Timeline for The prime numbers modulo $k$, are not periodic
Current License: CC BY-SA 3.0
9 events
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May 28, 2014 at 20:01 | comment | added | Joël | This answer correctly solves the question as initially posted, so I don't understand the down votes. | |
May 28, 2014 at 14:33 | history | edited | Jeff Strom | CC BY-SA 3.0 |
added 96 characters in body
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May 28, 2014 at 13:43 | history | edited | Jeff Strom | CC BY-SA 3.0 |
improved clarity
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May 28, 2014 at 4:17 | comment | added | The Masked Avenger | Actually the Bertrand-Finsler idea doesn't work that easily. It is still hard to imagine nk+2 and mk+3 being consecutive primes for small values of m, in particular for m much less than k, as that would suggest an unusually early occurrence of a large prime gap. | |
May 28, 2014 at 3:28 | comment | added | The Masked Avenger | Showing eventual nonperiodicity mod k for large k is harder, but one might look at prime free regions (near nk! perhaps) and possibly reach a contradiction. | |
May 28, 2014 at 3:24 | comment | added | The Masked Avenger | Here is another. By Bertrand and Finsler there are ( for k large) at least two primes between k and 2k. But then they have to be k+2 and k+3. | |
May 28, 2014 at 3:04 | comment | added | The Masked Avenger | Try this then: let p be a prime divisor of k. Then p mod k is 0 or p. No other prime has that residue mod k. | |
May 28, 2014 at 2:28 | comment | added | Igor Rivin | All that shows is that $p$ is NOT a prime divisor mod $k$... | |
May 28, 2014 at 2:06 | history | answered | Jeff Strom | CC BY-SA 3.0 |