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 $k > 1$.

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**.