Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.

Dr. Patrick Brown (PatrickBrown496 AT gmail.com) has communicated to me an attempt to prove my 2008 conjecture that $q^k < n$. In particular, Dr. Brown appears to have completed a proof for the inequality $q < n$. He accomplished this by proving the implication $k = 1 \Rightarrow q < n$. (

__Update (Feb 6 2016)__: Dr. Brown's preprint has appeared in the arXiv.)
In this post, we will investigate the implications of a proof for $k = 1$, in addition to Dr. Brown's claim that $q < n$.

Hereinafter, we will assume the Descartes-Frenicle-Sorli conjecture that $k = 1$.

We then have

$$\frac{\sigma(n^2)}{q} = D(n^2) = 2n^2 - \sigma(n^2) = \gcd\left(n^2, \sigma(n^2)\right).$$

If

$$\frac{\sigma(n^2)}{q} \leq q,$$

then $n^2 < \sigma(n^2) \leq q^2$, which would contradict $q < n$.

Hence

$$\frac{\sigma(n^2)}{q} > q.$$

We want to show that

$$\frac{\sigma(n^2)}{q} \neq n^2.$$

Assume that

$$\frac{\sigma(n^2)}{q} = n^2.$$

Then

$$\frac{\sigma(n^2)}{q} = D(n^2) = 2n^2 - \sigma(n^2) = n^2$$

which implies that

$$\sigma(n^2) = n^2.$$

This contradicts the fact that $n > \sqrt[3]{N} > {10}^{500}$.

Observe that the inequality

$$\frac{\sigma(n^2)}{q} > n^2$$

cannot hold because this, together with $q \geq 5$, will imply that $n^2$ is abundant, contradicting $I(n^2) < 2$.

Consequently, we have

$$\frac{\sigma(n^2)}{q} < n^2.$$

Claim:

$$\frac{\sigma(n^2)}{q} > n$$

Suppose that $\sigma(n^2)/q \leq n$. Then $\sigma(n^2) \leq qn < n^2$, a contradiction.