A number $N$ is said to be perfect when $\sigma(N)=2N$, where $\sigma$ is the classical sum-of-divisors function.
An odd perfect number $N = {q^k}{n^2}$ is said to be given in Eulerian form if $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Here is my question:
(1) If $N={q^k}{n^2}$ is an odd perfect number given in Eulerian form, is it true that $$\sigma(q^{k-1}) \mid \left(2n^2 - \sigma(n^2)\right)?$$ (2) If the answer to question (1) is YES, is it then true that $$\gcd\left(n^2,\sigma(n^2)\right) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}?$$