This question is an offshoot of this MSE post.
Let $\sigma$ be the classical sum-of-divisors function. An odd perfect number $N$ is said to be given in Eulerian form if $\sigma(N)=2N$ and $N={q^k}{n^2}$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Descartes, Frenicle, and subsequently Sorli conjectured that $k = 1$ always holds.
If the DFS conjecture on odd perfect numbers is true, then we have: $$q = \frac{\sigma(n^2)}{2n^2 - \sigma(n^2)} = -1 + 2\cdot\frac{n^2}{2n^2 - \sigma(n^2)}$$ $$\frac{q+1}{2} = \frac{n^2}{2n^2 - \sigma(n^2)}.$$
It follows that $$\sigma(n^2) = q\cdot(2n^2 - \sigma(n^2))$$ and $$n^2 = \left(\frac{q+1}{2}\right)\cdot(2n^2 - \sigma(n^2)).$$
Does it now follow that $$\gcd(n^2, \sigma(n^2)) = 2n^2 - \sigma(n^2)?$$