Search This Blog


On odd perfect numbers given in Eulerian form - Part 3

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.

Dr. Patrick Brown (PatrickBrown496 AT has communicated to me an attempt to prove my 2008 conjecture that $q^k < n$.  In particular, Dr. Brown appears to have completed a proof for the inequality $q < n$.  He accomplished this by proving the implication $k = 1 \Rightarrow q < n$. (Update (Feb 6 2016):  Dr. Brown's preprint has appeared in the arXiv.)

In this post, we will investigate the implications of a proof for $k = 1$, in addition to Dr. Brown's claim that $q < n$.

Hereinafter, we will assume the Descartes-Frenicle-Sorli conjecture that $k = 1$.

We then have
$$\frac{\sigma(n^2)}{q} = D(n^2) = 2n^2 - \sigma(n^2) = \gcd\left(n^2, \sigma(n^2)\right).$$

$$\frac{\sigma(n^2)}{q} \leq q,$$
then $n^2 < \sigma(n^2) \leq q^2$, which would contradict $q < n$.

$$\frac{\sigma(n^2)}{q} > q.$$

We want to show that 
$$\frac{\sigma(n^2)}{q} \neq n^2.$$

Assume that
$$\frac{\sigma(n^2)}{q} = n^2.$$

$$\frac{\sigma(n^2)}{q} = D(n^2) = 2n^2 - \sigma(n^2) = n^2$$
which implies that
$$\sigma(n^2) = n^2.$$
This contradicts the fact that $n > \sqrt[3]{N} > {10}^{500}$.

Observe that the inequality
$$\frac{\sigma(n^2)}{q} > n^2$$
cannot hold because this, together with $q \geq 5$, will imply that $n^2$ is abundant, contradicting $I(n^2) < 2$.

Consequently, we have
$$\frac{\sigma(n^2)}{q} < n^2.$$

$$\frac{\sigma(n^2)}{q} > n$$

Suppose that $\sigma(n^2)/q \leq n$.  Then $\sigma(n^2) \leq qn < n^2$, a contradiction.