## 3.6.17

### Verification of an equation from a recent preprint on odd perfect numbers (to appear in NNTDM)

The paper titled "Conditions Equivalent to the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers" has finally been accepted and is set to appear in the journal Notes on Number Theory and Discrete Mathematics (Volume 23, Number 2) this June or July, 2017.

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
$$\frac{6q}{6q-5}+\frac{6q-5}{3q}=\frac{3q^2 + 2q + 1}{q(q+1)}$$
$$\bigg(q = \frac{8}{3}\bigg) \lor \bigg(q = 5\bigg),$$
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
$$\bigg(q = \frac{25}{12}\bigg) \lor \bigg(q = 5\bigg),$$
Since $q \geq 5$, $q = 5$ holds, which again validates the result.