On GCDs and odd perfect numbers

Question

(Note: This question is closely related to this other one in MO.)

Let $N=q^k n^2$ be an odd perfect number with special prime $q$.

The index $i(q)$ of $N$ at the prime $q$ is then equal to $$i(q):=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}=\frac{D(n^2)}{s(q^k)}=\frac{s(n^2)}{D(q^k)/2}=\gcd(n^2,\sigma(n^2)),$$ where $D(x):=2x-\sigma(x)$ is the deficiency and $s(x):=\sigma(x)-x$ is the sum of aliquot divisors of $x \in \mathbb{N}$.

We show that

$$\gcd(\sigma(q^k),\sigma(n^2))=\gcd\bigg(i(q),\frac{n^2}{i(q)}\bigg).$$

Proof

We have $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2 q^k n^2$$ from which we obtain $$\sigma(q^k)=\frac{2 q^k n^2}{\sigma(n^2)}=\frac{2n^2}{\sigma(n^2)/q^k}=\frac{2n^2}{i(q)}$$ and $$\sigma(n^2)=\frac{2 q^k n^2}{\sigma(q^k)}=q^k \cdot \bigg(\frac{n^2}{\sigma(q^k)/2}\bigg)=q^k i(q)$$ so that we get $$\gcd(\sigma(q^k),\sigma(n^2))=\gcd\bigg(q^k i(q),\frac{2n^2}{i(q)}\bigg).$$ Now, since $\gcd(q,n)=\gcd(q^k,2n^2)=1$ and $i(q)$ is odd, we conclude that $$G:=\gcd(\sigma(q^k),\sigma(n^2))=\gcd\bigg(i(q),\frac{n^2}{i(q)}\bigg).$$

QED

Dandapat, Hunsucker and Pomerance proved in the year 1975 that $G > 1$.

Is my derivation of the formula for the GCD of $\sigma(q^k)$ and $\sigma(n^2)$ correct?

Answer 1

It appears that the derivation in the OP is correct.

Additionally, we have the following:

Here is a conditional proof that $$G = \gcd(\sigma(q^k),\sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)).$$

As derived in the OP, we have $$G = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

This is equivalent to $$G = \frac{1}{i(q)}\cdot\gcd\bigg(n^2, (i(q))^2\bigg) = \frac{1}{i(q)}\cdot\bigg(\gcd(n, i(q))\bigg)^2.$$

But we also have $$\gcd(n, i(q)) = \gcd\bigg(n, \gcd(n^2, \sigma(n^2))\bigg) = \gcd\bigg(\sigma(n^2), \gcd(n, n^2)\bigg) = \gcd(n, \sigma(n^2)).$$

Consequently, we obtain $$G = \frac{1}{i(q)}\cdot\bigg(\gcd(n, \sigma(n^2))\bigg)^2 = \frac{\bigg(\gcd(n, \sigma(n^2))\bigg)^2}{\gcd(n^2, \sigma(n^2))}.$$

In particular, we get $$\gcd(\sigma(q^k), \sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2))$$ if and only if $\gcd(n^2, \sigma(n^2)) = \gcd(n, \sigma(n^2)).$