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.
So we are now left with Case (3) and Case (4).
Now, let us consider the quantity
$$\Delta := p^k m^2 – (m^2 – p^k) – 1 = (m^2 + 1)(p^k – 1) = (m^2 + 1)(p – 1)s(p^k).$$
Since $p \equiv 1 \pmod 4$ and $m^2 + 1 \equiv 2 \pmod 4$, then $\Delta$ is not (?) a square. Note that $\Delta \equiv 0 \pmod 4$.
This implies that we may write
$$\Delta := p^k m^2 – (m^2 – p^k) – 1 = (m^2 + 1)(p^k – 1) = (m^2 + 1)(p – 1)s(p^k) = 2^u v$$
where $u \geq 2$, and $\gcd(2,v)=1$.
We thereby obtain
$$n – 1 = 2^r t + 2^u v = 2^r t + (m^2 + 1)(p – 1)s(p^k).$$
Note that
$$s(p^k) = \sigma(p^k) – p^k \equiv (k + 1) – 1 \equiv k \equiv 1 \pmod 4.$$
We also get
$$s(p^k) = \frac{n – {2^r t} – 1}{(m^2 + 1)(p – 1)}$$
from which it follows that
$$\frac{\sigma(m^2)}{p^k} = \frac{D(m^2)}{s(p^k)} = \frac{(2m^2 – \sigma(m^2))(m^2 + 1)(p – 1)}{n – {2^r t} – 1} = \gcd(m^2, \sigma(m^2))$$
$$= p\sigma(m^2) – 2(p – 1)m^2.$$
Here is my:
QUESTION: Is my proof that "$\Delta$ is not a square" logically sound? If it is not correct, how can it be mended so as to produce a valid argument?
Best Answer
According to WolframAlpha, for $p=41$ and $k=1$, one has $\Delta=40(m^2+1)$ which is a square when $$m=\frac 12\bigg|(10−\sqrt 3)(19+6\sqrt{10})^t−(10+\sqrt 3)(19−6\sqrt{10})^t\bigg|$$ where $t$ is a non-negative integer.
Also, $$10^{375}\lt m=\frac 12\bigg|(10−\sqrt 3)(19+6\sqrt{10})^t−(10+\sqrt 3)(19−6\sqrt{10})^t\bigg|$$ can hold since one has $$\begin{align}&\lim_{t\to\infty}\dfrac 12\bigg|(\sqrt{10} - 3) (19 + 6 \sqrt{10})^t-(\sqrt{10}+3) (19 - 6 \sqrt{10})^t\bigg| \\\\&=\lim_{t\to\infty}\dfrac{(19 + 6 \sqrt{10})^t}{2}\bigg|\sqrt{10}-3-(\sqrt{10}+3)\bigg(\dfrac{19 - 6 \sqrt{10}}{19+6\sqrt{10}}\bigg)^t\bigg| \\\\&=+\infty.\end{align}$$