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 (2018) 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

$$m^2 – bigg(lfloor{sqrt{m^2 – p^k} + 1}rfloorbigg)^2 = p^k + Bigg(4z – bigg(lfloor{sqrt{m^2 – p^k} + 1}rfloorbigg)^2Bigg).$$

So the only remaining question now is whether it could be proved that

$$Bigg(4z – bigg(lfloor{sqrt{m^2 – p^k} + 1}rfloorbigg)^2Bigg) = -y < 0$$

for some **positive** integer $y$?

In other words, is it possible to prove that it is always the case that

$$Bigg((m^2 – p^k) – bigg(lfloor{sqrt{m^2 – p^k} + 1}rfloorbigg)^2Bigg) < 0,$$

if $n = p^k m^2$ is an odd perfect number with special prime $p$?

If so, it would follow that

$$Bigg(m + lfloor{sqrt{m^2 – p^k} + 1}rfloorBigg)Bigg(m – lfloor{sqrt{m^2 – p^k} + 1}rfloorBigg) = p^k – y$$

which would imply that

$$Bigg(m + lfloor{sqrt{m^2 – p^k} + 1}rfloorBigg) mid (p^k – y)$$

from which it follows that

$$m < Bigg(m + lfloor{sqrt{m^2 – p^k} + 1}rfloorBigg) leq p^k – y < p^k.$$