## 6.12.15

### An improvement to "On a Conjecture of Dris Regarding Odd Perfect Numbers"

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
$$2 = \frac{\sigma(q^k)}{n} = \frac{\sigma(n)}{q}$$
is ruled out in this MSE post.

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

$$\frac{\sigma(q)}{n} < \frac{1}{2}$$
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.