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(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.
Jose Arnaldo Bebita Dris
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  • If I'm not mistaken, I think that it is false that if $\sigma(q^k)/2$ divides $n$, then $\sigma(q^k)/2$ is squarefree. Take $q=17,k=1$ and $n=9t$ where $t$ is an odd integer such that $\gcd(17,t)=1$. – mathlove Dec 25 '21 at 14:14
  • Thank you for your time and attention, @mathlove! If $\sigma(q^k)/2 \mid n$, then since $\sigma(q^k)/2 \mid n^2$ holds in general, it follows that $\sigma(q^k)/2$ is squarefree. (I got this fact from this tangentially related MSE question and the answer contained therein.) If you recall, it was you who referred me to this question via an answer to this question. – Jose Arnaldo Bebita Dris Dec 25 '21 at 14:21
  • The answer says that if $n$ is squarefree and $n\mid a^2$ then every prime of $n$ divides $a$ and therefore, $n$ divides $a$. It does not say that if $n\mid a^2$ and $n\mid a$, then $n$ is squarefree. – mathlove Dec 25 '21 at 14:28
  • Okay, I think I got it mixed up. So I think I should suppose that $\sigma(q^k)/2$ is squarefree, to begin with. Then since the condition $\sigma(q^k)/2 \mid n^2$ holds in general, it follows that $\sigma(q^k)/2 \mid n$. This then implies that $n \mid \sigma(n^2)$. Do you concur, @mathlove? – Jose Arnaldo Bebita Dris Dec 25 '21 at 14:28
  • Additionally, per your last comment @mathlove, the condition $\sigma(q^k)/2 \mid n$ implies $n \mid \sigma(n^2)$, but these two conditions are no longer equivalent. – Jose Arnaldo Bebita Dris Dec 25 '21 at 14:36
  • I am updating the question in a while to reflect these recent edits. Thanks, @mathlove! – Jose Arnaldo Bebita Dris Dec 25 '21 at 14:37
  • If you could write out an answer confirming the correctness of the above proof, I will be more than obliged to upvote then accept it, @mathlove! =) – Jose Arnaldo Bebita Dris Dec 25 '21 at 14:51
  • I meant to say: "Additionally, per your last comment @mathlove, the condition $\sigma(q^k)/2$ is squarefree implies $n \mid \sigma(n^2)$, but these two conditions are no longer equivalent." – Jose Arnaldo Bebita Dris Dec 25 '21 at 15:32

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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.

mathlove
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