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This question has been cross-posted from this MSE question and is an offshoot of this other MSE question.

(Note that MSE user mathlove has posted an answer in MSE, which I could not completely understand. I have therefore cross-posted this question in MO, hoping the sages here would be able to give some enlightenment. I hope this is okay, and that the question is research-level.)

Let $n = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

It was conjectured in Dris (2008) and Dris (2012) that the inequality $p^k < m$ holds.

Brown (2016) showed that the Dris Conjecture (that $p^k < m$) holds in many cases.

It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$. This means that $m^2 - p^k = 4z$, where it is known that $4z \geq {10}^{375}$. (See this MSE question and answer, where the case $m < p^k$ is considered.) Note that if $p^k < m$, then $$m^2 - p^k > m^2 - m = m(m - 1),$$ and that $${10}^{1500} < n = p^k m^2 < m^3$$ where the lower bound for the magnitude of the odd perfect number $n$ is due to Ochem and Rao (2012). This results in a larger lower bound for $m^2 - p^k$. Therefore, unconditionally, we have $$m^2 - p^k \geq {10}^{375}.$$ We now endeavor to disprove the Dris Conjecture.

Consider the following sample proof arguments:

Theorem 1 If $n = p^k m^2$ is an odd perfect number satisfying $m^2 - p^k = 8$, then $m < p^k$.

Proof

Let $p^k m^2$ be an odd perfect number satisfying $m^2 - p^k = 8$.

Then $$(m + 3)(m - 3) = m^2 - 9 = p^k - 1.$$

This implies that $(m + 3) \mid (p^k - 1)$, from which it follows that $$m < m + 3 \leq p^k - 1 < p^k.$$ We therefore conclude that $m < p^k$.

QED

Theorem 2 If $n = p^k m^2$ is an odd perfect number satisfying $m^2 - p^k = 40$, then $m < p^k$.

Proof

Let $p^k m^2$ be an odd perfect number satisfying $m^2 - p^k = 40$.

Then $$(m+7)(m-7)=m^2 - 49=p^k - 9,$$ from which it follows that $$(m+7) \mid (p^k - 9)$$ which implies that $$m < m+7 \leq p^k - 9 < p^k.$$

QED

Note that $49$ is not the nearest square to $40$ ($36$ is), but rather the nearest square larger than $40$.

With this minor adjustment in the logic, I would expect the general proof argument to work.

(Additionally, note that it is known that $m^2 - p^k$ is not a square, if $p^k m^2$ is an OPN with special prime $p$. See this MSE question and the answer contained therein.)

So now consider the equation $m^2 - p^k = 4z$. Following our proof strategy, we have:

Subtracting the smallest square that is larger than $m^2 - p^k$, we obtain

$$m^2 - \bigg(\lceil{\sqrt{m^2 - p^k}}\rceil\bigg)^2 = p^k + \Bigg(4z - \bigg(\lceil{\sqrt{m^2 - p^k}}\rceil\bigg)^2\Bigg).$$

So the only remaining question now is whether it could be proved that $$\Bigg(4z - \bigg(\lceil{\sqrt{m^2 - p^k}}\rceil\bigg)^2\Bigg) = -y < 0$$ for some positive integer $y$?

In other words, is it possible to prove that it is always the case that $$\Bigg((m^2 - p^k) - \bigg(\lceil{\sqrt{m^2 - p^k}}\rceil\bigg)^2\Bigg) < 0,$$ if $n = p^k m^2$ is an odd perfect number with special prime $p$?

