Timeline for Does 2^m = 3^n + r have finitely many solutions for every r?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2010 at 5:46 | comment | added | Dave R | I have extended the calculation to $0 \le r \le 100$. M(r) = 3 for r = 3 (mod 6); M(r) = 8 for r = 17, 19, 25, 35, 41, 43, 49, 59, 65, 67, 73, 83, 89, 91, 97; M(r) = 60 for r = 31, 53, 79, 85, 95; M(13) = 131584; M(23) = 1088; M(29) = 117; M(37) = 21951; M(47) = 65972; M(55) = 999; M(61) = 252; M(71) = 63; M(77) = 28. | |
| Mar 20, 2010 at 8:54 | vote | accept | Dave R | ||
| Mar 20, 2010 at 2:49 | vote | accept | Dave R | ||
| Mar 20, 2010 at 8:54 | |||||
| Mar 20, 2010 at 2:49 | vote | accept | Dave R | ||
| Mar 20, 2010 at 2:49 | |||||
| Mar 20, 2010 at 2:49 | vote | accept | Dave R | ||
| Mar 20, 2010 at 2:49 | |||||
| Mar 20, 2010 at 2:41 | answer | added | Felipe Voloch | timeline score: 15 | |
| Mar 20, 2010 at 2:41 | answer | added | Jonas Meyer | timeline score: 6 | |
| Mar 20, 2010 at 2:06 | history | edited | Dave R | CC BY-SA 2.5 |
Clarified meaning of M(r), I hope.
|
| Mar 20, 2010 at 2:00 | history | asked | Dave R | CC BY-SA 2.5 |