# If $q^k n^2$ is an odd perfect number with Euler prime $q$, are the following statements known to hold in general?

(Note: This question has been cross-posted from MO.)

Let $$\sigma$$ 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.

Is it possible to do better than this, apart from attempting a proof of (obviously) $$q^k < n$$?

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 $q \leq q^k < n$ 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 $q^k < n$.