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