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The following query is an offshoot of this post 1 and this post 2.

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.

The topic of odd perfect numbers likely needs no introduction.

Euler proved that a hypothetical odd perfect number (which is an odd $N$ satisfying $\sigma(N)=2N$), if one exists, must necessarily have the form $$N = p^k m^2$$ where $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

From post 1, it was proved by mathlove that, if $$C := \frac{p^k (m^2 - p^k)}{\sigma(m^2)}$$ then $$C \geqslant \frac{2024}{675} = 2.99\overline{851}.$$

From post 2, we know that $$m^2 - p^k = 2^r t,$$ where $\gcd(2,t)=1$ and $r \geq 2$.

Here is my question:

Does $\sigma(m^2)/p^k$ divide $m^2 - p^k$, if $p^k m^2$ is an odd perfect number with special prime $p$?

MY ATTEMPT

Let us first consider the multiplicative form of $$m^2 - p^k = 2^r t,$$ where $\gcd(2,t)=1$ and $r \geq 2$.

FredH gave an unconditional proof for the assertion that $m^2 - p^k$ is not a square in this answer. Additionally, note that $m^2 - p^k$ is not squarefree (basically because $4 \mid (m^2 - p^k)$). It follows (from this self-answer) that, since $\gcd(2,t)=1$, then $2^r$ is a square (and therefore, $r$ is even) and $t$ is squarefree. (It cannot be the other way around.)

So we can write $$\frac{m^2 - p^k}{2} = {t}\cdot{2^{r-1}}.$$ However, using the fact that the divisor sum $\sigma$ is multiplicative, we obtain $$\sigma(p^k)\sigma(m^2)=\sigma(p^k m^2)=\sigma(N)=2N=2p^k m^2$$ from which it follows that $$2=\frac{\sigma(m^2)}{p^k}\cdot\frac{\sigma(p^k)}{m^2}.$$ Thus, we get $$\dfrac{m^2 - p^k}{\dfrac{\sigma(m^2)}{p^k}\cdot\dfrac{\sigma(p^k)}{m^2}} = \frac{m^2 - p^k}{2} = {t}\cdot{2^{r-1}},$$ which is equivalent to $$2^r t = m^2 - p^k = \dfrac{m^2 - p^k}{\sigma(m^2)/{p^k}}\cdot\dfrac{m^2}{\sigma(p^k)/2}.$$

Now, assume to the contrary that $\sigma(m^2)/p^k \mid (m^2 - p^k)$. This implies that we have the system $$\begin{cases} 2^r = \dfrac{m^2 - p^k}{\sigma(m^2)/{p^k}} \\ t = \dfrac{m^2}{\sigma(p^k)/2} \\ \end{cases}.$$

Notice that $t$ is nothing but $H = i(p)$ in this question. Since $t = H = i(p)$ is squarefree, this implies that $\sigma(p^k)/2$ is not squarefree. In particular, from this preprint, we have the equation $J=1$, which essentially means that $$\gcd\Bigg(\sigma(p^k)/2,H\Bigg)=\gcd\Bigg(m,H\Bigg)=\gcd\Bigg(m^2,H\Bigg)=\frac{m^2}{\sigma(p^k)/2}=H,$$ by using the formulas computed by mathlove for $G$ and $I$. These equations are equivalent to the divisibility conditions $H \mid \sigma(p^k)/2$ and $H \mid m$, respectively. (These conditions imply that the divisibility constraint $H \mid \gcd\bigg(\sigma(p^k)/2,m\bigg)$ holds.) In particular, I know from the preprint that $J=1$ if and only if $m \mid \sigma(p^k)/2$.

Alas, this is where I get stuck!

Jose Arnaldo Bebita Dris
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2 Answers2

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Blimey!

Let $p^k m^2$ be an odd perfect number with special prime $p$.

Note that $$i(p) = \gcd(m^2,\sigma(m^2)) = \frac{\sigma(m^2)}{p^k} = \frac{m^2}{\sigma(p^k)/2}.$$

Consequently, $$\gcd\bigg(p,i(p)\bigg) = \gcd\bigg(p,\frac{m^2}{\sigma(p^k)/2}\bigg) = 1,$$ since $p$ is the special prime satisfying $\gcd(p,m)=1$.

