(Note: This post is an offshoot of this earlier MSE question.)
In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. We also let $D(x)=2x-\sigma(x)$ denote the deficiency of $x$.
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
From this paper in NNTDM, we have the equation
$$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right). \tag{1}\label{eq1}$$
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
$$\gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$
The following is copied verbatim from this MathOverflow answer to a closely related question:
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 $\gcd(n, \sigma(n^2)) = \gcd(n^2, \sigma(n^2))$.
Here is an attempt at proving the following conjecture:
CONJECTURE: If $q^k n^2$ is an odd perfect number with special prime $q$, then the inequation
$$\gcd(\sigma(q^k), \sigma(n^2)) \neq \gcd(n^2, \sigma(n^2))$$
holds.
MY ATTEMPT AT A PROOF
Let $N = q^k n^2$ be an odd perfect number with special prime $q$.
Assume to the contrary that $$\gcd(\sigma(q^k), \sigma(n^2)) = \gcd(n^2, \sigma(n^2)).$$
By the considerations above, we obtain
$$\gcd(n^2, \sigma(n^2)) = \gcd(n, \sigma(n^2)).$$
But note that we also have
$$\frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} = \gcd(n^2, \sigma(n^2)).$$
This implies that
$$\frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} = \gcd(n^2, \sigma(n^2)) = \gcd(n, \sigma(n^2)),$$
from which we get the divisibility constraint
$$\frac{n^2}{\sigma(q^k)/2} \mid n$$
which gives
$$\frac{n}{\sigma(q^k)/2} \mid 1$$
and then
$$n = \frac{\sigma(q^k)}{2}.$$
We obtain
$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \frac{D(n^2)}{\sigma(q^{k-1})} = \gcd(n^2, \sigma(n^2)) = i(q) = n = \frac{\sigma(q^k)}{2}.$$
It follows that
$$\sigma(n^2) = q^k n = \frac{q^k \sigma(q^k)}{2}.$$
In particular, note that we obtain the following equations from equation $\eqref{eq1}$, after multiplying through by $(q^k \sigma(q^k))/2$:
$$\frac{q^k \sigma(q^k)}{2}\cdot\frac{D(n^2)}{\sigma(q^{k-1})} = \frac{q^k \sigma(q^k)}{2}\cdot\gcd(n^2, \sigma(n^2)) = \frac{q^k \sigma(q^k)}{2}\cdot\gcd(n, \sigma(n^2)) = q^k n^2 = N = n\sigma(n^2).$$
Alas, this is where I get stuck. I was hoping somebody out there might have some bright ideas on how to prove either $k=1$ or $q^k < n$ (in order to arrive at a contradiction), under the assumption that
$$\gcd(\sigma(q^k),\sigma(n^2))=\gcd(n^2,\sigma(n^2)),$$
where $q^k n^2$ is an odd perfect number with special prime $q$?
POSTSCRIPT: Note that to prove $k=1$, it suffices to show that
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
Compare with what we have:
$$\frac{\sigma(n^2)}{q^k} = n \mid n^2.$$
UPDATE (October 30, 2020 – 1:53 PM Manila time) I have just posted what appears to be a complete proof of the conjecture.
Best Answer
solution-verification
Writing $\dfrac{n}{\sigma(q^k)/2} \mid 1$, you are implicitly using that $\dfrac{n}{\sigma(q^k)/2}$ is an integer, but I don't see why $\dfrac{n}{\sigma(q^k)/2}$ is an integer.
After assuming that $$\gcd(\sigma(q^k),\sigma(n^2))=\gcd(n^2,\sigma(n^2))$$ we have $$i(q)= \frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \frac{D(n^2)}{\sigma(q^{k-1})} = \gcd(n^2,\sigma(n^2))=\gcd(n,\sigma(n^2))$$ So, we can have $$0\lt\frac{n}{\sigma(q^k)/2} = \frac{\gcd(n,\sigma(n^2))}{n}\le\frac nn=1$$ from which we can at least say that $$\text{$\frac{n}{\sigma(q^k)/2}$ is an integer}\iff \gcd(n,\sigma(n^2))=n\iff n=\frac{\sigma(q^k)}{2}$$