This is a follow-up inquiry to this MSE question.
Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.
A number $M$ is said to be perfect if $\sigma(M)=2M$. For example, $6$ and $28$ are perfect since $$\sigma(6) = 1 + 2 + 3 + 6 = 2\cdot{6}$$ and $$\sigma(28) = 1 + 2 + 4 + 7 + 14 + 28 = 2\cdot{28}.$$
It is currently unknown if there are infinitely many even perfect numbers. It is also an open problem whether any odd perfect numbers exist. It is widely believed that there are no odd perfect numbers.
Euler proved that an odd perfect number $N$, if one exists, must necessarily have the so-called Eulerian form $$N = q^k n^2$$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Here are my initial inquiries:
INITIAL QUESTION #1
If the odd perfect number $N = q^k n^2$ is of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{n},$$ can $\sigma(q^k)/2$ be a square?
ATTEMPT #1 TO ANSWER MY INITIAL QUESTION #1
Suppose that the odd perfect number $N = q^k n^2$ is of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{n}.$$ Evidently, $\sigma(q^k)/2 = n$. Assume to the contrary that $\sigma(q^k)/2$ is a square. This implies (by a result of Broughan, Kevin A.; Delbourgo, Daniel; Zhou, Qizhi, Improving the Chen and Chen result for odd perfect numbers, Integers 13, Paper A39, 8 p. (2013). ZBL1284.11009.) that $k=1$. This contradicts Patrick A. Brown's result from 2016 that $q < n$, which together with the (trivial) estimate $2n = \sigma(q^k) < 2q^k$, implies that $k > 1$.
ATTEMPT #2 TO ANSWER MY INITIAL QUESTION #1
If $N = q^k n^2$ is an odd perfect number with special prime $q$, then MSE user FredH showed unconditionally that $n^2 - q^k$ is not a square.
Suppose to the contrary that $N = q^k n^2$ is of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{n},$$ where $\sigma(q^k)/2$ is a square. From ATTEMPT #1 to INITIAL QUESTION #1, we again have that $k=1$. It follows that we have $N = q((q+1)/2)^2$ must be an odd perfect number.
We now compute $$n^2 - q^k = n^2 - q = ((q+1)/2)^2 - q = ((q-1)/2)^2,$$ which contradicts FredH's result.
We therefore conclude that $\sigma(q^k)/2$ is not a square.
INITIAL QUESTION #2
If the odd perfect number $N = q^k n^2$ is of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{n},$$ can $\sigma(q^k)/2$ be squarefree?
MY ATTEMPT TO ANSWER INITIAL QUESTION #2
As before, under the conditions of the problem being considered, the equation $\sigma(q^k)/2 = n$ must hold. Suppose to the contrary that $\sigma(q^k)/2$ is squarefree. This contradicts the fact that $n$ must contain a square factor. (Steuerwald, Rudolf, Verschärfung einer notwendigen Bedingung für die Existenz einer ungeraden vollkommenen Zahl, Sitzungsber. Bayer. Akad. Wiss., Math.-Naturwiss. Abt. 1937, 69-72 (No. 2) (1937). ZBL0018.20301.)
We therefore conclude that $\sigma(q^k)/2$ is not squarefree.
FINAL QUESTION
It follows that $\sigma(q^k)/2$ is the product of a square and a squarefree number (both greater than $1$), if $N = \frac{q^k \sigma(q^k)}{2}\cdot{n}$ is an odd perfect number with special prime $q$. Can you then obtain an expression for $\sigma(q^k)/2$, apart from the trivial equation $$\frac{\sigma(q^k)}{2} = \frac{n^2}{\sigma(n^2)/q^k} = \frac{n^2}{n} = n?$$
