(Note: This has been cross-posted to MO.)
Note that $\gcd(\sigma(q^k),\sigma(n^2))=i(q)=\gcd(n^2,\sigma(n^2))$ if and only if $\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$. I have therefore undeleted this question, in view of this closely related MSE question: When does $\gcd(m,\sigma(m^2))$ equal $\gcd(m^2,\sigma(m^2))$? What are the exceptions?.
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 $$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right).$$
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 $\gcd(n, \sigma(n^2)) = \gcd(n^2, \sigma(n^2))$.
We now try to express $\gcd(n^2, \sigma(n^2))$ as a linear combination of $n^2$ and $\sigma(n^2)$:
To begin with, write $$\gcd(n^2, \sigma(n^2)) = \frac{\sigma(n^2)}{q^k} = \frac{D(n^2)}{\sigma(q^{k-1})} = \frac{(2n^2 - \sigma(n^2))(q - 1)}{q^k - 1}.$$
Now, using the identity $$\frac{A}{B}=\frac{C}{D}=\frac{A-C}{B-D},$$ where $B \neq 0$, $D \neq 0$, and $B \neq D$, we obtain $$\gcd(n^2, \sigma(n^2)) = \frac{\sigma(n^2) - (2n^2 - \sigma(n^2))(q - 1)}{q^k - (q^k - 1)}$$ from which we get $$\gcd(n^2, \sigma(n^2)) = \sigma(n^2) - (2n^2 - \sigma(n^2))(q - 1) = 2n^2 - q(2n^2 - \sigma(n^2)) = 2n^2 - qD(n^2),$$ or equivalently, $$\gcd(n^2, \sigma(n^2)) = 2(1 - q)n^2 + q\sigma(n^2).$$
Note that the linear combination for $\gcd(n^2, \sigma(n^2))$ was obtained without assuming the Descartes-Frenicle-Sorli Conjecture for odd perfect numbers that $k=1$.
Here is my question:
How can I then derive a linear combination for $\gcd(\sigma(q^k), \sigma(n^2))$ in terms of $\sigma(q^k)$ and $\sigma(n^2)$? Note that this should always be (unconditionally) possible by Bezout's Identity.
