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A curious inequality involving divisors of odd perfect numbers

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

Since $q^k n \mid N$ and $N$ is perfect, then $q^k n$ is deficient.  Therefore, $I(q^k n) < 2$, where $I(x)=\sigma(x)/x$ is the abundancy index of $x \in \mathbb{N}$.

It follows that
$$\frac{1}{2}\cdot\frac{\sigma(q^k)}{n} < \frac{q^k}{\sigma(n)}$$
$$\frac{1}{2}\cdot\frac{\sigma(n)}{q^k} < \frac{n}{\sigma(q^k)}.$$
Adding the last two inequalities, we get
$$\frac{\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}}{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}.$$

Since the arithmetic mean is never less than the harmonic mean, and since
$$\frac{\sigma(q^k)}{n} \neq \frac{\sigma(n)}{q^k},$$
then we have
$$\frac{2}{\frac{1}{\sigma(q^k)/n}+\frac{1}{\sigma(n)/q^k}}< \frac{\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}}{2},$$
so that we obtain
$$\frac{2}{\frac{q^k}{\sigma(n)}+\frac{n}{\sigma(q^k)}}< \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}.$$
This last inequality implies that
$$2 < \bigg(\frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}\bigg)^2$$
which further means that
$$\sqrt{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}.$$

Now, we also have that (for the case of two summands/factors)
$$\text{ Arithmetic Mean }\cdot\text{ Harmonic Mean } = \bigg(\text{ Geometric Mean }\bigg)^2.$$
(See this hyperlink for a proof.)

Therefore, we have
$$\frac{\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}}{\frac{n}{\sigma(q^k)}+\frac{q^k}{\sigma(n)}}=\bigg(\sqrt{\frac{\sigma(q^k)}{n}\cdot\frac{\sigma(n)}{q^k}}\bigg)^2 = I({q^k}n),$$
an equation which can be readily verified by an inspection of the complex fraction on the LHS.  We shall return to this relationship later.

From the inequality
$$\sqrt{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}$$
we claim that either
$$\frac{\sigma(q^k)}{n} < \sqrt{2} < \frac{\sigma(n)}{q^k}$$
$$\frac{\sigma(n)}{q^k} < \sqrt{2} < \frac{\sigma(q^k)}{n}$$
It suffices to prove one inequality, as the proof for the other one is very similar.

To this end, assume that
$$\sqrt{2} < \frac{\sigma(q^k)}{n}.$$
This implies that
$$\frac{n}{\sigma(q^k)} < \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
which further means that
$$\sqrt{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)} < \frac{q^k}{\sigma(n)} + \frac{\sqrt{2}}{2}.$$
We obtain
$$\frac{\sqrt{2}}{2} < \frac{q^k}{\sigma(n)},$$
from which we get
$$\frac{\sigma(n)}{q^k} < \sqrt{2}.$$
It follows that
$$\frac{\sigma(n)}{q^k} < \sqrt{2} < \frac{\sigma(q^k)}{n},$$
and the proof is done.

We state this result as a theorem in this blog post.

Theorem 1.  Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$.  Then we have the unconditional result
$$\sqrt{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}$$
from which it follows that either one of the following conditions hold:
(a)  $$\frac{\sigma(q^k)}{n} < \sqrt{2} < \frac{\sigma(n)}{q^k}$$
(b)  $$\frac{\sigma(n)}{q^k} < \sqrt{2} < \frac{\sigma(q^k)}{n}$$

As before, let $N=q^k n^2$ be an odd perfect number with Euler prime $q$. Suppose that the Descartes-Frenicle-Sorli conjecture that $k=1$ is true. From pages 7 to 8 of this preprint, under the assumption $k=1$, we have the following cases to consider:

$$\text{ Case I: } \sigma(q)/n < 1 < \sqrt{2} < \sigma(n)/q \Longrightarrow q < n$$
$$\text{ Case II: } 1 < \sigma(q)/n < \sqrt{2} < \sigma(n)/q < 2 \Longrightarrow n < q < n\sqrt{2}$$
$$\text{ Case III: } \sigma(n)/q < \sqrt{2} < \sigma(q)/n < \sqrt{3}+{10}^{-375} \Longrightarrow n < q < n\sqrt{3}$$