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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
May 28, 2014 at 13:43 history edited Jeff Strom CC BY-SA 3.0
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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