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

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

From this paper in NNTDM, we have the equation $$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right).$$

In particular, we know that the *index* $i(q)$ is an integer greater than $5$ by a result of Dris and Luca.

We now attempt to compute an expression for $\gcd\left(\sigma(q^k),\sigma(n^2)\right)$ in terms of $i(q)$.

First, since we have $$\sigma(q^k)\sigma(n^2) = \sigma({q^k}{n^2}) = \sigma(N) = 2N = 2{q^k}{n^2}$$ 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 $$\sigma(n^2) = \frac{2 q^k n^2}{\sigma(q^k)} = {q^k}\cdot\bigg(\frac{2n^2}{\sigma(q^k)}\bigg) = {q^k}{i(q)},$$ so that we get $$\gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg).$$

Now, since $\gcd(q, n) = \gcd(q^k, 2n^2) = 1$ and $i(q)$ is odd, we get $$\gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

Hence, we conclude that $G:=\gcd(\sigma(q^k),\sigma(n^2))=\gcd\bigg({n^2}/{i(q)}, i(q)\bigg)$.

*(Edited Sept 12 2017)* Here are my questions:

Original Question

I seem to recall that somebody (was it Pomerance [?] et. al) proved that $$G \neq 1.$$ Does anybody here happen to know a reference? Additionally, does $G \neq 1$ imply that $G = i(q)$?

Additional Question *(Added Sept 12 2017)*

Some authors have already considered the possibility that $i(q)$ may be a square. This would imply that $G$ is also a square. Would $G$ a square mean that $G = i(q)$ holds?

*(End Edit)*

Thank you.