(The first part of this blog post is taken from this MSE question, which was asked on June 12, 2021, and the answer contained therein, which was last updated on June 20, 2021.)
QUESTION
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 rule out Case (5) and Case (6), and under Case (1) and Case (2), we can prove that the inequality $m < p^k$ holds.
So we are now left with Case (3) and Case (4):
Under both cases left under consideration, we have
$$(m - 2^r)(m - t) < 0$$
$$m^2 + 2^r t < m(2^r + t)$$
$$m^2 + (m^2 - p^k) < m(2^r + t)$$
$$2m^2 < m(2^r + t) + p^k.$$
Since we want to prove $m < p^k$, assume to the contrary that $p^k < m$. We get
$$2m^2 < m(2^r + t) + p^k < m(2^r + t) + m < m(2^r + t + 1)$$
which implies, since $m > 0$, that
$$2m < 2^r + t + 1.$$
Here then is our question:
Will it be possible to derive a contradiction from the inequality
$$2m < 2^r + t + 1,$$
under Case (3) and Case (4) above, considering that $2m$ is large? (In fact, it is known that $m > {10}^{375}$.)
ANSWER BY mathlove
On OP's request, I am converting my comment into an answer.
$p^k\lt m$ is equivalent to $$m\lt\frac{1+\sqrt{1+2^{r+2}t}}{2}\tag7$$ since we have
$$p^k\lt m \iff m^2-2^rt\lt m \iff m^2-m-2^rt\lt 0 \iff m\lt\frac{1+\sqrt{1+2^{r+2}t}}{2}$$
$(7)$ is better than $2m\lt 2^r+t+1$ since
$$\frac{1+\sqrt{1+2^{r+2}t}}{2}\lt \frac{2^r+t+1}{2}\tag8$$
holds. To see that $(8)$ holds, note that
$$(8) \iff \sqrt{1+2^{r+2}t}\lt 2^r+t \iff 1+2^{r+2}t\lt 2^{2r}+2^{r+1}t+t^2 \iff (2^r-t)^2\gt 1 \iff |2^r-t|\gt 1$$
which does hold.
We can say that $$\bigg(\frac{1+\sqrt{1+2^{r+2}t}}{2}-t\bigg)\bigg(\frac{1+\sqrt{1+2^{r+2}t}}{2}-2^r\bigg)\lt 0\tag9$$
since
$$(9) \iff \bigg(\frac{1+\sqrt{1+2^{r+2}t}}{2}\bigg)^2-\frac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0 \iff \frac{1+\sqrt{1+2^{r+2}t}+2^{r+1}t}{2}-\frac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0$$
$$\iff 1+\sqrt{1+2^{r+2}t}+2^{r+1}t-(1+\sqrt{1+2^{r+2}t})(t+2^r)+2^{r+1}t\lt 0 \iff 2^{r+2}t-2^r-t+1\lt (t+2^r-1)\sqrt{1+2^{r+2}t}$$
$$\iff (2^{r+2}t-2^r-t+1)^2\lt (t+2^r-1)^2(1+2^{r+2}t) \iff 2^{r + 2} t (2^r - t - 1) (2^r - t + 1)\gt 0 \iff (2^r-t)^2\gt 1 \iff |2^r-t|\gt 1$$
which does hold.
It follows from $(7)(9)$ that if $p^k\lt m$ with $(m-t)(m-2^r)\lt 0$, then $$\min(t,2^r)\lt m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \max(t,2^r).$$
(The second part of this blog post is taken from this MO question.)
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.
