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This question is an offshoot of this post #1 and this post #2.

Let $N = q^k n^2$ be an odd perfect number be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Set $$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$ $$H := \gcd\bigg(n^2,\sigma(n^2)\bigg)$$ $$I := \gcd\bigg(n,\sigma(n^2)\bigg).$$

It is known (and fairly easy to prove) that $$GH = I^2.$$

I claim that: $$\sigma(q^k)/2 \text{ is squarefree} \implies \Bigg(\bigg(G \mid I\bigg) \land \bigg(I \mid H\bigg)\Bigg).$$

(Note that there are two unconditional proofs for $G \mid H$ in post #1.)

PROOF OF CLAIM

Suppose that $\sigma(q^k)/2$ is squarefree. Since $\sigma(q^k)/2 \mid n^2$ holds in general, this assumption implies that $\sigma(q^k)/2 \mid n$. From post #2, this is true if and only if $n \mid \sigma(n^2)$. We then obtain $$G = \sigma(q^k)/2$$ from Theorem B in this post. Since $$H = \frac{n^2}{\sigma(q^k)/2},$$ it follows that $$I = n.$$ We get $$\dfrac{I}{G} = \dfrac{H}{I} = \dfrac{n}{\sigma(q^k)/2},$$ which is an integer. We therefore conclude that $G \mid I$ and $H \mid I$. (In particular, note that this yields another proof for $G \mid H$, by transitivity.)

QED

Here is my inquiry:

Does $G \mid I$ and $I \mid H$ still hold if $\sigma(q^k)/2$ is not squarefree, where $q^k n^2$ is an odd perfect number with special prime $q$?

I highly suspect that the answer to my question is YES, since we were able to come up with unconditional proofs for $G \mid H$ in post #1.

Alas, this is where I get stuck, as I do not have a proof for $G \mid I$ and $I \mid H$ when $\sigma(q^k)/2$ is not squarefree.

Jose Arnaldo Bebita Dris
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    $I\mid H$ can be seen as a result of $a\mid b \implies \gcd(a,c) \mid \gcd(b,c)$. In this way, it has nothing to do with whether something unrelated is square-free. – Brian Moehring Feb 02 '22 at 07:54
  • Thank you very much for your time and attention, @BrianMoehring! I did not know about that GCD property. Do you have any further comments about $G \mid I$? I know that $G \mid I$ holds for as long as $\sigma(q^k)/2 \mid n$, even though $\sigma(q^k)/2$ might not be squarefree. That being said, please flesh out your last comment as an actual answer so that I can upvote and then accept. =) – Jose Arnaldo Bebita Dris Feb 02 '22 at 08:00
  • Blimey! I just realized that $GH = I^2$ and $I \mid H$ implies $G \mid I$. Thanks again for the hint, @BrianMoehring! =) – Jose Arnaldo Bebita Dris Feb 02 '22 at 08:17

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Since $I \mid n$ and $n \mid n^2$, it follows that $I \mid n^2$. Also, by definition $I \mid \sigma(n^2).$ Therefore, by the definitional property of the gcd, $$I \mid \gcd(n^2, \sigma(n^2)) = H.$$

As you have pointed out, if we additionally assume $GH=I^2$, then this also implies $G \mid I$, so we're finished with no reference to whether $\sigma(q^k)/2$ is squarefree or not.

Brian Moehring
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