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(Note: I asked this question in MSE this June 2018 but did not receive any responses there. I have therefore cross-posted it here, hoping that it gets answered.)

Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $z$ by $D(z):=2z-\sigma(z)$, and the sum of the aliquot divisors of $z$ by $s(z):=\sigma(z)-z$.

If $n$ is odd and $\sigma(n)=2n$, then $n$ is said to be an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the form $n = p^k m^2$, where $p$ is the special / Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Starting from the fundamental equality $$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)}$$ one can derive $$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)} = \gcd(m^2, \sigma(m^2))$$ so that we ultimately have $$\frac{D(m^2)}{s(p^k)} = \frac{2m^2 - \sigma(m^2)}{\sigma(p^k) - p^k} = \gcd(m^2, \sigma(m^2))$$ and $$\frac{s(m^2)}{D(p^k)/2} = \frac{\sigma(m^2) - m^2}{p^k - \frac{\sigma(p^k)}{2}} = \gcd(m^2, \sigma(m^2)),$$ whereby we obtain $$\frac{D(p^k)D(m^2)}{s(p^k)s(m^2)} = 2.$$ Note that we also have (Equation A) $$\frac{2D(m^2)s(m^2)}{D(p^k)s(p^k)} = \bigg(\gcd(m^2, \sigma(m^2))\bigg)^2.$$ Lastly, notice that we can easily get $$\sigma(p^k) \equiv k + 1 \equiv 2 \pmod 4$$ so that it remains to consider the possible equivalence classes for $\sigma(m^2)$ modulo $4$. Since $\sigma(m^2)$ is odd, we only need to consider two.

Here is my question:

Which equivalence class of $\sigma(m^2)$ modulo $4$ makes Equation A untenable?

I know that the answer must somehow depend on the equivalence class of $p$ and $k$ modulo $8$, but as I only know that $p \equiv k \equiv 1 \pmod 4$, I am stuck.

UPDATED September 19 2018 (Manila time) After considering various cases, I think I am able to prove that:

  1. If $p \equiv k \equiv 1 \pmod 8$, then $\sigma(m^2) \equiv 3 \pmod 4$ is impossible.
  2. If $p \equiv 1 \pmod 8$ and $k \equiv 5 \pmod 8$, then $\sigma(m^2) \equiv 1 \pmod 4$ is impossible.
  3. If $p \equiv 5 \pmod 8$ and $k \equiv 1 \pmod 8$, then $\sigma(m^2) \equiv 1 \pmod 4$ is impossible.
  4. If $p \equiv k \equiv 5 \pmod 8$, then $\sigma(m^2) \equiv 3 \pmod 4$ is impossible.
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  • $\begingroup$ That $p = 5$ and $k = 5$ is impossible is proved in page $4$ of the article titled "ON ODD PERFECT NUMBERS AND EVEN 3-PERFECT NUMBERS", by Cohen and Sorli. $\endgroup$ Commented Sep 19, 2018 at 9:20
  • $\begingroup$ The conjunction $$(p = 5) \land (k \neq 1)$$ holds unconditionally. (See this answer.) $\endgroup$ Commented May 14, 2023 at 11:13

4 Answers 4

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It is rare that complicating an expression leads to an insight. There are, of course, important counter-examples to this principle. But generally speaking, one seeks to simplify an expression, rather that add extra complications. Let me explain how I see some unnecessary complications in your post.

Let $n$ be an odd perfect number. As you mention, this means that $\sigma(n)=2n$.

Using the fact that $\sigma$ is a multiplicative function, with a very nice multiplicative definition, we can investigate how this equation behaves on prime factorizations. Since it involves the prime $2$ quite prominently, it is natural to look at this equality $2$-adically. Doing so, we discover (as Euler did) that $n$ must be of the form $n=p^k m^2$, where $p\equiv k\equiv 1\pmod{4}$ and $p$ is a prime, and $p\nmid m$.

Thus, we can remove all reference to the variable $n$, and we have $$ \sigma(p^k)\sigma(m^2)=2p^km^2. $$ We also have information about how the factors on the right side divide the factors on the left side. We have $2\mid \sigma(p^k)$, while $p^k\mid \sigma(m^2)$, and finally $m^2$ can have some of factors go into both parts.

