**EUREKA???**

__Introduction__
If $N=q^k n^2$ is an odd perfect number with Euler prime $q$, then the Descartes-Frenicle-Sorli conjecture predicts that $k=1$ always holds.

In this blog post, we will attempt to give an elementary proof for this conjecture. We will also derive the

*possible value(s)*for the Euler prime $q$!
We denote the abundancy index of the positive integer $x$ by

$$I(x) = \frac{\sigma(x)}{x}.$$

__Preliminaries__
First, we restate the following results from these earlier posts in this blog:

**Proposition 1**

Suppose that $N=q^k n^2$ is an odd perfect number with Euler prime $q$. Then $k=1$ implies

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

**Proposition 2**

Suppose that $N=q^k n^2$ is an odd perfect number with Euler prime $q$. Then

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

implies that $k=1$ and $q=5$.

**Main Results****Theorem**

Suppose that $N=q^k n^2$ is an odd perfect number with Euler prime $q$. Then

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

Additionally, $I(n^2) = 2 - \frac{5}{3q}$ if and only if $k=1$ and $q=5$.

**Proof**

Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$.

By the contrapositive of

**Proposition 1**,
$$I(n^2) > 2 - \frac{5}{3q} \Rightarrow k \neq 1.$$

By

**Proposition 2**, we have
$$I(n^2) \geq 2 - \frac{5}{3q} \Rightarrow \left\{k = 1 \land q = 5\right\}.$$

Now, assume that

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

is true. Then we have $k=1$ and $q=5$. In particular, we are sure that $k=1$ must hold.

Consequently,

$$I(n^2) \geq 2 - \frac{5}{3q} \Rightarrow k = 1$$

holds.

Note that

$$I(n^2) > 2 - \frac{5}{3q} \Rightarrow k = 1$$

also holds.

Putting it all together,

$$\left\{I(n^2) > 2 - \frac{5}{3q} \Rightarrow k \neq 1\right\} \land \left\{I(n^2) > 2 - \frac{5}{3q} \Rightarrow k = 1\right\} \Rightarrow I(n^2) \leq 2 - \frac{5}{3q}.$$

Notice that

$$I(n^2) = 2 - \frac{5}{3q} \Rightarrow \left\{k=1 \land q=5\right\}$$

holds.

Thus, it remains to rule out the case

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

It is easy to show that if

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

is true, then it cannot happen that both $k=1$ and $q=5$ are true. Thus, either $k \neq 1$ or $q \neq 5$ is true. This means that the implication

$$k = 1 \Rightarrow q \neq 5$$

is true.

However, Iannucci (1999)

**[Lemma 12, page 873]**proved that
$$q=5$$

implies

$$k=1.$$

Consequently, we have that $q \neq 5$. (Note that this implies that $q \geq 13$.)

(

**THIS BLOG POST IS CURRENTLY A WORK IN PROGRESS**.)