Timeline for On the nearest-square function and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2023 at 2:35 | comment | added | Jose Arnaldo Bebita | I have some closely related number-theoretic investigations going on over at this MO question. Would you mind taking a look, @mathlove? Thank you in advance for your time and attention! =) | |
| S Nov 26, 2020 at 3:07 | history | bounty ended | Jose Arnaldo Bebita | ||
| S Nov 26, 2020 at 3:07 | history | notice removed | Jose Arnaldo Bebita | ||
| Nov 26, 2020 at 3:07 | vote | accept | Jose Arnaldo Bebita | ||
| Nov 23, 2020 at 10:55 | answer | added | Pascal Ochem | timeline score: 3 | |
| Nov 22, 2020 at 12:35 | comment | added | mathlove | I've summarized the findings so far in an answer. Also, I deleted my comments saying that yours is not a research question since your main interest is now "is it true that $m\not=\left\lceil\sqrt{m^2-p^k}\right\rceil$ ?" which might be a research level question. | |
| Nov 22, 2020 at 12:35 | answer | added | mathlove | timeline score: 2 | |
| Nov 22, 2020 at 0:28 | comment | added | Jose Arnaldo Bebita | If you could summarize the findings so far in an answer, I would greatly appreciate it @mathlove. | |
| Nov 22, 2020 at 0:18 | comment | added | Jose Arnaldo Bebita | Since $m^2 - p^k$ is not a square, then $m^2 - p^k \neq (m-1)^2$. Note that $m^2 - p^k < (m-1)^2 = m^2 - 2m + 1$ is impossible since $m^2 - 2m < m^2 - p^k$. Therefore, the smallest square that is larger than $m^2 - p^k$ is $m^2$. OHH NO, @mathlove! =( | |
| Nov 21, 2020 at 15:39 | comment | added | mathlove | I don't know why you can say that the smallest square larger than $m^2-p^k$ is $(m-1)^2$. It is possible that $m^2-2m\lt (m-1)^2\lt m^2-p^k$. | |
| Nov 21, 2020 at 15:19 | comment | added | Jose Arnaldo Bebita | In particular, this means that, under the problematic case, $m^2 - p^k > m^2 - 2m$, so that the smallest square larger than $m^2 - p^k$ is $m^2 - 2m + 1 = (m - 1)^2$. Does this proof suffice, @mathlove? | |
| Nov 21, 2020 at 15:03 | comment | added | Jose Arnaldo Bebita | If indeed $m-a=\sqrt{m^2 - p^k}$, then $m^2 - 2am + a^2 = m^2 - p^k$. This implies that $p^k = 2am - a^2 = a(2m - a)$, from which it follows that $0 < a < 1$. We conclude that $p^k < 2m$. | |
| S Nov 20, 2020 at 1:46 | history | bounty started | Jose Arnaldo Bebita | ||
| S Nov 20, 2020 at 1:46 | history | notice added | Jose Arnaldo Bebita | Draw attention | |
| Nov 14, 2020 at 21:11 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
pointed the Ochem and Rao hyperlink to the actual publication URL
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| Nov 14, 2020 at 12:12 | comment | added | JoshuaZ | "We now endeavor to disprove the Dris Conjecture" I'm not sure what you mean by this. Wouldn't a disproof by nature need to mean one had an actual odd perfect number as a counterexample? | |
| Nov 13, 2020 at 12:32 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
corrected the erroneous year in Dris thesis from 2018 to 2008
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| Nov 12, 2020 at 8:32 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
corrected formula for smallest square that is larger than $m^2 - p^k$
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| Nov 12, 2020 at 7:26 | history | asked | Jose Arnaldo Bebita | CC BY-SA 4.0 |