(**Preamble:** We have asked this same question in [MSE](https://math.stackexchange.com/questions/4269988) two weeks ago, without getting any answers. We have therefore cross-posted it to MO, hoping that it gets answered here.)


The topic of [odd perfect numbers](https://en.wikipedia.org/wiki/Perfect_number#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](http://hdl.handle.net/10453/20034) conjectured that $k=1$ always holds. Dris conjectured that the inequality $p^k < m$ is true in his [M. Sc.](https://arxiv.org/abs/1204.1450) [thesis](https://cs.uwaterloo.ca/journals/JIS/VOL15/Dris/dris8.html).


[Brown (2016)](https://arxiv.org/abs/1602.01591) eventually produced a proof for the weaker inequality $p < m$. (Pomerance informed Brown that our past proofs for the estimate $p < m$ may be (inherently) flawed. See this [MO question](https://mathoverflow.net/questions/443950) for the details of Pomerance's rebuttal.)


Now, [recent](https://mathoverflow.net/questions/376268) [evidence](https://mathoverflow.net/questions/393738) 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)](http://nntdm.net/volume-26-2020/number-4/33-38/)). Note: I am already too sleepy to add the details, but Dris and San Diego's proof in NNTDM for the statement "$m^2 - p^k$ is **not a square**" is incomplete and flawed as originally published since our proof makes use of the estimate $p < m$. (Nonetheless, it is still *possible* to prove the same statement, essentially by using Acquaah and Konyagin's result in [(IJNT, 2012)](https://www.worldscientific.com/doi/abs/10.1142/s1793042112500935). Will submit a corrigendum for [(Dris and San Diego (2020))](http://nntdm.net/volume-26-2020/number-4/33-38/) to NNTDM as soon as possible.)


This implies that we may write
$$m^2 - p^k = 2^r t$$
where $r \geq 2$ and $2^r \neq t$. (I have removed the condition $\gcd(2,t)=1$ since we still have not completely ruled out the possibility that $m^2 - p^k$ may be a power of two. We do know that under the assumption that $k=1$ holds, then $m^2 - p^k$ is not a power of two when $p$ is a Fermat prime.)


It is trivial to prove that $m \neq 2^r$ and $m \neq t$, so that we consider the following cases:


$$\text{Case (1): } m > t > 2^r$$
$$\text{Case (2): } m > 2^r > t$$
$$\text{Case (3): } t > m > 2^r$$
$$\text{Case (4): } 2^r > m > t$$
$$\text{Case (5): } t > 2^r > m$$
$$\text{Case (6): } 2^r > t > m$$


Note that these six (6) cases may be summarized as:


$$\text{Case (A): } m > \max(2^r, t)$$
Cases (1) and (2) are included under this Case (A).


$$\text{Case (B): } \max(2^r, t) > m > \min(2^r, t)$$
Cases (3) and (4) are included under this Case (B).


$$\text{Case (C): } \min(2^r, t) > m$$
Cases (5) and (6) are included under this Case (C).




---


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.



On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part II

> 
> **QUESTION:** Is the condition $\left|2^r - t\right|=1$ also necessary for $p^k < m$ to hold, under Case **(3)** and Case **(4)**?
> 
> 
> 


The answer is no.


It is true that $p^k\lt m\implies |2^r-t|\not=1$.


*Proof* :


Since you ruled out (5) and (6), and showed that, under (1) or (2), $m < p^k$ holds, one can say that it is true that $p^k\lt m\implies$ (3) or (4).


Also, since, under (3) or (4), $2^r$ and $t$ cannot be consecutive integers, one can say that $|2^r-t|\not=1$.


As a result, one can say that
$$p^k\lt m\implies (3)\ \text{or}\ (4)\implies |2^r-t|\not=1.\quad\blacksquare$$


