo Bebita Dris, Feb 21 '22 at 11:08
That is, does the condition $G$ is squarefree always hold, since the unique representation for $H$ is given by $H = G \times J^2$? Please advise, @mathlove. — Jose Arnaldo Bebita Dris, Feb 21 '22 at 11:17
Do you agree then, that the following conditions hold under the assumption that $G$ is squarefree, @mathlove? (1) $J=1$ if and only if $H$ is squarefree. (2) $G=1$ if and only if $H$ is a square. (3) The remaining case is (indeed) when $G > 1$ and $J > 1$. — Jose Arnaldo Bebita Dris, Feb 21 '22 at 11:57
Thank you for your confirmation, @mathlove! I will accept your answer by tomorrow morning, as I will have to sleep for now. =) — Jose Arnaldo Bebita Dris, Feb 21 '22 at 12:27

score 0 · Answer 1 · answered Feb 20 '22 at 23:09

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.

score 0 · Accepted Answer · answered Feb 21 '22 at 05:06

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 $k\gt 1$.

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

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