## 11.11.15

### OPN Research - 11/11/2015

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

We denote the abundancy index of $x$ as
$$I(x) = \frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the sum of the divisors of $x$.

First, we prove the following claim:

Proposition 1.
$$I(n^2) \neq \frac{9}{5}$$

Proof.
Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, and assume further that $I(n^2) = 9/5$.

This implies that $I(q^k) = 2/I(n^2) = 10/9$.  Since $9 < 10 < \sigma(9) = 13$ and $\gcd(10, 9) = 1$, then the fraction $10/9$ is an abundancy outlaw. This contradicts $I(q^k) = 10/9$.

QED.

By Proposition 1, either $I(n^2) < 9/5$ or $I(n^2) > 9/5$ is true.

If $I(n^2) < 9/5$, we have the following result.

Proposition 2.
$$I(n^2) < \frac{9}{5} \Longleftrightarrow q = 5$$

Proof.
Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form.

First, we show that $I(n^2) < 9/5 \Longrightarrow q = 5$:
$$\frac{2}{I(q^k)} = I(n^2) < \frac{9}{5} \Longrightarrow \frac{10}{9} < I(q^k) < \frac{q}{q - 1} \Longrightarrow q < 10.$$
That $q = 5$ follows from the fact that $q$ is the Euler prime (i.e., $q \equiv 1 \pmod 4$).

Next, we show that $I(n^2) > 9/5 \Longrightarrow q \geq 13$:
$$\frac{9}{5} < I(n^2) = \frac{2}{I(q^k)} \Longrightarrow 1 + \frac{1}{q} \leq I(q^k) < \frac{10}{9} \Longrightarrow q > 9.$$
That $q \geq 13$ follows from the fact that $q$ is the Euler prime.

In particular, we have shown that
$$I(n^2) < \frac{9}{5} \Longleftrightarrow q = 5.$$

QED.

The following result appears as Lemma 12 in "The third largest prime divisor of an odd perfect number exceeds one hundred" by Iannucci.

Proposition 3.
$$q = 5 \Longrightarrow k = 1$$

(This post is currently a WORK IN PROGRESS.)