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.
