29.3.12

Sorli's Conjecture, March 2012 (Part II)

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then Sorli's conjecture predicts that $k = 1$.

We will be needing the following theorem from the author's M. Sc. thesis:

[Dris, 2008]  If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then $q^k < n^2$.

It follows that, if $n < q$, then $k = 1$.

Here, we prove the following result:

If $\sigma(n) < \sigma(q)$, then $n < q$ if and only if $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$.

Suppose $\sigma(n) < \sigma(q)$.

If $n < q$, then $\frac{1}{q} < \frac{1}{n}$.  It follows that $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$.

If $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$, then since $\frac{\sigma(q)}{q} < \frac{\sigma(n)}{n}$,
it follows that $n < q$.

QED.

In fact, we have the following result:

If $\sigma(n) < \sigma(q)$, then $n < q$.

Since $\sigma(n) < \sigma(q)$, $1 < \frac{\sigma(q)}{\sigma(n)}$.

But $\frac{\sigma(q)}{q} < \frac{\sigma(n)}{n}$, which implies that

$\frac{\sigma(q)}{\sigma(n)} < \frac{q}{n}$.

It follows that $n < q$.

QED.

As a bonus, we have:

If $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$, then $n < q$.

Now, in order to show that:

"If $n < q$, then $\sigma(n) < \sigma(q)$".

we need to rule out Case 2:  $n < q < \sigma(q) < \sigma(n)$, which is equivalent to $\frac{\sigma(q)}{\sigma(n)} < 1 < \frac{q}{n}$.