# On the inequality $m < p^k$ where $p^k m^2$ is an odd perfect number

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?

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

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 $$p^k\lt m$$.

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