If $q^k n^2$ is an odd perfect number with special prime $q$, is $\sigma(q^k)$ coprime to $\sigma(n^2)$? Let $q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Here is my:
QUESTION

If $q^k n^2$ is an odd perfect number with special prime $q$, is $\sigma(q^k)$ coprime to $\sigma(n^2)$?

The function $\sigma(x)=\sigma_1(x)$ is called the sum of divisors of $x$.
MY ATTEMPT FOR SOME SPECIFIC VALUES OF $q$, $k$, AND $n$
For example, there may be an odd perfect number with $q^k = 17$ and $(n/3)^2$ coprime to $3$.  Then $\sigma(n^2) = 17(n^2/9)$ is coprime to $\sigma(17) = 18$.
How about the general case?  My hunch is that the following conjecture ought to hold:

Conjecture: If $q^k n^2$ is an odd perfect number with special prime $q$, then $$\gcd(\sigma(q^k),\sigma(n^2))>1.$$

Initially, I thought that a proof of this "Conjecture" was in the following paper by Dandapat et al., but after an in-depth reading, it appears that I was mistaken.
I have therefore tagged this as a reference-request for a proof of this Conjecture.
 A: This post shows that $\gcd(\sigma(q^k),\sigma(n^2))=1$ would lead to $k=1$.
Let $N=q^k n^2$ be an odd perfect number with $q$ being the special prime, $q \nmid n$ and $k \equiv 1 \pmod 4$. Then from $\sigma(N)=2N$ and multiplicativity of $\sigma$ we get $\sigma(q^k)\sigma(n^2)=2q^kn^2$. Now clearly $q^k$ is coprime with $\sigma(q^k)=1+q+\dots+q^k$, hence $q^k \mid \sigma(n^2)$. We also have $2 \mid \sigma(q^k)$ because $q,k$ are odd, thus we can write
\begin{equation}
\left[\frac{\sigma(q^k)}{2}\right]\cdot \left[\frac{\sigma(n^2)}{q^k}\right]=n^2.
\end{equation}
Now assuming $\sigma(q^k)$ and $\sigma(n^2)$ are coprime, then also the two integers on the left side are coprime. That means both are perfect squares, and so in particular
$$
2m^2=\sigma(q^k)=1+q+\dots+q^k
$$
for some integer $m$. However this equation has no solution in integers for $q,k \geq 2$, so if there is an odd perfect number with this property, we must have $k=1$ (since $q \geq 5$ anyway).
A: Not a complete answer, I merely wanted to record some recent thoughts that occurred to me, that are directly related to this question.

Let $q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.
Suppose to the contrary that
$$\gcd(\sigma(q^k),\sigma(n^2))=1.$$
Since $q^k n^2$ is (odd) perfect, we obtain
$$\sigma(q^k)\sigma(n^2) = \sigma(q^k n^2) = 2q^k n^2.$$
Applying the $\sigma$-function to both sides of the last equation, we obtain
$$\sigma(\sigma(q^k))\sigma(\sigma(n^2)) = \sigma\bigg(\sigma(q^k)\sigma(n^2)\bigg) = \sigma(2 q^k n^2) = \sigma(2)\sigma(q^k)\sigma(n^2) = 3\sigma(q^k)\sigma(n^2)$$
so that
$$I(\sigma(q^k))I(\sigma(n^2)) = \frac{\sigma(\sigma(q^k))}{\sigma(q^k)}\frac{\sigma(\sigma(n^2))}{\sigma(n^2)} = 3.$$
Note that, trivially, we have $\sigma(q^k) > 1$ and $\sigma(n^2) > 1$, so that
$$I(\sigma(q^k)) > 1$$
and
$$I(\sigma(n^2)) > 1$$
both hold.
These inequalities imply that
$$I(\sigma(q^k)) > 1 \implies I(\sigma(n^2)) < 3$$
and
$$I(\sigma(n^2)) > 1 \implies I(\sigma(q^k)) < 3$$
so that we obtain
$$1 < I(\sigma(q^k)) < 3$$
and
$$1 < I(\sigma(n^2)) < 3.$$
Suppose to the contrary that
$$I(\sigma(q^k)) = 2$$
holds.  Then we obtain
$$I(\sigma(n^2)) = \frac{3}{I(\sigma(q^k))} = \frac{3}{2} = I(2).$$
Since $2$ is solitary, this last equation forces
$$\sigma(n^2) = 2,$$
which contradicts the fact that $\sigma(n^2)$ is odd.  (It also contradicts $n^2 > {10}^{750}$.)
Suppose to the contrary that
$$I(\sigma(n^2)) = 2$$
holds.  Then we obtain
$$I(\sigma(q^k)) = \frac{3}{I(\sigma(n^2))} = \frac{3}{2} = I(2).$$
Since $2$ is solitary, this last equation forces
$$\sigma(q^k) = 2,$$
which contradicts the fact that
$$\sigma(q^k) \geq q^k + 1 \geq 6$$
since $q \geq 5$ and $k \geq 1$.
Hence, neither $\sigma(q^k)$ nor $\sigma(n^2)$ could be perfect, if $\gcd(\sigma(q^k),\sigma(n^2))=1$.

