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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?

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  • $\begingroup$ Can you add some explanations about how $\frac{m^2}{2} < p^k$ contradicts $\sigma(m^2)/p^k > 5$? $\endgroup$
    – mathlove
    Commented Feb 23, 2020 at 10:22
  • $\begingroup$ @mathlove: Sure, hold on. Doing so now. $\endgroup$ Commented Feb 23, 2020 at 11:19
  • $\begingroup$ Thanks. I got it. $\endgroup$
    – mathlove
    Commented Feb 23, 2020 at 11:35

1 Answer 1

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This is a partial answer.

Does this proof argument conclusively show how to prove the Dris Conjecture that $p^k < m$?

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 $p^k\lt m$.

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  • $\begingroup$ Thank you for your answer, @mathlove! I appreciate it. =) $\endgroup$ Commented Feb 23, 2020 at 11:40

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