By rewriting this equality as $$ \frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)}, $$ and asserting that both sides are odd integers, we have exactly the same information, but now it involves fractions (an unnecessary complication).

Rather than introducing fractions, it would be much cleaner to simply write $\sigma(p^k)=2t_1$, $\sigma(m^2)=p^k t_2$, and note that $m^2=t_1 t_2$. By giving names to the cofactors, you can avoid using fractions. Notice that with these identifications, your "fundamental" equality is just asserting that $t_2=t_2$.

Now, since $p\nmid m$, then $$ \gcd(\sigma(m^2),m^2)=\gcd(p^k t_2,m^2)=\gcd(t_2,m^2)=\gcd(t_2,t_1t_2)=t_2. $$ What is fundamentally happening here is we are using the fact that $N=p^k m^2$ is a partial prime factorization. The introduction of a $\gcd$ in your post seems to be an unnecessary complication. It seems to simply be a stand-in for the quantity $t_2$.

At this point, the fractions in your post get even more complicated, via the introduction of the functions $D$ and $s$. These expressions are just rewriting the given quantities as alternative expressions. This added complexity adds nothing new, since you never use (or need) any properties $D$ and $s$ except their definitions (which lets you get rid of them, in terms of more fundamental quantities).

In particular, after simplifying both sides of Equation A separately, we should get $t_2^2=t_2^2$. So, that equation, in and of itself, is a tautology in terms of the previous information. It gives no new restrictions on the congruence classes of $\sigma(m^2)$. Any restrictions on those congruence classes must come from the previous information.

This might explain why a corrigendum was necessary. It is easy to make mistakes when a situation is over-complicated by irrelevant conditions.

Moreover, as useful test cases you might consider spoof OPNs. If you only want to focus on congruence conditions that apply to minor modifications of $\sigma$, then any result you prove should equally apply to spoofs. For example, the theorem by Chen and Luo (you cited) seems to apply equally well to spoofs. In particular, if you believe that you have shown that the special prime must be $5$ (which would be an enormous advance in odd perfect numbers), you should do what's called a "sanity check", where you test your theorem again one of the spoof OPNs, to see whether or not you made an error (and also to see if you really do use more than mere congruence considerations).

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  • $\begingroup$ Professor @PaceNielsen, kindly check my response to your comment below. $\endgroup$ Commented May 17, 2023 at 0:46
  • $\begingroup$ That $p=5$ and $k=5$ is impossible was essentially proved by Cohen and Sorli (2012). (See their Theorem 3 on page 3, and the discussion that follows from pages 4 to 5, particularly the 5th paragraph on page 5, which contains a list of impossible Eulerian components for an odd perfect number. In particular, $p^k = 5^5$ is in that list.) @PaceNielsen $\endgroup$ Commented May 17, 2023 at 1:07
  • $\begingroup$ Professor @PaceNielsen, given your answer from $7$ hours ago today (May 17, 2023 - 09:16 AM - Manila time), can you categorically state whether ALL results on spoofs (based on the computations you carried out for this paper) carry over to an "actual" OPN, assuming such a number exists? I was, of course, under the (previous) impression (on page 26, Section 6) that it was the other way around. Meaning, theoretical results... $\endgroup$ Commented May 17, 2023 at 1:36
  • $\begingroup$ (continued) on odd perfect numbers should be the ones that carry over to spoofs, correct? You seem to be implying that it goes both ways, Professor @PaceNielsen. $\endgroup$ Commented May 17, 2023 at 1:37
  • $\begingroup$ @JoseArnaldoBebitaDris You are correct. I misread your conditions modulo 16 (I misread the change between $p+k$ and $p-k$). Also, you are correct that $5^5$ cannot be a component in an OPN (because it leads to deficiency). I was misremembering the special prime in the spoof. I've changed my answer to fix these issues. $\endgroup$ Commented May 17, 2023 at 4:31
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The following assertion appears in Theorem 3.3 (page 7, equations (5) to (6)) of Odd multiperfect numbers by Shi-Chao Chen and Hao Luo:

