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

**of $x$.***sum of the divisors*
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?

MathOverflow user Michael Renardy found a flaw with the reasoning above. He says:

*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.*