Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
In a previous blog post, I showed that the following equation must hold:
$$GH = I^2$$
where $G, H,$ and $I$ are defined as
$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$
$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg)$$
$$I := \gcd\bigg(n,\sigma(n^2)\bigg).$$
Note that
$$H = i(q)$$
is the index of the odd perfect number $N$ at the prime $q$, and that
$$i(q) := \frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \frac{D(n^2)}{s(q^k)} = \frac{2s(n^2)}{D(q^k)}. \tag{1}$$
Note further that
$$i(q) = q\sigma(n^2) - 2(q - 1)n^2$$
so that it follows from this blog post that
$$\frac{qn^2}{i(q)} = \frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)} = \frac{q\sigma(q^k)}{2}.$$
Since $q \equiv k \equiv 1 \pmod 4$ and $q$ is prime, then
$$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4,$$
so that $\sigma(q^k)/2$ is an integer.
Hence, $H = i(q) \mid n^2 \mid N$, so that $H = \gcd\bigg(n^2,\sigma(n^2)\bigg)$ is a factor of the odd perfect number $N$.
Next, we show that $I \mid H$.
Since $a \mid b$ implies that $\gcd(a, c) \mid \gcd(b, c)$, then setting $a = n$, $b = n^2$, and $c = \sigma(n^2)$, we get that
$$n \mid n^2 \Rightarrow \gcd\bigg(n,\sigma(n^2)\bigg) = I \mid H = \gcd\bigg(n^2,\sigma(n^2)\bigg).$$
Since $n \mid n^2$ is true in general, we conclude that $I \mid H$.
Lastly, since $GH = I^2$ is true, then we obtain
$$\frac{H}{I} = \frac{I}{G}. \tag{2}$$
Because $I \mid H$, the LHS of Equation (2) is an integer. It follows that the RHS of Equation (2) is also an integer. Consequently, the divisibility constraint $G \mid I$ holds.
Hence, we have the following proposition:
THEOREM (02/02/2022): If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, and $G, H,$ and $I$ defined as follows
$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg),$$
$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg),$$
and
$$I := \gcd\bigg(n,\sigma(n^2)\bigg),$$
then we have the equation
$$G \times H = I^2$$
and the divisibility chain
$$G \mid I \mid H.$$
In particular, we have the following corollary:
COROLLARY (02/02/2022): If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, and $G, H,$ and $I$ defined as follows
$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg),$$
$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg),$$
and
$$I := \gcd\bigg(n,\sigma(n^2)\bigg),$$
then $G, H,$ and $I$ are factors of $N$.
Remark:
(1) Note that $G = H$ holds if and only if $H = I$.
(2) Suppose that $\sigma(q^k)/2$ is squarefree. We want to show that $G \neq H$. Assume to the contrary that $G = H$. The assumption that $\sigma(q^k)/2$ is squarefree implies that $\sigma(q^k)/2 \mid n$ (since $\sigma(q^k)/2 \mid n^2$ holds in general). But the condition $\sigma(q^k)/2 \mid n$ is equivalent to $n \mid \sigma(n^2)$, because of the equation
$$\frac{\sigma(n^2)}{n}=\frac{q^k n}{\sigma(q^k)/2}.$$
Hence, since $n \mid \sigma(n^2)$, then it follows that
$$G = \gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}$$
since we can compute
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{2n^2/\sigma(q^k)}=\frac{\sigma(q^k)}{2}\cdot\Bigg(\gcd\bigg(1,\frac{\sigma(n^2)}{n}\bigg)\Bigg)^2.$$
But we know from Equation (1) that
$$H = \frac{n^2}{\sigma(q^k)/2}.$$
It follows from the assumption $G = H$ that $\sigma(q^k)/2 = n$. This contradicts Steuerwald (1937) who showed that $n$ must contain a square factor.
(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. - See this MathOverflow answer for more information.)
Thus, we also have the following proposition:
THEOREM (02/02/2022): If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, $\sigma(q^k)/2$ is squarefree, and $G, H,$ and $I$ defined as follows
$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg),$$
$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg),$$
and
$$I := \gcd\bigg(n,\sigma(n^2)\bigg),$$
then we have $G < I < H$.
