I don't know if my question can be answered easily from your reasonings and knowledeges of the theory of odd perfect numbers. I wondered about it yesterday. Now this post is cross-posted on MathOverflow with identificator MO 362492, I add this information for all users.
Definitions and notation. We denote the sum of divisors function $\sum_{1\leq d\mid m}d$ as $\sigma(m)$ and the radical of an integer $m>1$ as $\operatorname{rad}(m)$, the product of distinct primes dividing $m$ with the definition $\operatorname{rad}(1)=1$. Both arithmetic functions are multiplicative functions. As reference for all users I add that Wikipedia has the articles for Perfect number, Euclid–Euler theorem and Radical of an integer.
For integers $m>1$ since the sum of divisors function $\sigma(m)$ is multiplicative, and we know that the fundamental theorem of arithmetic provide us the representation $m=\prod_{q\mid m}q^{e_q}$ ($q$ are the distinct primes dividing $m$ and $e_q$ the corresponding exponents in this unique factorization: $m$ can be represented as product of those prime powers $q^{e_q}$ and the representation is unique, see [2]) we get that $\sigma(m)$ has the representation $\prod_{q\mid m}\sigma(q^{e_q})$ (see [1], again see the notation of the support of this finite product in previous brackects).
The following statement is obvious invoking Euclid–Euler theorem for even perfect numbers, and I add it as motivation of the question.
Claim. If $n$ is an even perfect number, there exist two factors of the form $\sigma(q^{e_q})$ such that their product is a multiple of $\operatorname{rad}(n)$.
Here again to emphasize $\sigma(q^{e_q})$ represents those generic factors in the previous representation of $\sigma(m)$.
Proof. We conclude it, since for an even perfect number $n=2^{p-1}(2^p-1)$ we've $\omega(n)=2$, where $\omega(m)$ counts the number of distinct primes dividing an integer $m>1$ (I mean the Wikipedia Prime omega function), and $\sigma(2^{p-1})$ isn't a multiple of $\operatorname{rad}(n)$ and $\sigma(2^p-1)$ isn't a multiple of $\operatorname{rad}(n)$, thus I need the product of (at least) two factors of the form $\sigma(q^{e_q})$ from the product representation of $\sigma(n)$ to get a multiple of $\operatorname{rad}(n)$. $\square$
Notice thus that for an even perfect number the least multiple of $\operatorname{rad}(n)$ that we can build mulitpliying factors of the form $\sigma(q^{e_q})$ from the factorization of $\sigma(n)$, is in our case (as is implicit in the Proof) $\sigma(2^{p-1})\sigma(2^{q}-1)=2n$.
Question. I would like to know if it is feasible a discussion around what is the number (I mean the fewest number possible) of those factors of the form $\sigma(q^{e_q})$ that I need to build the least multiple of $\operatorname{rad}(n)$, where $n$ is an odd perfect number. Many thanks.
Here $n=\prod_{q\mid n}q^{e_q}$ is the canonical representation ([2]) of our odd perfect number $n$, compare if you need with Euler's theorem for odd perfect number, $\omega(n)$ also counts the number of all factors of the form $\sigma(q^{e_q})$. And I'm asking, how many factors of the form $\sigma(q^{e_q})$ do I need to build the least multiple of $\operatorname{rad}(n)$, on assumption that $n$ is an odd perfect number?
I don't know if my question is in the literature, in the case that it is in the literature feel free to answer my question as a reference request and I try to search and read it from the literature.
References:
[1] Tom M. Apostol, Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag (1976).
[2] The Wikipedia Fundamental theorem of arithmetic