Is the following statement true if $N = q^k n^2$ is an odd perfect number given in Eulerian form?

Question

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 $$\gcd\left(n^2,\sigma(n^2)\right) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}?$$

Answer 1

True for both (1) and (2) : $$\begin{align}\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}&=\frac{2n^2-\frac{2q^kn^2}{\sigma(q^k)}}{\frac{q^k-1}{q-1}}=\frac{2n^2-\frac{2q^kn^2(q-1)}{q^{k+1}-1}}{\frac{q^k-1}{q-1}} \\\\&=\frac{2n^2(q-1)(q^{k+1}-1)-2q^kn^2(q-1)^2}{(q^k-1)(q^{k+1}-1)}=\frac{2n^2(q-1)(q^k-1)}{(q^k-1)(q^{k+1}-1)} \\\\&=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))\end{align}$$