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