Timeline for Go I Know Not Whither and Fetch I Know Not What
Current License: CC BY-SA 3.0
16 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Sep 16, 2014 at 19:07 | vote | accept | Will Jagy | ||
| Sep 16, 2014 at 19:07 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| S Sep 16, 2014 at 14:35 | history | suggested | CommunityBot | CC BY-SA 3.0 |
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| Sep 16, 2014 at 14:29 | review | Suggested edits | |||
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| Sep 16, 2014 at 3:43 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Sep 16, 2014 at 3:18 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Sep 16, 2014 at 3:07 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Sep 16, 2014 at 2:48 | comment | added | Will Jagy | @FelipeVoloch, thanks. I will finish it with computer, I need to fiddle a bit for the prime $5$ because the 4th powers times coefficients get a bit too large for ordinary C++ integers. I see what you mean about your use of the word congruences, I think I noticed that in your comment at first, then i got distracted by David's answer. | |
| Sep 16, 2014 at 2:45 | comment | added | Felipe Voloch | That's what I meant by congruences and it's not automatic. By pure thought, if the value of polynomial is divisible by a suitably high power of three, then the variables are divisible by three. To figure out what suitably high means, requires a calculation which I won't do. | |
| Sep 16, 2014 at 2:39 | comment | added | Will Jagy | @FelipeVoloch, thanks. What about the first question, if the polynomial is divisble by $81,$ then all variables are divisible by $3?$ David suspects that is lots more work. | |
| Sep 16, 2014 at 2:30 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Sep 16, 2014 at 2:30 | answer | added | David E Speyer | timeline score: 12 | |
| Sep 16, 2014 at 2:30 | comment | added | Felipe Voloch | This is the norm of $a+b\sqrt{3}+c\sqrt{5}+d\sqrt{15}$ from the quartic field to the rationals, so it's a norm form. I don't think $1,\sqrt{3},\sqrt{5},\sqrt{15}$ form a basis of the ring of integers over the usual integers, so the congruences don't automatically follow from facts about norms, but it should not be hard to prove them that way. You can also use that to get information on numbers (integers?, rationals?) represented by it. I don't know how hard it would be to answer your last question. | |
| Sep 16, 2014 at 2:07 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Sep 16, 2014 at 1:58 | history | asked | Will Jagy | CC BY-SA 3.0 |