__ORIGINAL QUESTION__
(Note: This question has been cross-posted from MSE.)

Let $\sigma = \sigma_{1}$ be the classical sum-of-divisors function.

A number is said to be

*perfect*if $\sigma(N)=2N$.
If $q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$), are the following statements known to hold in general?

**(a)**$\gcd(n^2, \sigma(n^2))$ is

*large*.

**(b)**The

**deficiency**$D(n^2) = 2n^2 - \sigma(n^2)$ is

*large*.

**(c)**The

**index**$i(q^k) = \sigma(N/q^k)/q^k$ is

*large*.

Using the

*trivial*relationships:
$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = i(q^k) = \frac{2n^2}{\sigma(q^k)}$$

then if the Descartes-Frenicle-Sorli conjecture that $k = 1$ is true, it is possible to show that a lower bound for the quantities

**(a)**,**(b)**and**(c)**is given by $n/\sqrt{3}$. (Here, I have used Acquaah and Konyagin's estimate $q < n\sqrt{3}$. The inequality $q^k < n^2$ then gives the desired*large*numerical bound if we use known lower bounds for the odd perfect number $N = q^k n^2,$ latest of which are by Ochem and Rao.)
What happens when $k > 1$? I do know that

$$\frac{\sigma(N/q^k)}{q^k} = \sigma(n^2)/q^k \geq 315$$

by using a result of Broughan, Delbourgo, and Zhou.

$$\frac{\sigma(N/q^k)}{q^k} = \sigma(n^2)/q^k \geq 315$$

by using a result of Broughan, Delbourgo, and Zhou.

Is it possible to do better than this, apart from attempting a proof of (

*obviously*) $q^k < n$?

__RECENTLY POSTED ANSWER__
Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

In a recent preprint, Brown claims a complete proof for $q < n$, and a partial proof that the inequality $q^k < n$ holds under many cases. (See arXiv.)

In particular, since $q < q^k < n$ holds if Brown's proofs are correct (and

*completed*), then the resulting lower bound is
$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = i(q^k) = \frac{2n^2}{\sigma(q^k)} > \frac{8}{5}\cdot\frac{n^2}{q^k} > \frac{8}{5}\cdot{n}.$$

Notice that Brown's proofs hold unconditionally (i.e., even if $k=1$).

We then have the desired

**large**numerical lower bound
$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = \frac{\sigma(N/q^k)}{q^k} > \frac{8}{5}\cdot{n} > \frac{8}{5}\cdot{{10}^{500}},$$

which is an easy consequence of Ochem and Rao's $N > {10}^{1500}$ and the inequality $q^k < n$.