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5.6.23
A Proof That There Are No Odd Perfect Numbers
30.5.23
Last Presenter (From Taguig City) During the 2023 MSP-NCR Annual Convention, Held Via Zoom Last May 27 (Saturday)
I was the last presenter during the 2023 MSP-NCR Annual Convention, held via Zoom last May 27 (Saturday), and hosted by the University of Santo Tomas - Manila, Philippines.
Please do check out the Beamer slides in ResearchGate, if you are interested to know more about what ideas were communicated during my talk.
For the record, no questions were asked after I was done. (IIRC, some of the participants were graduate students from Polytechnic University of the Philippines and one of my former professors from De La Salle University.)
I finished presenting in less than fifteen minutes.26.5.23
Some New Results on Odd Perfect Numbers - Part IV
Continuing from this earlier blog post: (In particular, if $m$ is almost perfect, then $m$ must be a square (since $m$ is odd). Consequently, if the equation $\sigma(p^k)=2m$ holds (which is true if and only if $G=H=I$), then $\sigma(p^k)/2$ is also a square. This implies that $k=1$, since $$p^k + 1 \leq \sigma(p^k)=2m<2p$$ holds under the assumption $m<p$.)
25.5.23
22.5.23
Some New Results on Odd Perfect Numbers - Part III
- $$\left(\frac{5}{3}\right)^{\ln(4/3)/\ln(13/9)} \approx 1.4912789897463723558$$
- $$k = 1 \Rightarrow I(m) \geq \left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)} \geq \left(\frac{5}{3}\right)^{\ln(4/3)/\ln(13/9)}$$
- We have the (generic/sharp [?]) upper bound $$\frac{I(m^2)}{I(m)} \leq \frac{\zeta(2)}{\zeta(3)}.$$
- Reference: Comment by Erick Wong underneath an answer by Will Jagy to MSE question https://math.stackexchange.com/questions/1097803.
- If $I(m) < 2$, then $$\frac{2m}{m+D(m)} \leq I(m) < \frac{2m+D(m)}{m+D(m)},$$ where equality holds if and only if $I(m)=D(m)=1$.
- The inequality $p^k < m^2$, obtained by Dris in 09/2012 [JIS (2012)], means we can infer that the implication $m < p \Rightarrow k = 1$ holds.
Some New Results on Odd Perfect Numbers - Part II
- If $k=1$, then $I(m^2) > \zeta(2)$.
- If $p=5$, then we have the two cases:
- If $k = 1$, then $I(m^2) = 5/3 > \zeta(2)$.
- If $k \neq 1$, then $I(m^2) < \zeta(2)$.
7.5.23
Some New Results on Odd Perfect Numbers - A Summary
Let $N = p^k m^2$ be a hypothetical odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$. Note that, since $\gcd(p, m) = 1$ and $p$ is (the special) prime, then $p^k \neq m$. (This also follows from the fact that prime powers are deficient, contradicting $N$ is perfect.) By trichotomy, either $(p^k < m) \oplus (m < p^k)$ is true, where $\oplus$ denotes exclusive-OR. (That is, $A \oplus B$ is true if and only if exactly one of $A$ or $B$ holds.)
We will denote the classical sum of divisors of the positive integer $z$ by $\sigma(z)=\sigma_1(z)$, and the abundancy index of $z$ by $I(z)=\sigma(z)/z$. Furthermore, we will denote the deficiency of $z$ by $D(z)=2z-\sigma(z)$, and the aliquot sum of $z$ by $s(z)=\sigma(z)-z$.
Here is a summary of some new results on odd perfect numbers, which were realized by the author on May 5, 2023:
- If $p < m$, then the quantity $m^2 - p^k$ is not a square. (Kindly note the contrapositive.)
- If $m < p$, then the following statements hold:
- Descartes's conjecture holds (i.e. $k = 1$).
- Dris conjecture (i.e. $p^k < m$) is false.
- The quantity $m^2 - p^k$ is a square.
- The square root of the non-Euler part $m^2$ is almost perfect.
- Define the following GCDs:
- $G=\gcd\left(\sigma(p^k),\sigma(m^2)\right)$
- $H=\gcd\left(m,\sigma(m^2)\right)$
- $I=\gcd\left(m^2,\sigma(m^2)\right)$
- Dris proved in February 10, 2022 that the chain of divisibility conditions
- $G \mid H \mid I$
- $G=\gcd\left(\sigma(p^k)/2,\sigma(m^2)/p^k\right)=\sigma(p^k)/2=\gcd(G,I)$
- $H=\gcd\left(m,\sigma(m^2)/p^k\right)=m=\gcd(H,I)$.
- In particular, the divisibility constraint $\sigma(p^k)/2 \mid m$ holds.
- The divisibility condition $\sigma(p^k) \mid 2m$ is equivalent to $m \mid \sigma(m^2)$.
- ( To be continued$\ldots$ )
14.1.23
If $p^k m^2$ is an odd perfect number with special prime $p$ and $p = k$, then $\sigma(p^k)/2$ is not squarefree.
While researching the topic of odd perfect numbers, we came across the following implication, which we currently do not know how to prove:
> CONJECTURE: If $p^k m^2$ is an odd perfect number with special prime $p$ and $p = k$, then $\sigma(p^k)/2$ is not squarefree.
Here, $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$. (Note that both $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$ hold.)
OUR ATTEMPT
Here are the details of our search in the range $5 \leq p < 50$, $1 \leq k < 50$ using the following Sage Math Cell - Pari-GP scripts:
> (1) Searching for examples where $\sigma(p^k)/2$ is not squarefree
for(x=1, 50, for(y=1, 50, if((isprime(x)) && (Mod(x,4) == 1) && (Mod(y,4) == 1) && !(issquarefree(sigma(x^y)/2)),print(x," ",y," ",factor(sigma(x^y))))))
> Output:
> (2) Searching for examples where $\sigma(p^k)/2$ is squarefree
for(x=1, 50, for(y=1, 50, if((isprime(x)) && (Mod(x,4) == 1) && (Mod(y,4) == 1) && (issquarefree(sigma(x^y)/2)),print(x," ",y," ",factor(sigma(x^y))))))
> Output:
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As you can see, the Conjecture does appear plausible. However, computational searches are very far from a complete proof, though they certainly add to the evidence supporting the Conjecture.
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Here is our:
QUESTION: Does anybody here have any ideas on how to prove the Conjecture?