If $N = q^k n^2$ is an odd perfect number with special prime $q$, then can $N$ be of the form $q^k \cdot (\sigma(q^k)/2) \cdot {n}$?

Let $N = q^k n^2$ be an odd perfect number with special prime $q$.

Since $$\dfrac{\sigma(n^2)}{n}=\dfrac{q^k n}{\sigma(q^k)/2}$$ is the integer $q^k$ when $\sigma(n^2)=q^k n$ and $\sigma(q^k)/2 = n$, then we obtain $$q^k = \dfrac{\sigma(n^2)}{n}=\dfrac{q^k n}{\sigma(q^k)/2} = \dfrac{\sigma(n^2) - q^k n}{n - \sigma(q^k)/2},$$ where the rightmost quantity evaluates to $0/0$.

Thus, we get $$q^k = \dfrac{0}{0}, \tag{1}$$ which contradicts the fact that $$\dfrac{0}{0}$$ is indeterminate.

Hence, we conclude that an odd perfect number $N = q^k n^2$ cannot be of the form $$N = \Bigg(\dfrac{q^k \sigma(q^k)}{2}\Bigg)\cdot{n}.$$

Note further that $$q^k < \dfrac{\sigma(n^2)}{n}=\dfrac{q^k n}{\sigma(q^k)/2} = \dfrac{\sigma(n^2) - q^k n}{n - \sigma(q^k)/2} \implies \dfrac{\sigma(n^2)}{q^k} > n > \dfrac{\sigma(q^k)}{2} \implies q^k < 2n$$ and that $$\dfrac{\sigma(n^2)}{n}=\dfrac{q^k n}{\sigma(q^k)/2} = \dfrac{\sigma(n^2) - q^k n}{n - \sigma(q^k)/2} < q^k \implies \dfrac{\sigma(n^2)}{q^k} < n < \dfrac{\sigma(q^k)}{2} \implies n < q^k.$$