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**,
$$\frac{q}{2} < d \leq \frac{3q}{5} \Longrightarrow \frac{5}{3q} \leq \frac{1}{d} < \frac{2}{q} \Longrightarrow 2 - \frac{2}{q} < 2 - \frac{1}{d} = I(n^2) \leq 2 - \frac{5}{3q}.$$

Since we already have

$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1}$$

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

$$2 - \frac{2 - \frac{5}{3q}}{q} = \frac{6q^2 - 6q + 5}{3q^2} \geq \frac{5}{3}$$

(WolframAlpha computation here, and here), and

$$2 - \frac{2 - \frac{2}{q}}{q} = \frac{2q^2 - 2q + 2}{q^2} < 2.$$

(WolframAlpha computation here, and here).

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.**