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 $x = 1$.

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!