Timeline for Are perfect cubes possible?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 3, 2017 at 11:45 | vote | accept | CommunityBot | ||
| Jun 13, 2016 at 16:14 | comment | added | Milo Brandt | @Yakk No - notice that the only way to have one of $a+b$ or $c+d$ or $a+b+c+d$ be zero for non-negative integers only would be to have either $a=b=0$ or $c=d=0$ (or both). Both of these possibilities violate the condition that the numbers be distinct. | |
| Jun 13, 2016 at 13:39 | comment | added | Yakk | Is there a solution with non-negative integers only? | |
| Jun 13, 2016 at 12:50 | comment | added | ABcDexter | That's a very good answer, explanation is crystal clear and especially the mention of Hardy-Ramanujan numbers :) | |
| Jun 12, 2016 at 19:57 | comment | added | user17008 | +1. With this someone could easily make a program to generate thousands of solutions. | |
| Jun 12, 2016 at 19:19 | history | edited | Milo Brandt | CC BY-SA 3.0 |
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| Jun 12, 2016 at 19:12 | history | edited | Milo Brandt | CC BY-SA 3.0 |
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| Jun 12, 2016 at 19:06 | history | edited | Milo Brandt | CC BY-SA 3.0 |
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| Jun 12, 2016 at 19:00 | history | undeleted | Milo Brandt | ||
| Jun 12, 2016 at 19:00 | history | edited | Milo Brandt | CC BY-SA 3.0 |
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| Jun 12, 2016 at 18:38 | history | deleted | Milo Brandt | via Vote | |
| Jun 12, 2016 at 18:34 | history | undeleted | Milo Brandt | ||
| Jun 12, 2016 at 18:34 | history | edited | Milo Brandt | CC BY-SA 3.0 |
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| Jun 12, 2016 at 18:27 | history | deleted | Milo Brandt | via Vote | |
| Jun 12, 2016 at 18:27 | history | answered | Milo Brandt | CC BY-SA 3.0 |