This question has been cross-posted from this [MSE question](https://math.stackexchange.com/q/3902896) and is an offshoot of this [other MSE question](https://math.stackexchange.com/q/3555053).


(Note that [MSE user mathlove](https://math.stackexchange.com/u/78967) has posted an [answer in MSE](https://math.stackexchange.com/a/3903162/28816), 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)](https://arxiv.org/abs/1204.1450) and [Dris (2012)](https://cs.uwaterloo.ca/journals/JIS/VOL15/Dris/dris8.html) that the inequality $p^k < m$ holds.


[Brown (2016)](https://arxiv.org/abs/1602.01591) 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](https://math.stackexchange.com/q/3556316) and [answer](https://math.stackexchange.com/a/3557070/28816), 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)](https://www.ams.org/journals/mcom/2012-81-279/S0025-5718-2012-02563-4/S0025-5718-2012-02563-4.pdf). 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](https://math.stackexchange.com/q/3121498) and the [answer contained therein](https://math.stackexchange.com/a/3122247/28816).)


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
$$m < \Bigg(m + \lceil{\sqrt{m^2 - p^k}}\rceil\Bigg) \leq p^k - y < p^k.$$



On the nearest-square function and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number
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 $q^k > n$.


