Timeline for How big are the prime factors of $2^kp - 1$?
Current License: CC BY-SA 3.0
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| Feb 8, 2016 at 16:16 | comment | added | Gerhard Paseman | I'm slow today. 2*14 - 1= 27. It seems best to compute for a few small k and then go for a proof of impossibility. Gerhard "And Not Other Way Around" Paseman, 2016.02.08. | |
| Feb 8, 2016 at 15:39 | comment | added | Gerhard Paseman | Sort of, but I don't call it a proof yet because I haven't written it down. In general, things look like 2^kr= 1 + stu, where r is a prime dividing n and s,t, and u are (possibly trivial) prime powers. s t and u aren't powers of 2 or r, and congruence conditions usually make t and u equal to 1. Now Zsigmondy limits how big s can be, and trial shows no solutions with k large and nonzero. For yes answers, a small amount of computation is enough.. Gerhard "It All Falls Into Place" Paseman, 2016.02.08. | |
| Feb 8, 2016 at 10:24 | comment | added | Ant | nice! Can you sketch an outline on how you convinced yourself that it works for $5,7,8,9,11,13$ and that it does not for $2,3,4,6,10,12$? :-) | |
| Feb 8, 2016 at 4:45 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |