Timeline for Go I Know Not Whither and Fetch I Know Not What
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 16, 2014 at 19:09 | comment | added | Will Jagy | @Noam, I checked by computer, divisibility of the polynomial by 27 implies divisibility of all four variables by 3, same thing for 125 and 5. In the other direction, 9 and 25 are primitively represented. | |
| Sep 16, 2014 at 19:07 | vote | accept | Will Jagy | ||
| Sep 16, 2014 at 4:03 | comment | added | Noam D. Elkies | While ${\bf Z}(\sqrt{3},\sqrt{5})$ is not the full ring of integers $O_K$, we can recover $O_K$ by dividing some elements of ${\bf Z}(\sqrt{3},\sqrt{5})$ by $2$, so the distinction betwen ${\bf Z}(\sqrt{3},\sqrt{5})$ and $O_K$ shouldn't affect questions of divisibility by powers of $3$ and $5$. | |
| Sep 16, 2014 at 3:28 | comment | added | David E Speyer | @FelipeVoloch Nonetheless, sorry about that. | |
| Sep 16, 2014 at 3:18 | comment | added | GH from MO | It is good to be in good company, folks. | |
| Sep 16, 2014 at 2:46 | comment | added | Felipe Voloch | I am honored to be confused with GH. | |
| Sep 16, 2014 at 2:38 | comment | added | Will Jagy | Tee hee hee....Anyway, as a quite good analogy, if $x^2 - 15 y^2$ is divisible by $9$ then both $x,y$ are divisible by $3,$ same for $25$ and $5$. youtube.com/watch?v=hTGyOKdgRN8 | |
| Sep 16, 2014 at 2:34 | comment | added | David E Speyer | No, I didn't do that one. The primes $2$, $3$ and $5$ are going to be wonky since (as GH from MO points out) you aren't working with an integer basis for the ring of integers, and I didn't want to work that hard. | |
| Sep 16, 2014 at 2:34 | comment | added | Will Jagy | oh, very nice, David. I cannot tell whether you have explained the first thing, that when $f(a,b,c,d)$ is divisible by $81,$ all the variables are divisible by $3.$ | |
| Sep 16, 2014 at 2:30 | history | answered | David E Speyer | CC BY-SA 3.0 |