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(Pardon me for being somewhat stubborn, but this question will be the last for this week. This post is an offshoot of this one.)

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 $$GH = I^2.$$

It is also known (and also very easy to prove) that the divisibility chain $$G \mid I \mid H$$ holds.

Here is my:

INITIAL QUESTION: What is the value of $$\frac{I}{G}=\frac{H}{I}?$$

MY ATTEMPT

I claim that $$\dfrac{I}{G} = \dfrac{H}{I} = \dfrac{n}{\sigma(q^k)/2},$$ if and only if $\sigma(q^k)/2 \mid n$.

To this end, initially it must be the case that $$\sigma(q^k)/2 \mid n \iff n \mid \sigma(n^2),$$ basically because of the equation $$\frac{q^k n}{\sigma(q^k)/2} = \frac{\sigma(n^2)}{n}.$$
But we also have $$n \mid \sigma(n^2) \iff G = \sigma(q^k)/2$$ from Theorem B in this post. Since $H$ is also the index of $N$ at the prime $q$, then $$H = \frac{n^2}{\sigma(q^k)/2}.$$ Also, we obtain $$n \mid \sigma(n^2) \iff I = n.$$ We therefore obtain $$\dfrac{I}{G} = \dfrac{H}{I} = \dfrac{n}{\sigma(q^k)/2},$$ which is an integer, if and only if $\sigma(q^k)/2 \mid n$.

FINAL QUESTION: What is the value of $$\frac{I}{G}=\frac{H}{I}$$ if $\sigma(q^k)/2$ does not divide $n$?

Jose Arnaldo Bebita Dris
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    FYI, one can at least say that $\dfrac HI=\dfrac{n^2}{\gcd\bigg(n,\sigma(n^2)\bigg)\sigma(q^k)/2}$$=\dfrac{n^2}{\gcd\bigg(n\sigma(q^k)/2,\sigma(n^2)\sigma(q^k)/2\bigg)}$$=\dfrac{n^2}{\gcd\bigg(n\sigma(q^k)/2,n^2q^k\bigg)}$$=\dfrac{n}{\gcd\bigg(\sigma(q^k)/2,nq^k\bigg)}$$=\dfrac{n}{\gcd\bigg(\sigma(q^k)/2,n\bigg)}$. – mathlove Feb 03 '22 at 06:01
  • Thank you for your time and attention, @mathlove! Please flesh out your last comment as an actual answer, so that I can upvote. =) – Jose Arnaldo Bebita Dris Feb 03 '22 at 06:44
  • I noticed that, since $$H = \dfrac{n^2}{\sigma(q^k)/2},$$ then we obtain $$\gcd\bigg(n,\sigma(n^2)\bigg) = I = \dfrac{n}{\sigma(q^k)/2} \cdot \gcd\bigg(n,\sigma(q^k)/2\bigg).$$ – Jose Arnaldo Bebita Dris Feb 03 '22 at 06:53
  • Additionally, we also obtain $$\gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg) = G = \dfrac{\Bigg(\gcd\bigg(n,\sigma(q^k)/2\bigg)\Bigg)^2}{\sigma(q^k)/2}.$$ – Jose Arnaldo Bebita Dris Feb 03 '22 at 07:43
  • Care to explain the downvote? – Jose Arnaldo Bebita Dris Dec 23 '22 at 08:59

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On OP's request, I am converting my comment into an answer.

FYI, one can at least say that $$\begin{align}\dfrac HI&=\dfrac{n^2}{\gcd\bigg(n,\sigma(n^2)\bigg)\sigma(q^k)/2}=\dfrac{n^2}{\gcd\bigg(n\sigma(q^k)/2,\sigma(n^2)\sigma(q^k)/2\bigg)} \\\\&=\dfrac{n^2}{\gcd\bigg(n\sigma(q^k)/2,n^2q^k\bigg)}=\dfrac{n}{\gcd\bigg(\sigma(q^k)/2,nq^k\bigg)} \\\\&=\dfrac{n}{\gcd\bigg(\sigma(q^k)/2,n\bigg)}\end{align}$$

mathlove
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