The following query is an offshoot of this [post 1](https://math.stackexchange.com/questions/3920469) and this [post 2](https://math.stackexchange.com/questions/4270823).

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Denote the *classical sum of divisors* of the positive integer $x$ by $\sigma(x)=\sigma\_1(x)$.

The topic of [odd perfect numbers](https://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers) likely needs no introduction.

Euler proved that a *hypothetical odd perfect number* (which is an **odd** $N$ satisfying $\sigma(N)=2N$), if one exists, 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 $\gcd(p,m)=1$.

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From [post 1](https://math.stackexchange.com/questions/3920469), it was proved by [mathlove](https://math.stackexchange.com/users/78967) that, if
$$C := \frac{p^k (m^2 - p^k)}{\sigma(m^2)}$$
then
$$C \geqslant \frac{2024}{675} = 2.99\overline{851}.$$

From [post 2](https://math.stackexchange.com/questions/4270823), we know that
$$m^2 - p^k = 2^r t,$$
where $\gcd(2,t)=1$ and $r \geq 2$.

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Here is my question:

> 
> Does $\sigma(m^2)/p^k$ divide $m^2 - p^k$, if $p^k m^2$ is an odd perfect number with special prime $p$?
> 
> 
> 

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**MY ATTEMPT**

Let us first consider the multiplicative form of
$$m^2 - p^k = 2^r t,$$
where $\gcd(2,t)=1$ and $r \geq 2$.

[FredH](https://math.stackexchange.com/users/82711) gave an unconditional proof for the assertion that $m^2 - p^k$ is **not a square** in [this answer](https://math.stackexchange.com/a/3122247/28816). Additionally, note that $m^2 - p^k$ is **not squarefree** (basically because $4 \mid (m^2 - p^k)$). It follows (from this [self-answer](https://math.stackexchange.com/a/4387163/28816)) that, since $\gcd(2,t)=1$, then $2^r$ is a square (and therefore, $r$ is even) and $t$ is squarefree. (It cannot be the other way around.)

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So we can write
$$\frac{m^2 - p^k}{2} = {t}\cdot{2^{r-1}}.$$
However, using the fact that the divisor sum $\sigma$ is [multiplicative](https://en.wikipedia.org/wiki/Multiplicative_function), we obtain
$$\sigma(p^k)\sigma(m^2)=\sigma(p^k m^2)=\sigma(N)=2N=2p^k m^2$$
from which it follows that
$$2=\frac{\sigma(m^2)}{p^k}\cdot\frac{\sigma(p^k)}{m^2}.$$
Thus, we get
$$\dfrac{m^2 - p^k}{\dfrac{\sigma(m^2)}{p^k}\cdot\dfrac{\sigma(p^k)}{m^2}} = \frac{m^2 - p^k}{2} = {t}\cdot{2^{r-1}},$$
which is equivalent to
$$2^r t = m^2 - p^k = \dfrac{m^2 - p^k}{\sigma(m^2)/{p^k}}\cdot\dfrac{m^2}{\sigma(p^k)/2}.$$

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Now, assume to the contrary that $\sigma(m^2)/p^k \mid (m^2 - p^k)$. This implies that we have the system
$$\begin{cases}
2^r = \dfrac{m^2 - p^k}{\sigma(m^2)/{p^k}} \\
t = \dfrac{m^2}{\sigma(p^k)/2} \\
\end{cases}.$$

Notice that $t$ is nothing but $H = i(p)$ in this [question](https://math.stackexchange.com/questions/4386812). Since $t = H = i(p)$ is squarefree, this implies that $\sigma(p^k)/2$ is **not squarefree**. In particular, from [this preprint](https://arxiv.org/abs/2202.08116), we have the equation $J=1$, which essentially means that
$$\gcd\Bigg(\sigma(p^k)/2,H\Bigg)=\gcd\Bigg(m,H\Bigg)=\gcd\Bigg(m^2,H\Bigg)=\frac{m^2}{\sigma(p^k)/2}=H,$$
by using the formulas computed by mathlove for $G$ and $I$. These equations are equivalent to the divisibility conditions $H \mid \sigma(p^k)/2$ and $H \mid m$, respectively. (These conditions imply that the divisibility constraint $H \mid \gcd\bigg(\sigma(p^k)/2,m\bigg)$ holds.) In particular, I know from the [preprint](https://arxiv.org/abs/2202.08116) that $J=1$ if and only if $m \mid \sigma(p^k)/2$.

**Alas, this is where I get stuck!**