(Note: This question has been cross-posted to MO.)
The topic of odd perfect numbers likely needs no introduction.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
If $n$ is odd and $\sigma(n)=2n$, then we call $n$ an odd perfect number. Euler proved that a hypothetical odd perfect number must necessarily have the form $n = p^k m^2$ where $p$ is the special prime 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 conjectured that the inequality $p^k < m$ is true in his M. Sc. thesis, and Brown (2016) eventually produced a proof for the weaker inequality $p < m$.
Now, recent evidence suggests that $p^k < m$ may in fact be false.
THE ARGUMENT
Let $n = p^k m^2$ be an odd perfect number with special prime $p$.
Since $p \equiv k \equiv 1 \pmod 4$ and $m$ is odd, then $m^2 - p^k \equiv 0 \pmod 4$. Moreover, $m^2 - p^k$ is not a square (Dris and San Diego (2020)).
This implies that we may write $$m^2 - p^k = 2^r t$$ where $2^r \neq t$, $r \geq 2$, and $\gcd(2,t)=1$.
It is trivial to prove that $m \neq 2^r$ and $m \neq t$, so that we consider the following cases:
$$\text{Case (1): } m > t > 2^r$$ $$\text{Case (2): } m > 2^r > t$$ $$\text{Case (3): } t > m > 2^r$$ $$\text{Case (4): } 2^r > m > t$$ $$\text{Case (5): } t > 2^r > m$$ $$\text{Case (6): } 2^r > t > m$$
We can easily rule out Case (5) and Case (6), as follows:
Under Case (5), we have $m < t$ and $m < 2^r$, which implies that $m^2 < 2^r t$. This gives $$5 \leq p^k = m^2 - 2^r t < 0,$$ which is a contradiction.
Under Case (6), we have $m < 2^r$ and $m < t$, which implies that $m^2 < 2^r t$. This gives $$5 \leq p^k = m^2 - 2^r t < 0,$$ which is a contradiction.
Under Case (1) and Case (2), we can prove that the inequality $m < p^k$ holds, as follows:
Under Case (1), we have: $$(m - t)(m + 2^r) > 0$$ $$p^k = m^2 - 2^r t > m(t - 2^r) = m\left|2^r - t\right|.$$
Under Case (2), we have: $$(m - 2^r)(m + t) > 0$$ $$p^k = m^2 - 2^r t > m(2^r - t) = m\left|2^r - t\right|.$$
So we are now left with Case (3) and Case (4).
Under Case (3), we have: $$(m + 2^r)(m - t) < 0$$ $$p^k = m^2 - 2^r t < m(t - 2^r) = m\left|2^r - t\right|.$$
Under Case (4), we have: $$(m - 2^r)(m + t) < 0$$ $$p^k = m^2 - 2^r t < m(2^r - t) = m\left|2^r - t\right|.$$
Note that, under Case (3) and Case (4), we actually have $$\min(2^r,t) < m < \max(2^r,t).$$
But the condition $\left|2^r - t\right|=1$ is sufficient for $p^k < m$ to hold.
Our inquiry is:
QUESTION: Is the condition $\left|2^r - t\right|=1$ also necessary for $p^k < m$ to hold, under Case (3) and Case (4)?
Note that the condition $\left|2^r - t\right|=1$ contradicts the inequality $$\min(2^r,t) < m < \max(2^r,t),$$ under the remaining Case (3) and Case (4), and the fact that $m$ is an integer.
