(MO link: https://mathoverflow.net/questions/470088)
The following is a complete proof that $m^2 - p^k$ is not a square, if $p^k m^2$ is an odd perfect number with special prime $p$. (I apologize in advance for the somewhat lengthy post, I merely had to combine several posts together into one for ease of reference later on.)
Assume that the estimate $p < m$ holds. We want to show that the quantity $m^2 - p^k$ is not a square. Suppose to the contrary that $m^2 - p^k = s^2$. It follows that
$$(m + s)(m - s) = p^k.$$
Since $p$ is prime, we infer that we have the simultaneous equations
$$m + s = p^{k-u}$$
and
$$m - s = p^u,$$
where $u$ is an integer satisfying $0 \leq u \leq (k-1)/2$. It follows that we have the system
$$2s = p^{k-u} - p^u = p^u (p^{k-2u} - 1)$$
and
$$2m = p^{k-u} + p^u= p^u (p^{k-2u} + 1).$$
We claim that $\gcd(s,p)=1$.
Proof: Suppose otherwise. Then $\gcd(s,p) > 1$. By the definition of GCD, we have both $\gcd(s,p) \mid p$ and $\gcd(s,p) \mid s$. Hence, either $\gcd(s,p) = 1$ (which contradicts our assumption) or $\gcd(s,p) = p$, since the only possible factors of the prime $p$ are $1$ and itself. We infer that $p \mid s$. This implies that $p \mid s^2 = m^2 - p^k$, from which we conclude that $p \mid m$. But this contradicts $\gcd(p,m)=1$.
Since $p$ is the special prime, then $p \equiv 1 \pmod 4$ implies that $\gcd(2,p)=1$. Consequently, from the two simultaneous equations
$$2s = p^{k-u} - p^u = p^u (p^{k-2u} - 1)$$
and
$$2m = p^{k-u} + p^u= p^u (p^{k-2u} + 1)$$
we obtain that $u = 0$. This implies that
$$2s = p^k - 1$$
and
$$2m = p^k + 1$$
which is equivalent to
$$2s + 1 = p^k$$
and
$$2m - 1 = p^k$$
or, expressed differently, as
$$s = \frac{p^k - 1}{2}$$
and
$$m = \frac{p^k + 1}{2}.$$
Notice that the resulting equation $m = (p^k + 1)/2$ implies that $m < p^k$. Under the assumption $p < m$, then we obtain $k > 1$. Since $k \equiv 1 \pmod 4$, then we know that $k \geq 5$. We can now use a proof by anonymous MSE user FredH to show that $m^2 - p^k$ is not a square (under the assumption $p < m$), as follows:
Since $N = p^k m^2$ is (odd) perfect, then we have the defining equation
$$\sigma(N) = 2N,$$
from which it follows that
$$\sigma(p^k)\sigma(m^2) = 2p^k m^2,$$
since the divisor sum $\sigma$ is a multiplicative function.
We know that $\sigma(p^k) = (p^{k+1} - 1)/(p - 1)$. Since we have shown that $m = (p^k + 1)/2$, then we have the equation
$$2(p^{k+1} - 1)\sigma(m^2) = p^k (p - 1)(p^k + 1)^2. \hspace{0.76in} (*)$$
FredH considered the $GCD$ of $p^{k+1} - 1$ with the right-hand side of Equation $(*)$:
$$p^{k+1} - 1 = \gcd\left(p^{k+1} - 1, p^k (p - 1)(p^k + 1)^2\right) \leq (p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2$$
where FredH used the fact that $\left(p^{k+1} - 1\right) \mid RHS$ and the property that
$$\gcd(x,yz) \leq \gcd(x,y)\gcd(x,z).$$
But FredH also noticed that $p^{k+1} - 1 = p(p^k + 1) - (p + 1)$, whence FredH did also find
$$\gcd(p^{k+1} - 1, p^k + 1) = \gcd(p + 1, p^k + 1),$$
which is $p + 1$ because $k$ is odd. Thus,
$$(p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2 = (p - 1)(p + 1)^2.$$
Hence, the inequality
$$p^{k+1} - 1 \leq (p - 1)(p + 1)^2$$
holds.