Let $n=\pi^{\alpha} M^2$ be an odd $2$-perfect number, with $\pi$ prime, $\gcd(\pi,M)=1$, and $\pi \equiv \alpha \equiv 1 \pmod 4$. Then \begin{align*} \sigma(M^2) \equiv 1 \pmod 4 &\iff \pi \equiv \alpha &\pmod 8, \\ \sigma(M^2) \equiv 3 \pmod 4 &\iff \pi \equiv \alpha + 4 &\pmod 8. \\ \end{align*}

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Further to this recent corrigendum in NNTDM, which corrects an oversight in Some modular considerations regarding odd perfect numbers – Part II, we realized that we do in fact have the following biconditionals:

In this article, we consider the various possibilities for $p$ and $k$ modulo $16$, and show conditions under which the respective congruence classes for $\sigma(m^2)$ (modulo $8$) are attained, if $p^k m^2$ is an odd perfect number with special prime $p$. We can prove that:

  • $\sigma(m^2) \equiv 1 \pmod 8$ holds if and only if $p + k \equiv 2 \pmod {16}$
  • $\sigma(m^2) \equiv 3 \pmod 8$ holds if and only if $p - k \equiv {12} \pmod {16}$
  • $\sigma(m^2) \equiv 5 \pmod 8$ holds if and only if $p + k \equiv {10} \pmod {16}$
  • $\sigma(m^2) \equiv 7 \pmod 8$ holds if and only if $p - k \equiv 4 \pmod {16}$.

If $p=5$, then these logical equivalences take the form

  • $\sigma(m^2) \equiv 1 \pmod 8$ holds if and only if $k \equiv {13} \pmod {16}$
  • $\sigma(m^2) \equiv 3 \pmod 8$ holds if and only if $k \equiv 9 \pmod {16}$
  • $\sigma(m^2) \equiv 5 \pmod 8$ holds if and only if $k \equiv 5 \pmod {16}$
  • $\sigma(m^2) \equiv 7 \pmod 8$ holds if and only if $k \equiv 1 \pmod {16}$.

If $k=1$, then these logical equivalences take the form

  • $\sigma(m^2) \equiv 1 \pmod 8$ holds if and only if $p \equiv 1 \pmod {16}$
  • $\sigma(m^2) \equiv 3 \pmod 8$ holds if and only if $p \equiv 13 \pmod {16}$
  • $\sigma(m^2) \equiv 5 \pmod 8$ holds if and only if $p \equiv 9 \pmod {16}$
  • $\sigma(m^2) \equiv 7 \pmod 8$ holds if and only if $p \equiv 5 \pmod {16}$.

In particular, we obtain the following assertion:

PROPOSITION. Let $p^k m^2$ be an odd perfect number with special prime $p$. If the conjunction $p=5$ and $k=1$ is true, then the congruence $\sigma(m^2) \equiv 7 \pmod 8$ holds.

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  • $\begingroup$ Thank you for your time and attention, Professor @PaceNielsen. I suggest you reconsider your (logical) implications, too. We can have $p + k \equiv 2 \pmod {16}$ when $k = 1$, which happens under the assumption that $\sigma(p^k)/2$ is a square. (In case you wonder, we used Chen and Luo, and then "lifted" the argument from modulo $8$ to $16$.) $\endgroup$ Commented May 17, 2023 at 0:40
  • $\begingroup$ As a "sanity check" (per a suggestion by @PaceNielsen in his answer from around $22$ hours ago (before May 18, 2023)), if $p=5$ and $k=1$, then we obtain $$\sigma_{-1}(m^2) = I(m^2) = \sigma(m^2)/m^2 = \frac{2}{I(p^k)} = \frac{2}{\sigma_{-1}(p^k)} = \frac{2p}{p+1} = \frac{2\cdot{5}}{5+1} = \frac{2\cdot{5}}{2\cdot{3}} = \frac{5}{3},$$ whereupon we obtain $3\sigma(m^2)=5m^2$. (Note that $m^2$ is odd, hence $m^2 \equiv 1 \pmod 8$, therefore $5m^2 \equiv 5 \pmod 8$. Lastly, notice that, modulo $8$, we have $3^{-1} \equiv 3 \pmod 8$.) ... $\endgroup$ Commented May 18, 2023 at 3:48
  • $\begingroup$ Ergo? We conclude that $\sigma(m^2) \equiv 15m^2 \equiv 7 \pmod 8$, under the given assumption. $\endgroup$ Commented May 18, 2023 at 3:51
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This answer is a direct response to Professor Pace Nielsen's suggestion to perform a "sanity check" and complements this other answer by providing a proof of the following biconditionals:

