Let $q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Here is my:
QUESTION
If $q^k n^2$ is an odd perfect number with special prime $q$, is $\sigma(q^k)$ coprime to $\sigma(n^2)$?
The function $\sigma(x)=\sigma_1(x)$ is called the sum of divisors of $x$.
MY ATTEMPT FOR SOME SPECIFIC VALUES OF $q$, $k$, AND $n$
For example, there may be an odd perfect number with $q^k = 17$ and $(n/3)^2$ coprime to $3$. Then $\sigma(n^2) = 17(n^2/9)$ is coprime to $\sigma(17) = 18$.
How about the general case? My hunch is that the following conjecture ought to hold:
Conjecture: If $q^k n^2$ is an odd perfect number with special prime $q$, then $$\gcd(\sigma(q^k),\sigma(n^2))>1.$$
Initially, I thought that a proof of this "Conjecture" was in the following paper by Dandapat et al., but after an in-depth reading, it appears that I was mistaken.
I have therefore tagged this as a reference-request for a proof of this Conjecture.
Best Answer
This post shows that $\gcd(\sigma(q^k),\sigma(n^2))=1$ would lead to $k=1$.
Let $N=q^k n^2$ be an odd perfect number with $q$ being the special prime, $q \nmid n$ and $k \equiv 1 \pmod 4$. Then from $\sigma(N)=2N$ and multiplicativity of $\sigma$ we get $\sigma(q^k)\sigma(n^2)=2q^kn^2$. Now clearly $q^k$ is coprime with $\sigma(q^k)=1+q+\dots+q^k$, hence $q^k \mid \sigma(n^2)$. We also have $2 \mid \sigma(q^k)$ because $q,k$ are odd, thus we can write \begin{equation} \left[\frac{\sigma(q^k)}{2}\right]\cdot \left[\frac{\sigma(n^2)}{q^k}\right]=n^2. \end{equation} Now assuming $\sigma(q^k)$ and $\sigma(n^2)$ are coprime, then also the two integers on the left side are coprime. That means both are perfect squares, and so in particular $$ 2m^2=\sigma(q^k)=1+q+\dots+q^k $$ for some integer $m$. However this equation has no solution in integers for $q,k \geq 2$, so if there is an odd perfect number with this property, we must have $k=1$ (since $q \geq 5$ anyway).