3
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Can you add some explanations about how $\frac{m^2}{2} < p^k$ contradicts $\sigma(m^2)/p^k > 5$? $\endgroup$ – mathlove Feb 23 '20 at 10:22
  • $\begingroup$ @mathlove: Sure, hold on. Doing so now. $\endgroup$ – Arnie Bebita-Dris Feb 23 '20 at 11:19
  • $\begingroup$ Thanks. I got it. $\endgroup$ – mathlove Feb 23 '20 at 11:35
1
$\begingroup$

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 $p^k\lt m$.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer, @mathlove! I appreciate it. =) $\endgroup$ – Arnie Bebita-Dris Feb 23 '20 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.