2
$\begingroup$

(Preamble: This question is tangentially related to this earlier post.)

Denote the classical sum of divisors of the positive integer $x$ to be $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. Finally, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$.

The topic of odd perfect numbers likely needs no introduction.

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

Dris conjectured in (Dris (2008)) and (Dris (2012)) that the inequality $p^k < m$ always holds. Brown was the first one to prove in a preprint (Brown (2016)) that the weaker inequality $p < m$ holds in general. However, recent evidence suggests that the Dris Conjecture that $p^k < m$ may in fact be false. (If we could rule out or show that $m < p^k$ follows from the remaining two cases in the linked question, then we will have a disproof for both the Dris Conjecture and the Descartes-Frenicle-Sorli Conjecture that $k=1$.)

Here is my question:

If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m)$?

MY ATTEMPT

Since $p$ is prime, we have $$1<I(p^k)=\frac{\sigma(p^k)}{p^k}=\frac{p^{k+1}-1}{p^k (p - 1)}<\frac{p^{k+1}}{p^k (p - 1)}=\frac{p}{p-1}.$$ Now, because $p$ is a prime satisfying $p \equiv 1 \pmod 4$, we have the lower bound $p \geq 5$, whereupon we obtain the upper bound $$I(p^k)<\frac{p}{p-1} \leq \frac{5}{4}.$$ Note that, since $p^k m^2$ is perfect and $\gcd(p,m)=1$, then we have $$2=I(p^k m^2)=I(p^k)I(m^2) \iff I(m^2) = \frac{2}{I(p^k)}.$$ This implies that we have the lower bound $$I(m^2) > \frac{2}{(5/4)} = \frac{8}{5},$$ from which we finally get $$1 < I(p^k) < \frac{5}{4} < \frac{8}{5} < I(m^2) < 2,$$ since $m^2$ is a proper factor of the perfect number $p^k m^2$, and is therefore deficient.

But we know that $$I(p^k) < \dfrac{5}{4} < \sqrt{\dfrac{8}{5}}< \sqrt{I(m^2)} < I(m) < I(m^2) < 2$$

In particular, we have $$0 < 2 - I(m) < 2 - I(p^k) < 1$$ $$0 < \dfrac{D(m)}{m} < \dfrac{D(p^k)}{p^k} < 1$$ $$0 < D(m) < \frac{m}{p^k}\cdot{D(p^k)} < m.$$

We therefore have the implication $$m < p^k \implies D(m) < D(p^k).$$

Alas, this is where I get stuck.

$\endgroup$
3
  • 1
    $\begingroup$ FYI, using $I(m) < I(m^2)$, one can get $$\dfrac{D(p^k)}{2-\dfrac{2}{I(p^k)}}\le m\implies D(m)\gt D(p^k).$$ $\endgroup$
    – mathlove
    Jul 10, 2021 at 7:04
  • 1
    $\begingroup$ Similarly, using $(I(m^2))^{\frac{\ln(4/3)}{\ln(13/9)}} < I(m)$, one can get$$m\le \dfrac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^{\frac{\ln(4/3)}{\ln(13/9)}}}\implies D(m)\lt D(p^k)$$ which is better than $m < p^k \implies D(m) < D(p^k)$. $\endgroup$
    – mathlove
    Jul 11, 2021 at 7:34
  • $\begingroup$ Thank you for your time and attention, @mathlove! I would be highly interested to see a proof of your second implication. =) $\endgroup$ Jul 11, 2021 at 7:54

2 Answers 2

1
$\begingroup$

Too long to comment :

Let $c:=\dfrac{\ln(4/3)}{\ln(13/9)}$. Using $(I(m^2))^{c} < I(m)$, one can get $$m\le \dfrac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^{c}}\implies D(m)\lt D(p^k)\tag1$$ which is better than $$m < p^k \implies D(m) < D(p^k).$$

Proof :

$(I(m^2))^c\lt I(m)$ is equivalent to $$\bigg(\frac{\sigma(m^2)}{m^2}\bigg)^c\lt\frac{\sigma(m)}{m}\iff \sigma(m)\gt m\bigg(\frac{\sigma(m^2)}{m^2}\bigg)^c=m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$$ Using $\sigma(m)\gt m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$, we get

$$D(p^k)-D(m)=D(p^k)-2m+\sigma(m)\gt D(p^k)-2m+m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$$

So, we can say that if $$D(p^k)-2m+m\bigg(\dfrac{2}{I(p^k)}\bigg)^c\ge 0\tag2$$ then $D(m)\lt D(p^k)$ where note that $$(2)\iff m\le \frac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^c}$$ So, we can say that $(1)$ holds.

To see that $(1)$ is better than $m < p^k \implies D(m) < D(p^k)$, it is sufficient to prove that $$p^k\lt \frac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^{c}}\tag3$$

We have $$\begin{align}(3)&\iff p^k\lt \frac{p^{k+1}-2p^k+1}{(p-1)\bigg(2-\bigg(\dfrac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\bigg)} \\\\&\iff p^k(p-1)\bigg(2-\bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\bigg)\lt p^{k+1}-2p^k+1 \\\\&\iff 2-\bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\lt \frac{p^{k+1}-2p^k+1}{p^k(p-1)} \\\\&\iff \bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\gt 2-\frac{p^{k+1}-2p^k+1}{p^k(p-1)} \\\\&\iff \bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\gt \frac{p^{k+1}-1}{p^k(p-1)} \\\\&\iff \frac{2p^k(p-1)}{p^{k+1}-1}\gt \bigg(\frac{p^{k+1}-1}{p^k(p-1)}\bigg)^{1/c} \\\\&\iff 2\gt \bigg(\frac{p^{k+1}-1}{p^{k}(p-1)}\bigg)^{(c+1)/c} \\\\&\iff 2^{c/(c+1)}\gt \frac{p^{k+1}-1}{p^{k}(p-1)} \\\\&\iff 2^{c/(c+1)}p^k(p-1)-p^{k+1}+1\gt 0 \\\\&\iff p^k\underbrace{\bigg((2^{c/(c+1)}-1)p-2^{c/(c+1)}\bigg)}_{\text{positive}}+1\gt 0\end{align}$$ which does hold. So, $(3)$ holds.

$\endgroup$
0
$\begingroup$

Following mathlove's hint in the comments, I obtain:

Assume that $$\dfrac{D(p^k)}{2 - \dfrac{2}{I(p^k)}} \leq m.$$

This implies that $$\dfrac{D(p^k)}{2 - I(m^2)} \leq m$$ $$\implies \dfrac{D(p^k)}{\dfrac{D(m^2)}{m^2}} \leq m$$ $$\implies \dfrac{m^2 D(p^k)}{D(m^2)} \leq m$$ $$\implies mD(p^k) \leq D(m^2)$$ $$\implies D(p^k) \leq \dfrac{D(m^2)}{m}$$

But we obtain from $$I(m) < I(m^2) \iff 2 - I(m^2) < 2 - I(m) \iff \dfrac{D(m^2)}{m^2} < \dfrac{D(m)}{m} \iff \dfrac{D(m^2)}{m} < D(m).$$

We conclude that $$\dfrac{D(p^k)}{2 - \dfrac{2}{I(p^k)}} \leq m \implies D(p^k) < D(m).$$

QED

Additional Considerations

When the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds, then $$\dfrac{D(p^k)}{2 - \dfrac{2}{I(p^k)}} = \dfrac{p - 1}{2 - \dfrac{2p}{p+1}} = \dfrac{p^2 - 1}{2}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.