(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 $D(m)$?


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.

  • 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


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.


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


Additional Considerations

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


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