(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.
  • $\begingroup$ A similar investigation off this hyperlink reports that $p \equiv m^2 \equiv 5 \pmod {10}$ does not hold. $\endgroup$ Sep 19 '18 at 9:15
  • $\begingroup$ 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. $\endgroup$ Sep 19 '18 at 9:20
  • $\begingroup$ May I know why this question was downvoted? Some form of feedback, hopefully constructive, would go a long way towards improving future questions/posts. $\endgroup$ Aug 27 '20 at 9:48

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 $\pi \equiv \alpha \equiv \pmod 4$. Then $$\sigma(M^2) \equiv 1 \pmod 4 \iff \pi \equiv \alpha \pmod 8,$$ $$\sigma(M^2) \equiv 3 \pmod 4 \iff \pi \equiv \alpha + 4 \pmod 8.$$


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.