## 29.3.12

### Sorli's Conjecture, March 2012

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form
(i.e., $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$), then

$$I(q) = \frac{\sigma(q)}{q} \leq \frac{6}{5} < \sqrt{\frac{5}{3}} < I(n) = \frac{\sigma(n)}{n}.$$

This implies that:

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

There are three cases to consider:

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

This is equivalent to the case $q < \sigma(q) < n < \sigma(n)$ alluded to before.

Multiplying through by $\sigma(n)$:

$$\sigma(q) < q{I(n)} < \sigma(n)$$

Dividing through by $q$:

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

Considering the reciprocals:

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

Multiplying through by $\sigma(q)$:

$$\sigma(q) < n{I(q)} < \sigma(n)$$

Dividing through by n:

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

Hence:

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

But:

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

Therefore:

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

Case 2: $\frac{\sigma(q)}{\sigma(n)} < 1 < \frac{q}{n}$

This is equivalent to the case $n < q < \sigma(q) < \sigma(n)$ alluded to before.

Multiplying through by $\sigma(n)$:

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

Dividing through by $q$:

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

Considering the reciprocals:

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

Multiplying through by $\sigma(q)$:

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

Dividing through by $n$:

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

Therefore:

$$1 < I(q) < \frac{\sigma(q)}{n} < I(n) < 2,$$
$$1 < I(q) < \frac{\sigma(n)}{q} < I(n) < 2.$$

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

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

Therefore:  $\sigma(q) < \sigma(n)$.
(No contradictions under Case 2 so far.)

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

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

Therefore:  $n < q$.
(No contradictions under Case 2 so far.)

Case 3:  $1 < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n}$

This is equivalent to the case $n < \sigma(n) \leq q < \sigma(q)$ alluded to before.

Multiplying through by $\sigma(n)$:

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

Dividing through by $q$:

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

Considering the reciprocals:

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

Multiplying through by $\sigma(q)$:

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

Dividing through by $n$:

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

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

But: $n < \sigma(n) \leq q < \sigma(q)$.

Therefore: $\frac{\sigma(n)}{q} \leq 1 < I(q) < I(n) < \frac{\sigma(q)}{n}$.