Since $k \geq 5$, we obtain
$$p^5 < p^{k+1} - 1 \leq (p - 1)(p + 1)^2 < p^4,$$
which is a contradiction.
Hence, we now have the implication
$$p < m \Rightarrow m^2 - p^k \text{ is not a square.}$$
In other words, we have the contrapositive
$$m^2 - p^k \text{ is a square} \Rightarrow m < p.$$
Now, suppose to the contrary that $m^2 - p^k$ is a square. This implies that $m < p$. Since $p^k < m^2$, we then have the implication $m < p \Rightarrow k = 1$. Therefore, $k = 1$. But we know (from the considerations above) that
$$m^2 - p^k \text{ is a square } \iff m = (p^k + 1)/2.$$
Since $k = 1$, we infer that $m = (p + 1)/2$, or in other words, $p = 2m - 1$. From Acquaah and Konyagin's results, we have the unconditional estimate $p < m \sqrt{3}$. This implies that $2m - 1 = p < m \sqrt{3}$, from which we infer that
$$m(2 - \sqrt{3}) < 1$$
which contradicts the fact that $\omega(m) > 4$. (In fact, we do know that $m > {10}^{375}$, by using Ochem and Rao's lower bound $N > {10}^{1500}$ for the magnitude of an odd perfect number $N$, together with $p^k < m^2$.)
We conclude that $m^2 - p^k$ is not a square. (Note that, in the case of even perfect numbers $M = 2^{q-1}(2^q - 1)$, the quantity
$$2^{q-1} - (2^q - 1) = 1 - 2^{q-1}$$
is likewise not a square.)
---
Now, here goes the part where I am a bit unsure about its logical tightness:
Does the following statement necessarily hold? "Since $m^2 - p^k$ is not a square, then it is between two (consecutive) squares."
If so, WLOG we may assume that
$$(m - a)^2 < m^2 - p^k < (m - a + 1)^2$$
for some positive integer $a$. We may likewise assume that $m > a$.
If I am not mistaken, these assumptions will then yield a proof for the inequality
$$m < p^k < 2am$$
except for the (*problematic*) case $a=1$, where we can only derive
$$p^k < 2m - 1.$$
Either way, I think the inequalities can be summarized as
$$p^k < 2am$$
for some positive integer $a$.
Here then, is my question for this post (which, I hope, is acceptable to the MathOverflow community):
Does this "proof" for the inequality $p^k < 2am$ hold water? If not, where does the argument break?
POSTED ANSWER (April 27, 2024 - Before 18:00 PM Manila time)
Let $p^k m^2$ be an odd perfect number with special prime $p$.
This answer says something about the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers (i.e. the prediction $k=1$).
---
Assume that $p^k < 2m - 1$. We will show that $k \neq 1$.
Then
$$\frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} < \frac{2pm - p - 1}{2(p - 1)}$$
which is less than $m$ if $p > 2m - 1$. By assumption, we have
$$p \leq p^k < 2m - 1$$
which means that
$$m < \frac{\sigma(p^k)}{2}.$$
We thus obtain
$$p^k < 2m - 1 < \sigma(p^k) - 1 < \sigma(p^k)$$
which implies that $k \neq 1$ holds, under the assumption $p^k < 2m - 1$.
---
Assume that $p^k > 2m - 1$. We will show that $k \neq 1$.
Then
$$\frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} > \frac{2pm - p - 1}{2(p - 1)}$$
which is more than $m$ if $p < 2m - 1$.
Note that
$$\frac{\sigma(p^k)}{2} \geq \frac{p^k + 1}{2} > m$$
does hold, under the assumption $p^k > 2m - 1$.
By assumption, we have
$$2m - 1 < p^k$$
which means that
$$p < 2m - 1 < p^k,$$
or $k \neq 1$ holds, under the assumption $p^k > 2m - 1$.
