On odd perfect numbers and a GCD - Part VI

(Note: This post is closely related to this earlier MSE question.)

Let $N = q^k n^2$ be an odd perfect number with special/Eulerian prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Suppose that $\sigma(q^k)/2$ is squarefree. Then since the condition "$\sigma(q^k)/2$ divides $n^2$" holds in general, we know that $\sigma(q^k)/2$ divides $n$.

The condition "$\sigma(q^k)/2$ is squarefree" is equivalent to implies the condition "$n$ divides $\sigma(n^2)$". (See this MO question and the answer contained therein for the details.)

But the condition "$n$ divides $\sigma(n^2)$" implies that $\gcd(\sigma(q^k),\sigma(n^2))=\sigma(q^k)/2$.

(Again, see the details in the MO question above.)

Furthermore, it is known that, if $n$ divides $\sigma(n^2)$, then we have $$\gcd(n,\sigma(n^2))=n.$$ In general, we know that $$\gcd(n^2,\sigma(n^2))=\frac{n^2}{\sigma(q^k)/2}$$ holds.

Note that, since it is known that $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\Bigg(\gcd(n,\sigma(n^2))\Bigg)^2}{\gcd(n^2,\sigma(n^2))},$$ then if $n$ divides $\sigma(n^2)$, we have $$\gcd(\sigma(q^k),\sigma(n^2))=\gcd(n^2,\sigma(n^2)) \iff \gcd(n,\sigma(n^2)) = \gcd(n^2,\sigma(n^2))$$ $$\iff n = \frac{n^2}{\sigma(q^k)/2} \iff \frac{\sigma(q^k)}{2} = n.$$

But this contradicts $\sigma(q^k)/2$ being squarefree and (Steuerwald, 1937) who showed that $n$ must contain a square factor. (See this MO answer for more information.)

The contradiction thus obtained implies that we have the following theorem:

THEOREM: Let $q^k n^2$ be an odd perfect number with special prime $q$. Suppose that $\sigma(q^k)/2$ is squarefree. It follows that $$\gcd(\sigma(q^k),\sigma(n^2)) \neq \gcd(n^2,\sigma(n^2)).$$

Here is my:

QUESTION: Is this proof correct?

REFERENCE

    R. Steuerwald, "Verschärfung einer notwendigen Bedingung für die Existenz einer
    ungeraden vollkommenen Zahl," S.-B. Math.-Nat. Abt. Bayer. Akad. Wiss., 1937, pp. 68-73.

I have not found any errors in your proof, so I think that your proof is correct.

I think that your proof can be simplified a bit as follows :

Suppose that $\sigma(q^k)/2$ is squarefree.

Then, we know that $\sigma(q^k)/2\mid n,n\mid \sigma(n^2)$ and $\gcd(\sigma(q^k),\sigma(n^2))=\sigma(q^k)/2$ hold.

In general, we know that $\gcd(n^2,\sigma(n^2))=\dfrac{n^2}{\sigma(q^k)/2}$ holds.

So, we have $$\begin{align}\gcd(\sigma(q^k),\sigma(n^2))=\gcd(n^2,\sigma(n^2))&\iff \sigma(q^k)/2=\frac{n^2}{\sigma(q^k)/2} \\\\&\iff n^2=\bigg(\frac{\sigma(q^k)}{2}\bigg)^2 \\\\&\iff n=\frac{\sigma(q^k)}{2}\end{align}$$ So, supposing that $\gcd(\sigma(q^k),\sigma(n^2))=\gcd(n^2,\sigma(n^2))$ implies that $n$ is squarefree, which contradicts that $n$ is not squarefree.