## 12.4.12

### Sorli's Conjecture, April 2012

If $N = {q^k}{n^2}$ is an odd perfect number with Euler prime $q$,
then Sorli's conjecture predicts that $k = 1$.

From previous posts, we have the implication "$n < q \rightarrow k = 1$".

Here, we show other statements that imply $n < q$.

Claim #1:  If $q = \sigma(n)$, then $n < q$.

Proof of Claim #1:
Suppose that $q = \sigma(n)$.  Then

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

From previous posts, $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n} \rightarrow n < q$.

Claim #2:  If $\sigma(q) = \sigma(n)$,
then $n < q$.

Proof of Claim #2:
Suppose that $\sigma(q) = \sigma(n)$.  Then

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

From previous posts, $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n} \rightarrow n < q$.