Further considerations are in the following questions:
If $N = q^k n^2$ is an odd perfect number with special prime $q$, then must $\sigma(n^2)$ be abundant? MSE question 4223529 Edit: (September 11, 2021 - 1:44 PM Manila time) - Finally proved that $\sigma(n^2)$ must be deficient.
If $N = q^k n^2$ is an odd perfect number with special prime $q$, then must $\sigma(q^k)$ be deficient? MSE question 3831043

Added (August 14, 2021 - 15:49 PM Manila time): Here we will be proving the following claim:

CLAIM: Let $q^k n^2$ be an odd perfect number with special prime $q$, satisfying $\gcd(\sigma(q^k),\sigma(n^2))=1$.


*

*$$\Bigg(\sigma(n^2) \text{ is abundant}\Bigg) \implies \Bigg(I(\sigma(q^k)) < I(\sigma(n^2))\Bigg) \implies \Bigg(\sigma(q^k) \text{ is deficient}\Bigg)$$

*$$\Bigg(\sigma(q^k) \text{ is abundant}\Bigg) \implies \Bigg(I(\sigma(n^2)) < I(\sigma(q^k))\Bigg) \implies \Bigg(\sigma(n^2) \text{ is deficient}\Bigg).$$
Proof: Recall that both $\sigma(q^k)$ and $\sigma(n^2)$ are not perfect, if $\gcd(\sigma(q^k),\sigma(n^2)) = 1$.
Suppose to the contrary that $I(\sigma(q^k)) = I(\sigma(n^2))$.
Then this implies that
$$(I(\sigma(q^k)))^2 = I(\sigma(q^k))I(\sigma(n^2)) = 3,$$
which implies $I(\sigma(q^k)) = \sqrt{3}$.  This contradicts the fact that $I(\sigma(q^k))$ is rational.
Hence, we know that either $I(\sigma(q^k)) < I(\sigma(n^2))$, or $I(\sigma(n^2)) < I(\sigma(q^k))$, must hold.
Suppose that $I(\sigma(q^k)) < I(\sigma(n^2))$.
This implies that $(I(\sigma(q^k)))^2 < I(\sigma(q^k))I(\sigma(n^2)) = 3$, from which it follows that $I(\sigma(q^k)) < \sqrt{3} < 2$.  Hence, $\sigma(q^k)$ is deficient, if $I(\sigma(q^k)) < I(\sigma(n^2))$.  By the contrapositive, since $\sigma(q^k)$ is not perfect if $\gcd(\sigma(q^k),\sigma(n^2))=1$, then if $\sigma(q^k)$ is abundant, we have that $I(\sigma(n^2)) < I(\sigma(q^k))$.
Suppose that $I(\sigma(n^2)) < I(\sigma(q^k))$.
This implies that $(I(\sigma(n^2)))^2 < I(\sigma(q^k))I(\sigma(n^2)) = 3$, from which it follows that $I(\sigma(n^2)) < \sqrt{3} < 2$.  Hence, $\sigma(n^2)$ is deficient,
if $I(\sigma(n^2)) < I(\sigma(q^k))$.  By the contrapositive, since $\sigma(n^2)$ is not perfect if $\gcd(\sigma(q^k),\sigma(n^2))=1$, then if $\sigma(n^2)$ is abundant, we have that $I(\sigma(q^k)) < I(\sigma(n^2))$.
QED.

SANITY CHECK: The Claim proves that $\sigma(q^k)$ and $\sigma(n^2)$ cannot be both abundant, which is true, because otherwise
$$4 = 2^2 < I(\sigma(q^k))I(\sigma(n^2)) = 3,$$
which is a contradiction.
Hence, we conclude that if $q^k n^2$ is an odd perfect number with special prime $q$ satisfying $\gcd(\sigma(q^k),\sigma(n^2))=1$, then either one of the following cases hold:

*

*$\sigma(q^k)$ is deficient and $\sigma(n^2)$ is deficient

*$\sigma(q^k)$ is abundant and $\sigma(n^2)$ is deficient

*$\sigma(q^k)$ is deficient and $\sigma(n^2)$ is abundant - This is ruled out by the proof for a deficient $\sigma(n^2)$, as mentioned below.

Alas, this is where I get stuck!

UPDATED ON September 11, 2021 - 2:01 PM (Manila time) In this answer to a closely related MSE question, it is proved that, unconditionally, $\sigma(n^2)$ must be deficient.
This implies that $I(\sigma(n^2)) < 2$.
Since $\gcd(\sigma(q^k),\sigma(n^2))=1$ by assumption, then we have
$$I(\sigma(q^k))I(\sigma(n^2))=3$$
which, together with the upper bound $I(\sigma(n^2)) < 2$, implies that
$$I(\sigma(q^k))I(\sigma(n^2))=3<2I(\sigma(q^k)).$$
Finally, we obtain
$$I(\sigma(q^k)) > \frac{3}{2}.$$
