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27.4.24
(Undeleted from MathOverflow) - If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a$, where $a < m$.
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: For the remainder of this post, assuming that $m$ 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 $m<p$.
26.7.22
Some Recent Updates to My Past Mathematics@StackExchange Posts
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n)=1$.
The recent answer to this MSE question titled "On odd perfect numbers and a GCD - Part V" appears to have successfully completed a proof for the inequation
$$\gcd(n,\sigma(n^2)) \neq \gcd(n^2,\sigma(n^2)).$$
If the proof holds water, then this shows that there cannot be an odd perfect number $N' = p^j m^2$ with special prime $p$ satisfying $p \equiv j \equiv 1 \pmod 4$ and $\gcd(p, m)=1$, of the form
$$N' = \frac{p^j \sigma(p^j)}{2}\cdot{m}.$$
You may refer to this MSE question (and the answer contained therein) for more information.
25.7.22
Is there any other known relationship between even perfect numbers and odd perfect numbers, apart from their multiplicative forms? | A proposed proof (?) for the nonexistence of odd perfect numbers
(Note 2: This was cross-posted from MO, because it was not well-received there. Will delete the MO post in a few.)
Observe that an even perfect number $M = (2^p - 1)\cdot{2^{p - 1}}$ and an odd perfect number $N = q^k n^2$ have similar multiplicative forms. (Indeed, it is conjectured that $k=1$, and this prediction goes back to Descartes ($1638$).)
Here is my initial question:
INITIAL QUESTION
Is there any other known relationship between even perfect numbers and odd perfect numbers, apart from their multiplicative forms?
I ask because of a related query here.
Indeed, if we could show that $$\left|2^r - t\right| = 1,$$
for $r$ and $t$ satisfying $n^2 - q^k = 2^r t$ and $\gcd(2, t)=1$, then we would have a proof for $n < q^k$, which together with Brown's estimate $q < n$ ($2016$) would yield a refutation of Descartes's Conjecture that $k = 1$.
Note that, for even perfect numbers, we actually have
$$(2^p - 1) - 2^{p-1} = 2^{p-1} - 1 = \bigg(2^{\frac{p-1}{2}} + 1\bigg)\bigg(2^{\frac{p-1}{2}} - 1\bigg) = ab,$$
where the factorization works whenever $M \neq 6$. (Note that $\gcd(a, b) = 1$.)
We compute that
$$\left|a - b\right| = 2.$$
DISCUSSION
In the hyperlinked MO question, the following (summarized) cases were considered for odd perfect numbers, which we now examine for even perfect numbers:
Case 1:
$$2^{\frac{p-1}{2}} < \min(a,b) = 2^{\frac{p-1}{2}} - 1$$
Notice that Case 1 clearly does not hold.
Case 2:
$$\min(a,b) = 2^{\frac{p-1}{2}} - 1 < 2^{\frac{p-1}{2}} < \max(a,b) = 2^{\frac{p-1}{2}} + 1$$
Notice that Case 2 clearly holds.
It follows (from mimicking the resulting inequality $q^k < n\cdot{\left|2^r - t\right|}$ for odd perfect numbers) that
$$2^p - 1 < {2^{\frac{p-1}{2}}}\cdot{\left|a - b\right|} = {2^{\frac{p-1}{2}}}\cdot{2} = {2^{\frac{p+1}{2}}}.$$
This last inequality implies that
$$p < -\frac{2\bigg(\log(2) - \log(\sqrt{2} + \sqrt{6})\bigg)}{\log(2)} \approx 1.89997,$$
which is a contradiction to $p \geq 2$.
Case 3:
$$\max(a,b) = 2^{\frac{p-1}{2}} + 1 < 2^{\frac{p-1}{2}}$$
Notice that Case 3 clearly does not hold.
Note that for even perfect numbers, we do have
$$2^{\frac{p-1}{2}} < 2^p - 1$$
which "mimics" the conjecture $n < q^k$ for odd perfect numbers above.
Here is my final question for this post:
FINAL QUESTION
Does the exhaustion of all possible cases in the DISCUSSION section essentially disprove the existence of odd perfect numbers?
4.3.22
My ResearchGate Profile
The link to my ResearchGate profile is here.
Here is a preview, as of March 4, 2022 - 9:09 PM (Manila time):