Timeline for Global character of ABC/Szpiro inequalities
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 29 at 5:25 | answer | added | Kirti Joshi | timeline score: 16 | |
| Apr 1 at 22:29 | answer | added | Kirti Joshi | timeline score: 56 | |
| Mar 31 at 14:19 | comment | added | Timothy Chow | @Nico See the last line of Will Sawin's answer. | |
| Mar 30 at 19:53 | answer | added | Will Sawin | timeline score: 43 | |
| Mar 30 at 19:50 | comment | added | Nico | @LuisFerroni Do you have a link? | |
| Mar 30 at 6:02 | history | became hot network question | |||
| Mar 29 at 19:58 | vote | accept | Jon23 | ||
| Mar 29 at 18:59 | answer | added | Peter Scholze | timeline score: 124 | |
| Mar 28 at 19:48 | answer | added | CHUAKS | timeline score: 20 | |
| Mar 26 at 10:03 | comment | added | Jon23 | The idea seems to be that for a given large $n$, for each $p$ one has an ABC inequality with a certain constant coming from the triple $(1,p^n,1+p^n)$, and that one can construct some other triple such that its ABC inequality cannot be obtained by summing those of its prime factors, the large $n$ presumably inversely related to the $\epsilon$ in the usual ABC statement, or something of the sort, but I cannot work out the argument. | |
| Mar 25 at 14:27 | history | edited | Jeremy Rouse | CC BY-SA 4.0 |
Corrected misspelling in title of question
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| Mar 25 at 12:17 | history | asked | Jon23 | CC BY-SA 4.0 |