Timeline for Numbers of the form $2^ma + 2^nb$ where $\text{gcd}(a,b) = 1$ [closed]
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2019 at 20:04 | history | closed |
Lucia Wojowu Ben Barber Jan-Christoph Schlage-Puchta Pace Nielsen |
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| Jan 16, 2019 at 17:34 | answer | added | Wojowu | timeline score: 0 | |
| Jan 16, 2019 at 16:54 | comment | added | Gerhard Paseman | Ah. d=31, a is 3 mod d and b is 5 mod d. We have 2^m times 3 run through 3,6,12,24,17 and for b we have 5,10,20,9,18. Gerhard "Likes The Power Of Small" Paseman, 2019.01.16. | |
| Jan 16, 2019 at 16:45 | comment | added | Gerhard Paseman | Upon further reflection, an example like Seva's should work. If d is a prime like a Mersenne prime where 2 is far from being a primitive root, there should be coprime a and b such that the divisibility relation fails. I am guessing d=17, a=3 and b=11, but that doesn't work, and I don't see how to save it., d=127 should yield an example. Gerhard "It's About Powers Of Two" Paseman, 2019.01.16. | |
| Jan 16, 2019 at 16:36 | comment | added | Gerhard Paseman | You will need d coprime to a and b. Otherwise an odd prime p could divide d and a, and you are out of luck. If d is coprime to both a and b, a version of the Chinese remainder theorem might work. Gerhard "The Question Needs More Work" Paseman, 2019.01.16. | |
| Jan 16, 2019 at 16:20 | review | Close votes | |||
| Jan 17, 2019 at 20:04 | |||||
| Jan 16, 2019 at 16:11 | answer | added | Seva | timeline score: 2 | |
| Jan 16, 2019 at 15:49 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |