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

Question

(Preamble: This post is an offshoot of this earlier MSE 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. (For the details of the proof, we refer the interested reader to this preprint.)

So we are now left with Case (3) and Case (4).

Under Case (3), we have: $$(m + 2^r)(m - t) < 0$$ $$m^2 - 2^r t < m(t - 2^r)$$ $$p^k < m(t - 2^r),$$ whence we do not obtain a contradiction, since $t - 2^r$ is positive under this case.

Under Case (4), we have: $$(m - 2^r)(m + t) < 0$$ $$m^2 - 2^r t < m(2^r - t)$$ $$p^k < m(2^r - t),$$ whence we do not obtain a contradiction, since $2^r - t$ is positive under this case.

Here then is our question:

Will it be possible to improve on the bounds $$p^k < m(t - 2^r)$$ under Case (3), and $$p^k < m(2^r - t)$$ under Case (4)? If YES, can you show how? If NOT, can you explain why?

PostScript: I think that both bounds may be combined as $$p^k < m\left|{2^r - t}\right|$$ but I may be wrong.

Answer 1

Too long to comment: This is not an answer to the original question, just some more thoughts that recently occurred to me, as I was pondering more on this problem.

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We will prove the following here: Under Case (1) and Case (2), we have $$p^k > m\left|{2^r - t}\right|.$$

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Under Case (1): $m > t > 2^r$

$$(m - t)(m + 2^r) > 0$$ $$p^k = m^2 - 2^r t > m(t - 2^r) = m\left|{2^r - t}\right|$$

Under Case (2): $m > 2^r > t$

$$(m - 2^r)(m + t) > 0$$ $$p^k = m^2 - 2^r t > m(2^r - t) = m\left|{2^r - t}\right|$$

QED

Answer 2

Too long to comment: This is not really an answer, just some more thoughts that recently occurred to me, while I was pondering more on this problem.

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As shown in the original question, we have $$p^k < m\left|{2^r - t}\right|$$ under both Case (3) and Case (4).

But we have the equation $$m^2 - 2^r t = p^k.$$

This implies that we have the inequality $$m^2 - 2^r t < m\left|{2^r - t}\right|$$ $$m^2 - m\left|{2^r - t}\right| - 2^r t < 0$$

This is a quadratic inequality, which can be solved as $$m_1 < m < m_2$$ where $$m_1 = \frac{\left|{2^r - t}\right| - \sqrt{\bigg(\left|{2^r - t}\right|\bigg)^2 + 4\cdot(2^r t)}}{2} = \frac{\left|{2^r - t}\right| - \left|{2^r + t}\right|}{2}$$ and $$m_2 = \frac{\left|{2^r - t}\right| + \sqrt{\bigg(\left|{2^r - t}\right|\bigg)^2 + 4\cdot(2^r t)}}{2} = \frac{\left|{2^r - t}\right| + \left|{2^r + t}\right|}{2}.$$

Since $2^r t$ is positive, and $2^r \geq 4$, then $t$ is likewise positive.

This means that $m_2$ is nothing but $$m_2 = \frac{(2^r + t) + \left|{2^r - t}\right|}{2} = \max(2^r,t)$$ so that $$m < m_2 = \max(2^r,t),$$ which does not improve on the known upper bounds for $m$ under Case (3) and Case (4).

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Now, let us consider the values of $$m_1 = \frac{\left|{2^r - t}\right| - \left|{2^r + t}\right|}{2} = \frac{\left|{2^r - t}\right| - (2^r + t)}{2}$$ under Case (3) and Case (4).

Case (3): $t > m > 2^r$ $$m_1 = \frac{\left|{2^r - t}\right| - (2^r + t)}{2} = \frac{(t - 2^r) - (2^r + t)}{2} = -2^r,$$ implying that $m > -2^r$, which is trivial.

Case (4): $2^r > m > t$ $$m_1 = \frac{\left|{2^r - t}\right| - (2^r + t)}{2} = \frac{(2^r - t) - (2^r + t)}{2} = -t,$$ implying that $m > -t$, which is trivial.