Recall that we have from [Dris, Dagal (2021) - discussion of the proof of Theorem 2.1 from pp. 13 to 15] that

$$I(m) \geq \left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}$$

which is equivalent to

$$I(m^2) \leq \left(I(m)\right)^{\ln(13/9)/\ln(4/3)},$$

from which we get

$$\frac{I(m^2)}{I(m)} \leq \left(I(m)\right)^{\ln(13/12)/\ln(4/3)}.$$

This last inequality implies that

$$1 > \frac{I(m)}{I(m^2)} \geq \left(I(m)\right)^{\ln(4/3)/\ln(13/12)}$$

$$> \Bigg(\left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}\Bigg)^{\ln(4/3)/\ln(13/12)}$$

$$= \left(I(m^2)\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}.$$

But in general, we know that $I(m^2) > 8/5$, so that

$$1 > \frac{I(m)}{I(m^2)} > \left(I(m^2)\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}$$

$$> \left(\frac{8}{5}\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))} > 3$$

which is a contradiction.

This seems to prove that there are, in fact, no odd perfect numbers.

We will stop here for the time being.