Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.


Descartes (1638), Frenicle (1657), and subsequently [[Sorli (2003) - Conjecture 3, Chapter 5 on page 89]](https://opus.lib.uts.edu.au/bitstream/10453/20034/2021/02Whole.pdf) conjectured that $k=1$ always holds for an odd perfect number $N = p^k m^2$.


It is fairly easy to [show](https://math.stackexchange.com/questions/4262319) that if $\sigma(p^k)/2$ is prime, then $k = 1$.


Moreover, [Broughan, Delbourgo, and Zhou (2013)](https://www.emis.de/journals/INTEGERS/papers/n39/n39.pdf) show that if $\sigma(p^k)/2$ is a square, then $k = 1$.


An interesting scenario holds when $\sigma(p^k)/2$ is squarefree. Indeed, this assumption implies that
$$H = \gcd(m^2,\sigma(m^2)) = \frac{m^2}{\sigma(p^k)/2} = G \times J^2$$
is **not squarefree**, where $G$ and $J$ are defined as
$$G = \gcd(\sigma(p^k),\sigma(m^2)) = \dfrac{\bigg(\gcd(\sigma(p^k)/2,m)\bigg)^2}{\sigma(p^k)/2}$$
and
$$J = \dfrac{m}{\gcd(\sigma(p^k)/2,m)}.$$
(For the case under consideration, that $\sigma(p^k)/2$ is squarefree, we do in fact have
$$G = \sigma(p^k)/2$$
and
$$J = \dfrac{m}{\sigma(p^k)/2}$$
since $\sigma(p^k)/2 \mid m^2$, and therefore $\sigma(p^k)/2 \mid m$, holds in general.)


Here is my initial question:


**FIRST INQUIRY**



> 
> Can you show that $\sigma(p^k)/2$ is squarefree likewise implies that $k=1$?
> 
> 
> 


The reason for this inquiry is because I currently [know](https://math.stackexchange.com/questions/4335854) that $k=1$ likewise implies that $H$ is **not squarefree**.


**LAST INQUIRY**



> 
> If $p^k m^2$ is an odd perfect number with special prime $p$, then under what other conditions on $\sigma(p^k)/2$ does $k=1$ follow?
> 
> 
> 



If $p^k m^2$ is an odd perfect number with special prime $p$, then under what other conditions on $\sigma(p^k)/2$ does $k=1$ follow?
This is a partial response, as it does not directly answer the original question that was asked. Additionally, what follows are actually some remarks that would be too long to fit in the *Comments* section.




---


This answer builds on the results in [this MSE question from January 03, 2023](https://math.stackexchange.com/questions/4609954), and shows that
$$H = \frac{\sigma(m^2)}{p^k}$$
is **not a square**, if $\sigma(p^k)/2$ is squarefree.




---


Throughout [this paper](https://arxiv.org/abs/2202.08116), we implicitly rely on the simple equality
$$\sigma(m^2) = \frac{2p^k m^2}{\sigma(p^k)}. \tag{1}$$
Unfortunately, this seems to introduce fractions. To avoid that, we can use prime factorizations, as follows. Write the prime factorization of $m$ as
$$m = {q\_1}^{a\_1} \cdots {q\_n}^{a\_n},$$
for some unique odd primes $3 \leq q\_1 < \ldots < q\_n$, and for some positive integer exponents $a\_1, \ldots, a\_n$. Since both sides of $(1)$ are integers, and since $p \equiv k \equiv 1 \pmod 4$ with $p$ prime, we know that
$$\sigma(p^k) = 2 {q\_1}^{b\_1} \cdots {q\_n}^{b\_n}$$
for some nonnegative integers $0 \leq b\_i \leq 2a\_i$. Thus, we have
$$\sigma(m^2) = p^k {q\_1}^{2a\_1 - b\_1} \cdots {q\_n}^{2a\_n - b\_n}.$$


Suppose that $\sigma(p^k)/2$ is squarefree. This assumption implies that
$$\sigma(p^k) = 2 {q\_1}^{b\_1} \cdots {q\_n}^{b\_n}$$
for $0 \leq b\_i \leq 1$.


Since
$$\frac{\sigma(p^k)}{2} \geq \frac{p^k + 1}{2} \geq 3,$$
then $b\_i = 1$ for at least one $i$.


This means that
$$H = \frac{\sigma(m^2)}{p^k} = {q\_1}^{2a\_1 - b\_1} \cdots {q\_n}^{2a\_n - b\_n}$$
where $2a\_i - b\_i = 2a\_i - 1 \equiv 1 \pmod 2$ for at least one $i$.


Consequently, $H$ is **not a square**, if $\sigma(p^k)/2$ is squarefree.


