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.)
