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](https://arxiv.org/abs/2202.08116), 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](https://arxiv.org/abs/2202.08116) are that $b\_i < 2a\_i$ for at least one index $i$, and that $q$ is different from the $p\_i$'s.




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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
$$I := \gcd\left(n,\sigma(n^2)\right) = \gcd\left({p\_1}^{a\_1} \cdots {p\_m}^{a\_m}, q^k {p\_1}^{2a\_1 - b\_1} \cdots {p\_m}^{2a\_m - b\_m}\right)$$
$$= {p\_1}^{\min(a\_1,2a\_1 - b\_1)} \cdots {p\_m}^{\min(a\_m,2a\_m - b\_m)}.$$




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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 $H = G \times J^2$.)




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**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.



Given that $H = \frac{n^2}{\sigma(q^k)/2} = G \times J^2$, where $q^k n^2$ is an odd perfect number, then what is the value of $\gcd(G, J)$?
**This is a partial answer, which uses the ideas in my earlier comments.**




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We compute
$$\gcd(G,J)={p\_1}^{\min\left(\min(b\_1,2a\_1 - b\_1),2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1)\right)} \cdots {p\_m}^{\min\left(\min(b\_m,2a\_m - b\_m),2a\_m - b\_m - \min(a\_m,2a\_m - b\_m)\right)}$$
$$={p\_1}^{\min\left(2a\_1 - b\_1,b\_1,2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1)\right)} \cdots {p\_m}^{\min\left(2a\_m - b\_m,b\_m,2a\_m - b\_m - \min(a\_m,2a\_m - b\_m)\right)}$$
$$={p\_1}^{\min\left(b\_1,\min(2a\_1 - b\_1,2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1))\right)} \cdots {p\_m}^{\min\left(b\_m,\min(2a\_m - b\_m,2a\_m - b\_m - \min(a\_m,2a\_m - b\_m))\right)}$$
$$={p\_1}^{\min\left(b\_1,2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1)\right)} \cdots {p\_m}^{\min\left(b\_m,2a\_m - b\_m - \min(a\_m,2a\_m - b\_m)\right)}.$$


**Alas! This is the furthest I could go with this technique.**