Since $p$ is prime, and $k$ is a positive integer, it follows that $$\gcd\bigg(p^k,\sigma(m^2)/{p^k}\bigg) = 1.$$

Now, suppose to the contrary that $$i(p) = \sigma(m^2)/p^k \mid \bigg(m^2 - p^k\bigg).$$ Since $\sigma(p^k)/2$ is an integer for $p \equiv k \equiv 1 \pmod 4$, then $i(p) \mid m^2$. The condition $i(p) \mid \bigg(m^2 - p^k\bigg)$ together with $i(p) \mid m^2$ then implies that $i(p) \mid p^k$. This means that we have $$\gcd\bigg(p^k,\sigma(m^2)/{p^k}\bigg) = \gcd\bigg(p^k,i(p)\bigg)= i(p) \geq 3.$$ This contradicts $$\gcd\bigg(p^k,\sigma(m^2)/{p^k}\bigg) = 1.$$

I therefore conclude that $\sigma(m^2)/p^k$ DOES NOT DIVIDE $m^2 - p^k$.

Jose Arnaldo Bebita Dris
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This is not a direct answer to the question, but I merely wanted to point out that there is an error in your proof that $r$ is even and $t$ is squarefree.

You have $m^2 - p^k = 2^r t$ where $\gcd(2,t)=1$ and $r \geqslant 2$, and you know that $m^2 - p^k$ is neither a square nor squarefree. Then, you used the fact that if $2^rt$ is neither a square nor squarefree, then $2^rt$ can be represented as the product of a square larger than $1$ and a squarefree integer larger than $1$. The fact itself is correct. Using the fact, it seems that you thought that there are only two possibilities $(2^r,t)=(\text{square},\text{squarefree}),(\text{squarefree},\text{square})$ from which you concluded that $r$ is even and $t$ is squarefree.

You are missing the other possibilities. For example, you are missing the following cases :

  • $(2^{r-1},2t)=(\text{square},\text{squarefree})$ where $r$ is odd and $t$ is squarefree.

  • $(2^{r-2},2^2t)=(\text{squarefree},\text{square})$ where $r=3$ and $t$ is a square.

More importantly, if you want to represent $2^rt$ as $(\text{square})\times (\text{squarefree})$, then I think that you need to know how to represent $t$ as $(\text{square})\times (\text{squarefree})$ including the cases where $t$ is either a square or squarefree. I'll write all the possibilities in the following :

  • If $r$ is even and $t$ is squarefree, then $2^rt$ can be represented as $\underbrace{2^r}_{\text{square}}\times \underbrace{t}_{\text{squarefree}}$.

  • If $r$ is odd and $t$ is squarefree, then $2^rt$ can be represented as $\underbrace{2^{r-1}}_{\text{square}}\times \underbrace{2t}_{\text{squarefree}}$.

  • If $r$ is odd and $t$ is a square, then $2^rt$ can be represented as $\underbrace{2^{r-1}t}_{\text{square}}\times \underbrace{2}_{\text{squarefree}}$.

  • If $r$ is even and $t=PQ^2$ is neither squarefree nor a square where $P$ is squarefree, then $2^rt$ can be represented as $\underbrace{2^rQ^2}_{\text{square}}\times \underbrace{P}_{\text{squarefree}}$.

  • If $r$ is odd and $t=PQ^2$ is neither squarefree nor a square where $P$ is squarefree, then $2^rt$ can be represented as $\underbrace{2^{r-1}Q^2}_{\text{square}}\times \underbrace{2P}_{\text{squarefree}}$.

The possibility that we can represent $2^rt$ as $\underbrace{2^r}_{\text{square}}\times \underbrace{t}_{\text{squarefree}}$ is just one of these possibilities. How to represent $2^rt$ as $(\text{square})\times (\text{squarefree})$ depends on how to represent $t$ as $(\text{square})\times (\text{squarefree})$.

mathlove
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