Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Let $\sigma(z)$ denote the classical sum of divisors of the positive integer $z$.

Define the GCDs
$$G = \gcd(\sigma(p^k),\sigma(m^2))$$
$$H = \gcd(m,\sigma(m^2))$$
and
$$I = \gcd(m^2,\sigma(m^2)).$$

Recall that we must have
$$\frac{\sigma(p^k)}{2}\cdot\frac{\sigma(m^2)}{p^k} = m^2,$$
and that
$$I = \gcd(m^2,\sigma(m^2)) = \frac{\sigma(m^2)}{p^k}.$$

One can show that
$$G = \gcd(\sigma(p^k)/2,\sigma(m^2)/p^k) = \frac{\left(\gcd(\sigma(p^k)/2, m)\right)^2}{\sigma(p^k)/2}$$
and
$$H = \gcd(m,\sigma(m^2)/p^k) = \frac{m}{\sigma(p^k)/2}\cdot\gcd(\sigma(p^k)/2,m),$$
whence we obtain
$$G \times I = H^2.$$
(Notice further that the divisibility constraints $G \mid H$ and $H \mid I$ must hold.)

Trivially, we must have $H \mid m$ and $G \mid \sigma(p^k)/2$.

This means that there exist positive integers $x$ and $y$ such that
$$m = Hx$$
and
$$\sigma(p^k)/2 = Gy.$$

We obtain
$$m^2 = H^2 x^2$$
$$\sigma(p^k)/2 = Gy$$
$$\frac{m^2}{\sigma(p^k)/2} = I = \frac{H^2}{G}\cdot\frac{x^2}{y},$$
from which it follows that
$$y = x^2.$$

It can be shown that
$$x = \frac{\sigma(p^k)/2}{\gcd(\sigma(p^k)/2,m)}.$$

Here is my question:
> Is it possible to rule out $x > 1$?

Notice that I am trying to prove that the divisibility constraint $\sigma(p^k)/2 \mid m$ holds, which is true if and only if $x = 1$.

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**Updated (January 03, 2024)**

I am getting
$$\sigma(p^k)/2 = Gx^2$$
$$m = Hx$$
$$GJ = H$$
where
$$J = \frac{m}{\gcd(\sigma(p^k)/2,m)},$$
so that		   
$$\frac{\sigma(p^k)}{2x^2} = G$$
$$\frac{m}{x} = H$$
$$\frac{\sigma(p^k)}{2x^2}\cdot{J} = \frac{m}{x}$$
and therefore
$$J = \frac{m}{\sigma(p^k)/2}\cdot\frac{\sigma(p^k)/2}{\gcd(\sigma(p^k)/2,m)}.$$
I am not too sure whether this, indeed, forces $\sigma(p^k)/2 \mid m$. Alas, this is where I get stuck!