Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
By the definition of a perfect number $N$, we have $\sigma(N)=2N$. Since $\gcd(q,n)=1$ and because the divisor sum $\sigma$ is a multiplicative function, it follows that $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2 q^k n^2.$$
Now, just like when solving equations in terms of one variable, one should solve the equality $$\sigma(q^k)\sigma(n^2)=2 q^k n^2$$ in terms of one of the expressions.
If one solves for $n$, that introduces a square-root, and takes one out of the integers. If one solves for $q$, it introduces a $k$th root, and also takes one out of the integers. One should not solve for $\sigma(q^k)$, since that quantity can already be re-expressed in terms of $q$ and $k$ as $$\sigma(q^k)=\frac{q^{k+1} - 1}{q - 1},$$ if needed. So, the more natural quantity to solve for is $\sigma(n^2)$.
Indeed, throughout this paper, we implicitly rely on the simple equality $$\sigma(n^2) = \frac{2q^k n^2}{\sigma(q^k)}. \tag{1}$$ Unfortunately, this seems to introduce fractions. To avoid that, we can use prime factorizations, as follows. Write the prime factorization of $n$ as $$n = {p_1}^{a_1} \cdots {p_m}^{a_m},$$ for some unique odd primes $3 \leq p_1 < \ldots < p_m$, and for some positive integer exponents $a_1, \ldots, a_m$. Since both sides of $(1)$ are integers, and since $q \equiv k \equiv 1 \pmod 4$ with $q$ prime, we know that $$\sigma(q^k) = 2 {p_1}^{b_1} \cdots {p_m}^{b_m}$$ for some nonnegative integers $0 \leq b_i \leq 2a_i$. Thus, we have $$\sigma(n^2) = q^k {p_1}^{2a_1 - b_1} \cdots {p_m}^{2a_m - b_m}.$$
Note that the only other facts about odd perfect numbers that we use in this paper are that $b_i < 2a_i$ for at least one index $i$, and that $q$ is different from the $p_i$'s.
With this information, we immediately see that $$G := \gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\left(2 {p_1}^{b_1} \cdots {p_m}^{b_m},q^k {p_1}^{2a_1 - b_1} \cdots {p_m}^{2a_m - b_m}\right)$$ $$= {p_1}^{\min(b_1,2a_1 - b_1)} \cdots {p_m}^{\min(b_m,2a_m - b_m)},$$ $$H := \gcd\left(n^2,\sigma(n^2)\right) = \gcd\left({p_1}^{2a_1} \cdots {p_m}^{2a_m}, q^k {p_1}^{2a_1 - b_1} \cdots {p_m}^{2a_m - b_m}\right)$$ $$= {p_1}^{2a_1 - b_1} \cdots {p_m}^{2a_m - b_m},$$ and $$I := \gcd\left(n,\sigma(n^2)\right) = \gcd\left({p_1}^{a_1} \cdots {p_m}^{a_m}, q^k {p_1}^{2a_1 - b_1} \cdots {p_m}^{2a_m - b_m}\right)$$ $$= {p_1}^{\min(a_1,2a_1 - b_1)} \cdots {p_m}^{\min(a_m,2a_m - b_m)}.$$
Here is our:
QUESTION: What is the (simplified) prime factorization for $\gcd(G,J)$, where $$G = \gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\left(2 {p_1}^{b_1} \cdots {p_m}^{b_m},q^k {p_1}^{2a_1 - b_1} \cdots {p_m}^{2a_m - b_m}\right)$$ $$= {p_1}^{\min(b_1,2a_1 - b_1)} \cdots {p_m}^{\min(b_m,2a_m - b_m)},$$ and $$J = \frac{H}{I} = \frac{I}{G} = \frac{n}{\gcd\left(\sigma(q^k)/2,n\right)}?$$
(Note that we have the identity $H = G \times J^2$.)
OUR ATTEMPT
We realized that, since the prime factorizations for $H$ and $I$ are given as follows: $$H = {p_1}^{2a_1 - b_1} \cdots {p_m}^{2a_m - b_m}$$ and $$I = {p_1}^{\min(a_1,2a_1 - b_1)} \cdots {p_m}^{\min(a_m,2a_m - b_m)},$$ then we have $$J = \frac{H}{I} = \frac{n}{\gcd\left(\sigma(q^k)/2,n\right)} = \frac{{p_1}^{2a_1 - b_1} \cdots {p_m}^{2a_m - b_m}}{{p_1}^{\min(a_1,2a_1 - b_1)} \cdots {p_m}^{\min(a_m,2a_m - b_m)}}$$ $$= {p_1}^{2a_1 - b_1 - \min(a_1,2a_1 - b_1)} \cdots {p_m}^{2a_m - b_m - \min(a_m,2a_m - b_m)}.$$ Therefore, we have $$\gcd\left(G,J\right)=\gcd\left({p_1}^{\min(b_1,2a_1 - b_1)} \cdots {p_m}^{\min(b_m,2a_m - b_m)},{p_1}^{2a_1 - b_1 - \min(a_1,2a_1 - b_1)} \cdots {p_m}^{2a_m - b_m - \min(a_m,2a_m - b_m)}\right)$$ $$={p_1}^{\min\left(\min(b_1,2a_1 - b_1),2a_1 - b_1 - \min(a_1,2a_1 - b_1)\right)} \cdots {p_m}^{\min\left(\min(b_m,2a_m - b_m),2a_m - b_m - \min(a_m,2a_m - b_m)\right)}.$$
We guess it all boils down to evaluating $$\min\left(\min(b,2a - b),2a - b - \min(a,2a - b)\right)$$ for arbitrary integers $a$ and $b$. Alas, this is where we get stuck!
We do know that $$\min(b,2a-b) + 2a-b = 2\min(a,2a-b)$$ holds; however, this does not seem to help in this scenario.
