This question is an offshoot of this earlier one and this other question as well.
Let $n = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m)=1$. Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.
Dris proved that $p^k < m^2$ and conjectured that $p^k < m$. The first inequality, together with Ochem and Rao's lower bound for the magnitude of an odd perfect number that $p^k m^2 = n > {10}^{1500}$, implies that $m > {10}^{375}$.
Now, following the discussion in the hyperlinked questions, we have the (Diophantine) equation
$$m^2 – p^k = 4z.$$
We obtain
$$m^2 – 1 = p^k + (4z – 1).$$
The last equation is equivalent to
$$(m+1)(m-1) = p^k + (4z – 1)$$
which implies that
$${10}^{375} – 1 < m – 1 = \frac{p^k + (4z – 1)}{m + 1}$$
from which we obtain
$$({10}^{375} – 1)(m + 1) < p^k + (4z – 1).$$
The last inequality implies that
$$m < ({10}^{375} – 1)m < p^k + [(4z – 1) – ({10}^{375} – 1)] < p^k$$
provided that
$$m^2 – p^k = 4z < {10}^{375}< m.$$
But the inequality
$$m^2 – p^k < m$$
together with the inequality
$$m < p^k$$
will imply that
$$\frac{m^2}{2} < p^k,$$
contradicting Dris and Luca's lower bound of $\sigma(m^2)/p^k > 5$.
Added in response to a comment from MSE user mathlove
Since
$$\sigma(p^k)\sigma(m^2)=\sigma(p^k m^2)=\sigma(n)=2n=2 p^k m^2,$$
$\sigma(m^2)/p^k > 5$ implies that $\sigma(p^k)/m^2 < 2/5$, from which it follows that
$$p^k < \sigma(p^k) < \frac{2m^2}{5}.$$
As already noted above, this contradicts
$$\frac{m^2}{2} < p^k.$$
Here is my question:
Does this proof argument conclusively show how to prove the Dris Conjecture that $p^k < m$? If not, how can it be mended to produce a logically sound proof?
Best Answer
This is a partial answer.
No, it doesn't.
What you've done is as follows :
(1) Suppose that $m^2 - p^k < {10}^{375}$.
(2) Then, $m^2 - p^k \lt m$.
(3) Also, $m < p^k$.
(4) Finally, $\frac{m^2}{2} < p^k$ which is a contradiction.
In short, what you've got is
"Supposing that $m^2 - p^k < {10}^{375}$ gives a contradiction."
Therefore, what you can say is $m^2 - p^k \ge {10}^{375}$, not $p^k\lt m$.