## 20.7.16

### If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$ implies that $k=1$ and $q=5$.

(Note:  This blog post is essentially an elucidation of the answer to this MSE question.)

Here, we prove the following proposition.

Theorem.  If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$$
implies that
$$k=1$$
and
$$q=5.$$

Proof.  Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

Since $N$ is perfect, we have
$$2N = 2q^k n^2 = \sigma(q^k n^2) = \sigma(q^k)\sigma(n^2) = \left(\frac{q^{k+1} - 1}{q - 1}\right)\cdot\sigma(n^2).$$
It follows that
$$\frac{\sigma(n^2)}{n^2} = \frac{2{q^k}\left(q - 1\right)}{q^{k+1} - 1} = \frac{2q^{k+1} - 2q^k}{q^{k+1} - 1} = \frac{2q^{k+1} - 2}{q^{k+1} - 1} - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right)$$
$$= 2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right).$$

By assumption, we have
$$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}.$$

This inequality is equivalent to
$$2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right) \geq 2 - \frac{5}{3q},$$
which, in turn, is equivalent to
$$6q\left(q^k - 1\right) \leq 5(q^{k+1} - 1).$$
This last inequality simplifies to
$$q^{k+1} - 6q + 5 \leq 0,$$
which cannot be true when $k > 1$.  Thus, we know that $k=1$.

Hence, we have
$$q^2 - 6q + 5 \leq 0.$$
But this inequality implies that
$$1 \leq q \leq 5,$$
which, together with $q \geq 5$, implies that $q=5$.

QED.