If so, it would follow that $$\Bigg(m + \lceil{\sqrt{m^2 - p^k}}\rceil\Bigg)\Bigg(m - \lceil{\sqrt{m^2 - p^k} }\rceil\Bigg) = p^k - y$$ which would imply that $$\Bigg(m + \lceil{\sqrt{m^2 - p^k}}\rceil\Bigg) \mid (p^k - y)$$ from which it follows that $$m < \Bigg(m + \lceil{\sqrt{m^2 - p^k}}\rceil\Bigg) \leq p^k - y < p^k.$$

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  • $\begingroup$ "We now endeavor to disprove the Dris Conjecture" I'm not sure what you mean by this. Wouldn't a disproof by nature need to mean one had an actual odd perfect number as a counterexample? $\endgroup$ – JoshuaZ Nov 14 '20 at 12:12
  • $\begingroup$ Thank you for your comment, @JoshuaZ! Yes, of course, if you want then you could go "Assume that $p^k m^2$ is an odd perfect number with special prime $p$, satisfying $m^2 - p^k = 4z$ and $p^k < m$." You will still get the same conclusion that $m < p^k$. It is in this sense that "we endeavor(ed) to disprove the Dris Conjecture (that $p^k < m$)". $\endgroup$ – Arnie Bebita-Dris Nov 14 '20 at 20:07
  • $\begingroup$ If indeed $m-a=\sqrt{m^2 - p^k}$, then $m^2 - 2am + a^2 = m^2 - p^k$. This implies that $p^k = 2am - a^2 = a(2m - a)$, from which it follows that $0 < a < 1$. We conclude that $p^k < 2m$. $\endgroup$ – Arnie Bebita-Dris Nov 21 '20 at 15:03
  • $\begingroup$ In particular, this means that, under the problematic case, $m^2 - p^k > m^2 - 2m$, so that the smallest square larger than $m^2 - p^k$ is $m^2 - 2m + 1 = (m - 1)^2$. Does this proof suffice, @mathlove? $\endgroup$ – Arnie Bebita-Dris Nov 21 '20 at 15:19
  • $\begingroup$ I don't know why you can say that the smallest square larger than $m^2-p^k$ is $(m-1)^2$. It is possible that $m^2-2m\lt (m-1)^2\lt m^2-p^k$. $\endgroup$ – mathlove Nov 21 '20 at 15:39
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Middle of page 6 of https://arxiv.org/pdf/1312.6001v10.pdf

" we always have $0 < n−\lceil\sqrt{n^2−q^k}\rceil$ "

No, this requires that $q^k\ge 2n-1$, an helpful assumption when the goal is to prove $q^k > n$.

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You are asking if $m\lt p^k$ can be proved in the following way :

We have $$\Bigg(m + \left\lceil{\sqrt{m^2 - p^k}}\right\rceil\Bigg)\Bigg(m - \left\lceil{\sqrt{m^2 - p^k} }\right\rceil\Bigg) = p^k +4z - \left\lceil{\sqrt{m^2 - p^k}}\right\rceil^2$$ which implies $$\bigg(m + \left\lceil{\sqrt{m^2 - p^k}}\right\rceil\bigg) \mid \bigg(p^k +4z - \left\lceil{\sqrt{m^2 - p^k}}\right\rceil^2\bigg)$$ from which it follows that $$m < m + \left\lceil{\sqrt{m^2 - p^k}}\right\rceil \leq p^k +4z - \left\lceil{\sqrt{m^2 - p^k}}\right\rceil^2 < p^k.\quad\square$$

This is not correct since this does not work when $m =\left\lceil{\sqrt{m^2 - p^k} }\right\rceil$.

(If it is true that $m \not=\left\lceil{\sqrt{m^2 - p^k} }\right\rceil$, then your method works.)


In the comments, you are trying to prove $m \not=\left\lceil{\sqrt{m^2 - p^k} }\right\rceil$ in the following way :

Suppose that $m=\left\lceil{\sqrt{m^2 - p^k} }\right\rceil$. Then, there is an $a\in[0,1)$ such that $m-a=\sqrt{m^2-p^k}$. Squaring the both sides, we get $p^k=2am-a^2$ which implies $p^k\lt 2m$ to have $m^2-p^k\gt (m-1)^2-1$. So, we see that the smallest square larger than $m^2-p^k$ is $(m-1)^2$, which is a contradiction.$\quad\square$

This is not correct since it is possible that $(m-1)^2-1\lt (m-1)^2\lt m^2-p^k$.

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