(Note: This question has been cross-posted to MO.)
Do any papers on odd perfect numbers approach the problem using the following equation?
$$N - (q^k + n^2) + 1 = \sigma(q^{k-1})(q-1)(n+1)(n-1)$$ where $N={q^k}{n^2}$ is an odd perfect number given in Eulerian form.
I do know that the equations $$\sigma(n^2) = {q^k}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$ and $$n^2 = {\frac{\sigma(q^k)}{2}}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$ are true.
From these two equations, we can prove that $$\gcd(n^2,\sigma(n^2)) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}.$$
Here, $\gcd(a,b)$ is the greatest common divisor of $a$ and $b$, and $\sigma(x)$ is the sum of the divisors of $x$.