This question is related to this post1 and this post2. Hereinafter, we will let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$, and $\gcd(y,z)$ will denote the greatest common divisor of the (positive) integers $y$ and $z$.
Let $m=q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
From the hyperlinked posts, we have
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))}=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\sigma(n^2)/q^k}=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{2n^2/\sigma(q^k)}=\frac{\sigma(q^k)}{2}\cdot{\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{n^2}}.$$
Since the LHS is an integer, the RHS is also an integer. But $\sigma(q^k)/2$ is an integer. Therefore,
$$\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{n^2} = \frac{\gcd\bigg(n^2,(\sigma(n^2))^2\bigg)}{n^2} = \gcd\Bigg(1, \bigg(\frac{\sigma(n^2)}{n}\bigg)^2\Bigg)$$
is an integer, which is just equal to $1$, and this further means that $n \mid \sigma(n^2)$.
Hence, we have the following proposition:
THEOREM A: If $q^k n^2$ is an odd perfect number with special prime $q$, then
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}.$$
In particular, this Theorem implies the following Corollary:
COROLLARY: If $q^k n^2$ is an odd perfect number with special prime $q$, then
$$\gcd(\sigma(q^k),\sigma(n^2)) \geq 3.$$
Here are my:
QUESTIONS: Is this argument logically sound? If it is not, how can it be mended so as to produce a valid proof?
EDIT: (September 26, 2021 – 2:16 PM Manila time) From the comments, I realized that what I actually have is the following modification to THEOREM A (note the additional divisibility condition on $\sigma(n^2)$):
THEOREM B: If $q^k n^2$ is an odd perfect number with special prime $q$, satisfying $n \mid \sigma(n^2)$, then
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}.$$
Best Answer
I don't see why you can say that $\dfrac{(\gcd(n,\sigma(n^2)))^2}{n^2}$ is an integer from $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}\cdot\frac{(\gcd(n,\sigma(n^2)))^2}{n^2}.$$ Something like $75=\dfrac{54}{2}\cdot\dfrac{5^2}{3^2}$ might happen where $\dfrac{5^2}{3^2}$ is not an integer. I think that to prove that $\dfrac{(\gcd(n,\sigma(n^2)))^2}{n^2}$ is an integer, you have to prove that $\dfrac{\sigma(q^k)}2$ divides $\gcd(\sigma(q^k),\sigma(n^2))$.
It is true that $\dfrac{\sigma(q^k)}{2}\mid \sigma(q^k)$, but I don't see why you can say that $$\sigma(q^k)\mid \gcd(\sigma(q^k),\sigma(n^2))$$
$A$ does not always divide $\gcd(A,B)$. For example, take $A=7,B=2$ for which $\gcd(A,B)=1$ holds.