Here, we outline an improvement to On a Conjecture of Dris Regarding Odd Perfect Numbers:
Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form. We want to show that
$$\frac{\sigma(n)}{q} \ne \frac{\sigma(q^k)}{n}.$$
Assume that
$$n\sigma(n) = q\sigma(q^k).$$
Since $\gcd(q,n) = 1$, this means that $n \mid \sigma(q^k)$ and $q \mid \sigma(n)$, so that
$$\frac{\sigma(q^k)}{n}$$
and
$$\frac{\sigma(n)}{q}$$
are equal positive integers.
But $\sigma(q^k)$ is even and $n$ is odd. We therefore have:
$$2 \le \frac{\sigma(q^k)}{n} = \frac{\sigma(n)}{q}$$
The case
Therefore, we obtain
$$4 \le \frac{\sigma(q^k)}{n} = \frac{\sigma(n)}{q}$$
Multiplying both sides of the inequality and equation by $\frac{\sigma(n)}{q^k}$, we get
$$4\cdot{\frac{\sigma(n)}{q^k}} \le {\frac{\sigma(n)}{q}}\cdot{\frac{\sigma(n)}{q^k}} = {\frac{\sigma(q^k)}{n}}\cdot{\frac{\sigma(n)}{q^k}} < 2$$
Multiplying both sides of the inequality and equation by $\frac{\sigma(q)}{n}$, we obtain
$$4\cdot{\frac{\sigma(q)}{n}} \le {\frac{\sigma(q^k)}{n}}\cdot{\frac{\sigma(q)}{n}} = {\frac{\sigma(q)}{n}}\cdot{\frac{\sigma(n)}{q}} < 2$$
Consequently, we have
$$\frac{\sigma(n)}{q^k} < \frac{1}{2}$$
and
$$\frac{\sigma(q)}{n} < \frac{1}{2}$$
and
$$\frac{\sigma(q)}{n} < \frac{1}{2}$$
from which it follows that
$$\frac{\sigma(n)}{q^k} < \frac{1}{2} < 4 \le \frac{\sigma(q^k)}{n}$$
and
$$\frac{\sigma(q)}{n} < \frac{1}{2} < 4 \le \frac{\sigma(n)}{q}.$$
We conclude that we must have
$$q < n < q^k$$
so that $k > 1$.
THIS POST IS CURRENTLY A WORK IN PROGRESS.
and
$$\frac{\sigma(q)}{n} < \frac{1}{2} < 4 \le \frac{\sigma(n)}{q}.$$
We conclude that we must have
$$q < n < q^k$$
so that $k > 1$.
THIS POST IS CURRENTLY A WORK IN PROGRESS.