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] conjectured that $k=1$ always holds for an odd perfect number $N = p^k m^2$.

It is fairly easy to show that if $\sigma(p^k)/2$ is prime, then $k = 1$.

Moreover, Broughan, Delbourgo, and Zhou (2013) 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 $$G = \dfrac{\bigg(\gcd(\sigma(p^k)/2,m)\bigg)^2}{\sigma(p^k)/2}$$ 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 $\sigma(p^k)/2 \mid m^2$ 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 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?

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] conjectured that $k=1$ always holds for an odd perfect number $N = p^k m^2$.

It is fairly easy to show that if $\sigma(p^k)/2$ is prime, then $k = 1$.

Moreover, Broughan, Delbourgo, and Zhou (2013) 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 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?

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] conjectured that $k=1$ always holds for an odd perfect number $N = p^k m^2$.

It is fairly easy to show that if $\sigma(p^k)/2$ is prime, then $k = 1$.

Moreover, Broughan, Delbourgo, and Zhou (2013) 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 $$G = \dfrac{\bigg(\gcd(\sigma(p^k)/2,m)\bigg)^2}{\sigma(p^k)/2}$$ and $$J = \dfrac{m}{\gcd(\sigma(p^k)/2,m)}.$$

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 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?

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] conjectured that $k=1$ always holds for an odd perfect number $N = p^k m^2$.

It is fairly easy to show that if $\sigma(p^k)/2$ is prime, then $k = 1$.

Moreover, Broughan, Delbourgo, and Zhou (2013) 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 $$G = \dfrac{\bigg(\gcd(\sigma(p^k)/2,m)\bigg)^2}{\sigma(p^k)/2}$$ 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 $\sigma(p^k)/2 \mid m^2$ 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 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?