If $p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then the following equivalences hold:

  • $\sigma(m^2) \equiv 1 \pmod 8 \iff p + k \equiv 2 \pmod {16}$
  • $\sigma(m^2) \equiv 3 \pmod 8 \iff p - k \equiv 12 \pmod {16}$
  • $\sigma(m^2) \equiv 5 \pmod 8 \iff p + k \equiv 10 \pmod {16}$
  • $\sigma(m^2) \equiv 7 \pmod 8 \iff p - k \equiv 4 \pmod {16}$.

Illustration of proof for one biconditional (as the other proof arguments are more or less similar):

Assume that $\sigma(m^2) \equiv 1 \pmod 8$. Note that this implies that $\sigma(m^2) \equiv 1 \pmod 4$. By Chen and Luo $\ldots$. Please do not downvote, this post is currently under construction. I will return to it again tomorrow morning.


Professor Pace Nielsen writes in a comment that (and I quote, "verbatim"):

"I agree that this quoted result passes all sorts of sanity checks. We have $$\left(\frac{\sigma(p^k)}{2p^k}\right)\cdot\left(\frac{\sigma(m^2)}{m^2}\right) = 1. \tag{*}$$ Since $m$ is odd, we know $m^2 \equiv 1 \pmod 8$. Thus, looking at the given equality (Equation $(*)$) modulo $8$, we know that the class of $\sigma(m^2)$ modulo $8$ will be the inverse of the class of $\sigma(p^k)/{2p^k}$ modulo $8$. The latter class is exactly half of the class of $\sigma(p^k)/p^k$ modulo $16$. These considerations make us expect a connection between $\sigma(m^2)$ modulo $8$, and information about $p$ and $k$ modulo $16$." $\ldots$

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    $\begingroup$ I agree that this quoted result passes all sorts of sanity checks. We have$\frac{\sigma(p^a)}{2p^a}\frac{\sigma(m^2)}{m^2} = 1$. Since $m$ is odd, we know $m^2\equiv 1\pmod{8}$. Thus, looking at the given equality modulo $8$, we know that the class of $\sigma(m^2)$ modulo 8 will be the inverse of the class of $\sigma(p^k)/2p^k$ modulo 8. The later class is exactly half the class of $\sigma(p^k)/p^k$ modulo 16. These considerations make us expect a connection between $\sigma(m^2)$ modulo 8, and information about $p$ and $k$ modulo 16.... $\endgroup$ Commented May 18, 2023 at 16:07
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    $\begingroup$ What doesn't make sense is your use of ζ(2). To do a sanity check here, you need to ask yourself: How am I using ζ(2)? What happens if I used some other constant like 1.5? What parts of my argument would change using a different constant? If my result is new, what is the new information that I used to get to that new result (i.e., boil it down to the most basic information you need)? If I believe that I've proved the special prime has to be 5, what information did I use that does not apply to spoofs (where the special prime can be 13)? etc.. $\endgroup$ Commented May 18, 2023 at 16:35
  • $\begingroup$ Thank you for your list of questions that I have to answer, @PaceNielsen, from your comment here around $7$ hours ago (May 19, 2023 - 07:37 AM - Manila time). I will keep them in mind, as I write my "corrigendum". That being said, I hereby refer you to a question by Igor Souza. I highly think, that Igor's post settles most, if not all, of your questions. $\ldots$ $\endgroup$ Commented May 18, 2023 at 23:38
  • $\begingroup$ (continued) That being said, my profuse thanks for your comments, @PaceNielsen. Although I would have to say, I did already know which path I had to take, back in the year $1999$. ^_^ $\endgroup$ Commented May 18, 2023 at 23:40
  • $\begingroup$ (1) $\endgroup$ Commented May 18, 2023 at 23:54

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