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

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 1

On OP's request, I am converting my comment into an answer.

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• $p^k\lt m$ is equivalent to $$m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\tag7$$ since we have$$\begin{align}p^k\lt m&\iff m^2-2^rt\lt m \\\\&\iff m^2-m-2^rt\lt 0 \\\\&\iff m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\end{align}$$

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• $(7)$ is better than $2m\lt 2^r+t+1$ since $$\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \frac{2^r+t+1}{2}\tag8$$ holds.
  
  To see that $(8)$ holds, note that $$\begin{align}(2)&\iff \sqrt{1+2^{r+2}t}\lt 2^r+t \\\\&\iff 1+2^{r+2}t\lt 2^{r+1}+2^{r+1}t+t^2 \\\\&\iff (2^r-t)^2\gt 1 \\\\&\iff |2^r-t|\gt 1\end{align}$$ which does hold.

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• We can say that $$\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-t\bigg)\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-2^r\bigg)\lt 0\tag9$$ since $$\begin{align}(9)&\iff \bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\bigg)^2-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0 \\\\&\iff \frac{1+\sqrt{1+2^{r+2}t}+2^{r+1}t}{2}-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0 \\\\&\iff 1+\sqrt{1+2^{r+2}t}+2^{r+1}t-(1+\sqrt{1+2^{r+2}t})(t+2^r)+2^{r+1}t\lt 0 \\\\&\iff 2^{r+2}t-2^r-t+1\lt (t+2^r-1)\sqrt{1+2^{r+2}t} \\\\&\iff (2^{r+2}t-2^r-t+1)^2\lt (t+2^r-1)^2(1+2^{r+2}t) \\\\&\iff 2^{r + 2} t (2^r - t - 1) (2^r - t + 1)\gt 0 \\\\&\iff (2^r-t)^2\gt 1 \\\\&\iff |2^r-t|\gt 1\end{align}$$ which does hold.

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• It follows from $(7)(9)$ that if $p^k\lt m$ with $(m-t)(m-2^r)\lt 0$, then $$\min(t,2^r)\lt m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \max(t,2^r)$$