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The topic of odd perfect numbers likely needs no introduction.

Let $\sigma=\sigma_{1}$ denote the classical sum of divisors. Denote the abundancy index by $I(x)=\sigma(x)/x$.

An odd perfect number $N$ is said to be given in Eulerian form if $$N = p^k m^2$$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

The question is as in the title:

Is it possible to prove $I(m^2) > \zeta(2) \approx 1.644934$, if $p^k m^2$ is an odd perfect number with special prime $p$?

MY ATTEMPT

By basic considerations, since $p$ is the special prime and satisfies $p \equiv 1 \pmod 4$, then $p \geq 5$ holds, from which it follows that $$I(p^k)=\dfrac{\sigma(p^k)}{p^k}=\dfrac{p^{k+1}-1}{p^k (p-1)}<\dfrac{p^{k+1}}{p^k (p-1)}=\dfrac{p}{p-1} \leq \frac{5}{4} \iff I(m^2)=\frac{2}{I(p^k)}>\dfrac{2(p-1)}{p} \geq \frac{8}{5}.$$

Now, I was thinking of attempting to improve this trivial lower bound to $$I(m^2) > \zeta(2) \approx 1.644934.$$

But I know that $$\zeta(2) = \prod_{\rho}{\bigg({\rho}^2 + {\rho} + 1\bigg)},$$ where $\rho$ runs over all primes. (I am not really too sure though, if that is really how I should define $\zeta(2)$. Any way, I just based my definition off this answer to a closely related MSE question.)

Update (September 18, 2020 - 6:16 PM Manila time) I was wrong, the correct formula for $\zeta(2)$ should have been $$\zeta(2) = \prod_{\rho}{\dfrac{{\rho}^2}{(\rho - 1)(\rho + 1)}},$$ as correctly pointed out by mathlove.

Note that we can write $$m = \prod_{i=1}^{\omega(m)}{{\rho_i}^{\alpha_i}}$$ so that we have $$m^2 = \prod_{i=1}^{\omega(m)}{{\rho_i}^{2\alpha_i}}$$ and therefore $$\sigma(m^2) = \sigma\Bigg(\prod_{i=1}^{\omega(m)}{{\rho_i}^{2\alpha_i}}\Bigg) = \prod_{i=1}^{\omega(m)}{\sigma\bigg({\rho_i}^{2\alpha_i}\bigg)}$$ from which we get $$I(m^2) = \dfrac{\displaystyle\prod_{i=1}^{\omega(m)}{\sigma\bigg({\rho_i}^{2\alpha_i}\bigg)}}{\displaystyle\prod_{i=1}^{\omega(m)}{{\rho_i}^{2\alpha_i}}}.$$

This is where I get stuck. I currently do not see a way to force the inequality $$I(m^2) > \prod_{\rho}{\bigg({\rho}^2 + {\rho} + 1\bigg)},$$ where $\rho$ runs over all primes, from everything that I have written so far.

Jose Arnaldo Bebita Dris
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  • Since $\alpha_i \geq 1$ for each $i$, then $$I(m^2) \geq \displaystyle\prod_{i=1}^{\omega(m)}{I\bigg({\rho_i}^2\bigg)}$$ which seems closer ("for lack of a better word") to $$\displaystyle\prod_{\rho}{\Bigg({\rho}^2 + {\rho} + 1\Bigg)}.$$ – Jose Arnaldo Bebita Dris Sep 18 '20 at 07:35
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    It is wrong that $\zeta(2) = \displaystyle\prod_{\rho}{({\rho}^2 + {\rho} + 1)}$. Since we have $\zeta(s)=\displaystyle\prod_{\rho}(1-{\rho}^{-s})^{-1}$ (see here), we get $\zeta(2)=\displaystyle\prod_{\rho}\dfrac{{\rho}^2}{({\rho}-1)({\rho}+1)}$. – mathlove Sep 18 '20 at 10:00
  • Thanks for your correction, @mathlove! Notwithstanding, would it still be possible to prove the conjecture $I(m^2) > \zeta(2)$? – Jose Arnaldo Bebita Dris Sep 18 '20 at 10:23
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    By the argument you give, this question is equivalent to asking whether it is possible that $p=5$ and $k\geq 5$. I do not believe this is known (e.g. there is no known way to rule out an odd perfect number of the form $5^5\cdot m^2$ for some $m$ with $5\nmid m$) so this inequality would not be provable without a significant new idea. – Thomas Bloom Sep 18 '20 at 10:32
  • @ThomasBloom: Actually, Cohen and Sorli ruled out $5^5$ as a possible Eulerian component $p^k$ for an odd perfect number in page 4 of their paper titled On Odd Perfect Numbers and Even 3-Perfect Numbers. Does this mean that the proof for $I(m^2) > \zeta(2)$ is indeed within reach? =) – Jose Arnaldo Bebita Dris Sep 18 '20 at 10:50
  • Ah, yes. I doubt it, since one would have to rule out $5^k\cdot m^2$ (with $5\nmid m$) for all $k\geq 5$. I don't see how one would rule this out for absurdly large $k$. – Thomas Bloom Sep 18 '20 at 10:54
  • Thus, under the assumption $q=5$ and $k>1$, we have that $k \geq 9$ (since $k \equiv 1 \pmod 4$), whereupon we obtain $$I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(5^9)} \leq \frac{1953125}{1220703} \approx 1.60000016384.$$ Hence, my interest in a lower bound for $I(m^2)$ that is obtained independently of Cohen and Sorli's results. – Jose Arnaldo Bebita Dris Sep 18 '20 at 10:54

