## 18.7.16

### On conditions equivalent to the Descartes-Frenicle-Sorli conjecture on odd perfect numbers

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

The Descartes-Frenicle-Sorli conjecture predicts that $k=1$ always holds.

From the equation $\sigma(N)=2N$, we get that
$$\left(q^k + \sigma(q^{k-1})\right)\sigma(n^2) = \sigma(q^k)\sigma(n^2) = 2{q^k}{n^2}$$
so that we obtain
$$\frac{\sigma(n^2)}{q}\cdot{I(q^{k-1})} = 2n^2 - \sigma(n^2).$$
(Note that $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of $n^2$, and that
$$I(x) = \frac{\sigma(x)}{x}$$
is the abundancy index of $x$.)

Now suppose that
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
Trivially, we know that
$$\frac{\sigma(n^2)}{q} \mid \sigma(n^2).$$
Thus, we have
$$\frac{\sigma(n^2)}{q} \mid \left(2n^2 - \sigma(n^2)\right) = \frac{\sigma(n^2)}{q}\cdot{I(q^{k-1})}.$$
This implies that $I(q^{k-1})$ is an integer;  in other words, $k=1$.

The other direction
$$k=1 \Longrightarrow \frac{\sigma(n^2)}{q} \mid n^2$$
is trivial.

We therefore have the following lemma.

Lemma 1.  Suppose that $N = q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$k = 1 \Longleftrightarrow \frac{\sigma(n^2)}{q} \mid n^2.$$

Now assume that $k=1$.

By Lemma 1, we have
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
This implies that there exists an integer $d$ such that
$$n^2 = d \cdot \left(\frac{\sigma(n^2)}{q}\right).$$

Note that, from the equation $\sigma(N)=2N$, we obtain (upon setting $k=1$)
$$(q+1)\sigma(n^2) = \sigma(q)\sigma(n^2) = 2q{n^2}$$
from which we get
$$d = \frac{n^2}{\frac{\sigma(n^2)}{q}} = \frac{q + 1}{2}.$$

Notice that, when $k=1$, we can derive
$$\frac{5}{3} \leq I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} < 2$$
so that we have
$$\frac{q}{2} < d = \frac{q}{I(n^2)} \leq \frac{3q}{5}.$$

We state this latest result as our second lemma.

Lemma 2.  Suppose that $N = q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$k = 1 \Longrightarrow \frac{q}{2} < d = \frac{q}{I(n^2)} \leq \frac{3q}{5}.$$

Note that, when $k=1$, we have
$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} = \frac{2q+2}{q+1} - \frac{2}{q+1} = 2 - \frac{1}{\frac{q+1}{2}} = 2 - \frac{1}{d}$$

However, notice that we know, by Lemma 2,
$$\frac{q}{2} < d \leq \frac{3q}{5} \Longrightarrow \frac{5}{3q} \leq \frac{1}{d} < \frac{2}{q} \Longrightarrow 2 - \frac{2}{q} < 2 - \frac{1}{d} = I(n^2) \leq 2 - \frac{5}{3q}.$$

$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1}$$
this WolframAlpha computation validates that, indeed,
$$\frac{2q}{q+1} \leq 2 - \frac{5}{3q}$$
implies
$$q \geq 5.$$

Iterating the computations, we obtain
$$\frac{2 - \frac{2}{q}}{q} < \frac{1}{d} = \frac{I(n^2)}{q} \leq \frac{2 - \frac{5}{3q}}{q}$$
so that
$$2 - \frac{2 - \frac{5}{3q}}{q} \leq 2 - \frac{1}{d} = I(n^2) < 2 - \frac{2 - \frac{2}{q}}{q}.$$
But
$$2 - \frac{2 - \frac{5}{3q}}{q} = \frac{6q^2 - 6q + 5}{3q^2} \geq \frac{5}{3}$$
$$2 - \frac{2 - \frac{2}{q}}{q} = \frac{2q^2 - 2q + 2}{q^2} < 2.$$

Lastly, note that, after multiplying throughout
$$\frac{6q^2 - 6q + 5}{3q^2} \leq I(n^2) < \frac{2q^2 - 2q + 2}{q^2}$$
by $I(q) = \frac{q+1}{q}$, and asking WolframAlpha to solve the resulting inequality
$$\left(\frac{q+1}{q}\right)\cdot\left(\frac{6q^2 - 6q + 5}{3q^2}\right) \leq 2 = I(q)I(n^2) = I(qn^2) < \left(\frac{q+1}{q}\right)\cdot\left(\frac{2q^2 - 2q + 2}{q^2}\right)$$
we obtain
$$q \geq 5.$$

This blog post is currently a WORK IN PROGRESS.