Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n)=1$.
The recent answer to this MSE question titled "On odd perfect numbers and a GCD - Part V" appears to have successfully completed a proof for the inequation
$$\gcd(n,\sigma(n^2)) \neq \gcd(n^2,\sigma(n^2)).$$
If the proof holds water, then this shows that there cannot be an odd perfect number $N' = p^j m^2$ with special prime $p$ satisfying $p \equiv j \equiv 1 \pmod 4$ and $\gcd(p, m)=1$, of the form
$$N' = \frac{p^j \sigma(p^j)}{2}\cdot{m}.$$
You may refer to this MSE question (and the answer contained therein) for more information.