2 Answers2

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As noted, if $x=p^km^2$ is an odd perfect number with special prime $p$, so that $p\equiv k\equiv1\pmod{4}$ and $\gcd(p,m)=1$, it follows that $$I(m^2)=\frac{\sigma(m^2)}{m^2}=\frac{p^k}{\sigma(p^k)}\frac{\sigma(p^k)}{p^k}\frac{\sigma(m^2)}{m^2}=\frac{p^k}{\sigma(p^k)}\frac{\sigma(p^km^2)}{p^km^2}=\frac{p^k}{\sigma(p^k)}I(x)=2\frac{p^k}{\sigma(p^k)},$$ where of course $$\sigma(p^k)=\sum_{i=0}^kp^i=\frac{p^{k+1}-1}{p-1},$$ from which it follows that $$I(m^2)=2\frac{p^k(p-1)}{p^{k+1}-1}=2\frac{p-1}{p-\tfrac{1}{p^k}}.$$ The latter allows several simple lower bounds. For example $$2\frac{p-1}{p-\tfrac{1}{p^k}}>2\frac{p-1}{p}=2\left(1-\frac1p\right)\geq\frac85,$$ because $p\geq5$. In particular, if $p\neq5$ we see that $p\geq13$ and so $$I(m^2)>2\left(1-\frac1p\right)\geq2\left(1-\frac{1}{13}\right)=\frac{24}{13}>\zeta(2).$$ So it remains to show that the inequality holds if $p=5$. In this case we have $$I(m^2)=2\frac{p-1}{p-\tfrac{1}{p^k}}=\frac{8}{5-\tfrac{1}{5^k}}.$$ This is a strictly decreasing function of $k$, and for $k=1,5$ we find that $$I(m^2)=\frac{8}{5-\tfrac15}=\frac{5}{3}>\zeta(2),$$ $$I(m^2)=\frac{8}{5-\tfrac{1}{5^5}}=\frac{3125}{1953}<\zeta(2),$$ so your question is equivalent to asking whether there exists an odd perfect number $x$ of the form $x=5^km^2$ with $5\nmid m$ and $k\equiv1\pmod{4}$ and $k\geq5$.

Servaes
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Edited on May 14, 2023 (22:50 PM - Manila time)

Adding an answer to complement Servaes's response - These are actually some minor thoughts that would be too long to fit in the Comments section.

In what follows, I will let $N = p^k m^2$ denote an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Consider the equation $I(m^2) = \zeta(2)$. This equation does not hold, since $LHS$ is rational while $RHS$ is transcendental.

Summarizing my takeaways from Servaes's answer here:

  • Servaes concluded (on September 18, 2020 at 20:14 PM) that we have the implication $$\left((p = 5) \land (k \neq 1)\right) \implies I(m^2) < \zeta(2). \tag{1}$$
  • It remains to consider the inequality $I(m^2) < \zeta(2)$. Note that $$\frac{2(p - 1)}{p} < I(m^2) < \zeta(2)$$ implies that $$p < \frac{12}{12 - {\pi}^2} \approx 5.63275666,$$ from which we infer that $p=5$. Substituting $p=5$ into $$I(m^2) = \frac{2}{I(p^k)} = \frac{2p^k (p - 1)}{p^{k+1} - 1}$$ we get the inequality $$I(m^2) = \frac{8\cdot{5^k}}{5^{k+1} - 1} < \zeta(2).$$ Solving for $k$, we finally obtain $$k > \frac{2 \log{\pi} - \log(5{\pi}^2 - 48)}{\log 5} = 2\log_{5}(\pi)-\log_{5}(5{\pi}^2 - 48) \approx 1.23696690592779795755777497329666 > 1.$$ This means that we have the implication $$I(m^2) < \zeta(2) \implies \left((p = 5) \land (k \neq 1)\right). \tag{2}$$
  • Therefore, we can conclude (from Implications (1) and (2)) that we have the biconditional $$\left(I(m^2) < \zeta(2)\right) \iff \left((p = 5) \land (k \neq 1)\right).$$

I will stop here for the time being, as I have to rest.

Either way, note that $$I(m^2) > \zeta(2) \text{ improves on } I(m^2) > \frac{8}{5},$$ while $$I(m^2) < \zeta(2) \text{ improves on } I(m^2) < 2.$$

Jose Arnaldo Bebita Dris
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