The latest preprint for this paper is available online via the arXiv.

In this blog post, we will verify the equation

$$I(n^2)=2-\frac{5}{3q}=\frac{6q-5}{3q}$$

which is true if and only if $q=5$ and $k=1$, where $I(n^2)=\sigma(n^2)/n^2$ is the abundancy index of $n^2$ and $N=q^k n^2$ is an odd perfect number with Euler prime $q$.

To this end, suppose that

$$I(n^2)=2-\frac{5}{3q}=\frac{6q-5}{3q}.$$

This implies that

$$I(q^k)=\frac{2}{I(n^2)}=\frac{6q}{6q-5},$$

and that

$$q = 5 \land k = 1.$$

But $k=1$ implies that

$$I(q^k)+I(n^2)=I(q)+\frac{2}{I(q)}=\frac{q+1}{q}+\frac{2q}{q+1}$$

so that

$$I(q^k)+I(n^2)=\frac{(q+1)^2 + 2q^2}{q(q+1)}=\frac{3q^2 + 2q + 1}{q(q+1)}=3-\frac{q-1}{q(q+1)}.$$

Plugging in the values for $I(q^k)$ and $I(n^2)$ from above, and then solving for $q$, we obtain

$$\frac{6q}{6q-5}+\frac{6q-5}{3q}=\frac{3q^2 + 2q + 1}{q(q+1)}$$

$$\bigg(q = \frac{8}{3}\bigg) \lor \bigg(q = 5\bigg),$$

per this WolframAlpha computation.

Since $q \geq 5$, $q = 5$ holds, which validates the result.

As an additional validation, when $q=5$ and $k=1$, we get

$$I(q^k)+I(n^2)=I(q)+\frac{2}{I(q)}=I(5)+\frac{2}{I(5)}=\frac{6}{5}+\frac{5}{3}=\frac{43}{15}.$$

Solving the equation

$$\frac{6q}{6q-5}+\frac{6q-5}{3q}=\frac{43}{15}$$

gives

$$\bigg(q = \frac{25}{12}\bigg) \lor \bigg(q = 5\bigg),$$

per this WolframAlpha computation.

Since $q \geq 5$, $q = 5$ holds, which again validates the result.