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*.)