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(Preamble: We have asked this same question in MSE 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 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.

Brown (2016) 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 for the details of Pomerance's rebuttal.)

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)). 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). Will submit a corrigendum for (Dris and San Diego (2020)) 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.

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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 $$p^k\lt m\implies (3)\ \text{or}\ (4)\implies |2^r-t|\not=1.\quad\blacksquare$$

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You wrote

$$\implies (a + 1)p^k + b(a + 1) = p^k + b + ap^k + ab = p^k + c$$Comparing coefficients, we obtain$$a + 1 = 1 \text{ and } b(a + 1) = c$$

I don't think that this is correct since it is not true that

$$(a + 1)p^k + b(a + 1) = p^k + c\implies a+1=1$$

For example, if $$a=2,\quad b=2,\quad c=16,\quad p=5,\quad k=1$$ then $$(a + 1)p^k + b(a + 1) = p^k + c$$ holds.

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  • $\begingroup$ Do you think that this rules out any proof for the implication $$p^k < m \implies \left|2^r - t\right| = 1,$$ @mathlove? $\endgroup$ Commented Oct 30, 2021 at 6:32
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    $\begingroup$ @Jose Arnaldo Bebita Dris : No, I don't. Someone might be able to prove in another way that $p^k\lt m\implies |2^r-t|=1$. $\endgroup$
    – mathlove
    Commented Oct 30, 2021 at 6:46
  • $\begingroup$ Okay, thank you for your feedback, @mathlove! =) $\endgroup$ Commented Oct 30, 2021 at 6:54
  • $\begingroup$ We have decided to award the 100 bounty to your answer, @mathlove, as it helped to clarify some things for us. $\endgroup$ Commented Oct 30, 2021 at 10:39

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