## 11.9.13

### OPN Research - September 2013

In this post, we consider the problem of deriving bounds for the quantity

$$\frac{q^k}{n^2},$$

in terms of $n$, if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form.

We will be using the bounds obtained in the post from July 2013.

Case I:    $q^k < n$

Case I-A: $k = 1 \Longrightarrow q = q^k < n$

We have the chain of inequalities

$$\frac{1}{2}n < q^k < n < 2{q^k}$$

which implies that

$$\frac{1}{2n} < \frac{q^k}{n^2} < \frac{1}{n}.$$

Case I-B: $k \neq 1 \Longrightarrow q < q^k < n$

We have the chain of inequalities

$$\frac{n}{\sqrt{2}} < q^k < n < {q^k}\sqrt{2}$$

which implies that

$$\frac{1}{{\sqrt{2}}n} < \frac{q^k}{n^2} < \frac{1}{n}.$$

Case II:   $n < q^k$

Case II-A:$k = 1 \Longrightarrow n < q = q^k$

We have the chain of inequalities

$$\frac{q^k}{\sqrt{3}} < n < \sqrt[4]{\frac{108}{125}}{q^k} < \sqrt[4]{\frac{108}{125}}{\sqrt{3}}{n}$$

which implies that

$$\sqrt[4]{\frac{125}{108}}\frac{1}{n} < \frac{q^k}{n^2} < \frac{\sqrt{3}}{n}.$$

Case II-B:$k \neq 1 \Longrightarrow q < n < q^k$

We have the chain of inequalities

$$\frac{q^k}{2} < n < \sqrt[4]{\frac{125}{128}}{q^k} < 2{\sqrt[4]{\frac{125}{128}}}{n}$$

which implies that

$$\sqrt[4]{\frac{128}{125}}\frac{1}{n} < \frac{q^k}{n^2} < \frac{2}{n}.$$

We have therefore proven the following theorem (as we already know, from the theorems $q^k < n^2$ [Dris, 2012] and $N = {q^k}{n^2} > {10}^{1500}$ [Ochem and Rao, 2012], that $n > {10}^{375}$):

Theorem:  If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then $q^k = \circ(n^2)$.