Timeline for For any n and some prime p there is an elemnet in Zp* of order n
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 24, 2013 at 18:08 | comment | added | Gerhard Paseman | I will agree that an approach involving arithmetic progressions like that does lead to a prime p with the desired property, and may be shorter if not cleaner than an approach involving Bang or Zsigmondy. I interpret Katz's answer as saying that the "question behind the question" is about primes in such progressions; this intepretation I disagree with, and suggest for self study of this problem an alternative. However, the poster has accepted Katz's answer; perhaps a proof more than understanding is wanted. Gerhard "To Each Their Own Heaven" Paseman, 2013.06.24 | |
| Jun 24, 2013 at 0:37 | comment | added | Todd Trimble | But I'm not sure that's the cleanest approach, Gerhard. I would agree that Dirichlet's theorem in its full statement (involving densities) is a sledgehammer, but the infinitude of primes in this arithmetic progression can be proven in very elementary fashion. See for example proposition 3, here: math.stanford.edu/~dlitt/exposnotes/primes1mod4.pdf | |
| Jun 19, 2013 at 19:35 | vote | accept | Milena | ||
| Jun 19, 2013 at 19:10 | comment | added | Gerhard Paseman | More elementarily, the poster is asking for a (positive) integer b such that b^n - 1 is a multiple of some prime p while b^m - 1 is not such a multiple for m < n. A proof by Bang was given for n > 6 (pick b=2) and for fixed b and n all the allowed cases were characterized by Zsigmondy. Gerhard "Ask Me About System Design" Paseman, 2013.06.19 | |
| Jun 19, 2013 at 18:31 | history | answered | Mikhail Katz | CC BY-SA 3.0 |