## 3.6.12

### OPN Research - Sorli's Conjecture - June 2012

Let $N = {q^k}{n^2}$ be an odd perfect number (hereinafter abbreviated as OPN) given in Eulerian form (i.e. $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$).

From previous posts, we know that $n < q \Longrightarrow k = 1$.

We also know that:

$$I(q) \leq 6/5 < \sqrt{5/3} < I(n)$$

In particular, we know that:

$$\sigma(q)/\sigma(n) < q/n$$

Suppose that $\sigma(n) \leq \sigma(q)$.

It follows that:

$n < q$

and that:

$\sigma(n)/q < \sigma(q)/n$

However, from previous posts, we also know that:

$\sigma(n)/q < \sigma(q)/n \Longrightarrow n < q$

Hence, we have proved the following lemma:

Lemma 1:  Let $N = {q^k}{n^2}$ be an OPN given in Eulerian form.  If $\sigma(n) \leq \sigma(q)$, then $n < q$ if and only if $\sigma(n)/q < \sigma(q)/n$.

The contrapositive of:

$\sigma(n) \leq \sigma(q) \Longrightarrow n < q$

is:

$q < n \Longrightarrow \sigma(q) < \sigma(n)$

Therefore, if $q < n$, it follows that:

$\sigma(q)/n < \sigma(n)/q$

However, from previous posts, we also know that:

$\sigma(q)/n < \sigma(n)/q \Longrightarrow \sigma(q) < \sigma(n)$

Consequently, we have proved the following lemma:

Lemma 2: Let $N = {q^k}{n^2}$ be an OPN given in Eulerian form.  If $q < n$, then $\sigma(q) < \sigma(n)$ if and only if $\sigma(q)/n < \sigma(n)/q$.

Note that, if $\sigma(q) < \sigma(n)$, then the best inequality that we could get relating $q$ and $n$ is:

$(q + 1)/2 < n$

This is because the upper bound $I(n) < 2$ is rather crude.