## 8.7.16

### A proof of the Descartes-Frenicle-Sorli conjecture on odd perfect numbers?

(Note:  This has been cross-posted from MathOverflow.)

Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.  Denote the abundancy index of the positive integer $x$ by
$$I(x) = \frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the sum of the divisors of $x$.

Suppose that the Descartes-Frenicle-Sorli conjecture on odd perfect numbers is false.  (That is, assume that $k > 1$.)

Then since we have
$$\frac{q+1}{q} = I(q) < I(q^k) < \frac{q}{q-1} < \frac{5}{4}$$
$$\frac{8}{5} < \frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} < \frac{2}{I(q)} = \frac{2q}{q+1},$$
we consider the product
$$\left(I(q^k) - \frac{q+1}{q}\right)\left(I(n^2) - \frac{q+1}{q}\right)$$
which is positive.  Therefore,
$$2 + \left(\frac{q+1}{q}\right)^2 = I(q^k)I(n^2) + \left(\frac{q+1}{q}\right)^2 > \frac{q+1}{q}\cdot\left(I(q^k) + I(n^2)\right)$$
which implies that
$$\frac{2q}{q+1} + \frac{q+1}{q} > I(q^k) + I(n^2).$$
But we have
$$I(q) + I(n^2) = I(q) + \frac{2}{I(n^2)} = \frac{q+1}{q} + \frac{2q}{q+1}.$$
We therefore have
$$I(q) + I(n^2) = \frac{q+1}{q} + \frac{2q}{q+1} > I(q^k) + I(n^2).$$
This results to the contradiction $I(q) > I(q^k)$.  Therefore, $k = 1$.

My question is:  What are the consequences of the truth of the Descartes-Frenicle-Sorli conjecture on odd perfect numbers?

Added on July 08, 2016
MathOverflow user Michael Renardy found a flaw with the reasoning above.  He says:

You seem to prove $I(n^2) < 2q/(q+1)$ (second line of inequalities), but a few lines onward you set $I(n^2) = 2q/(q+1)$.  You cannot have it both ways.