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Nov 17, 2018 at 21:03 vote accept Daniel Soltész
Nov 14, 2018 at 19:44 answer added GH from MO timeline score: 4
Nov 14, 2018 at 16:24 history edited GH from MO CC BY-SA 4.0
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Nov 13, 2018 at 17:50 answer added Gerhard Paseman timeline score: 1
Nov 13, 2018 at 12:05 comment added Daniel Soltész @GerhardPaseman I am fine with something "tossed together" and doing some work myself.
Nov 12, 2018 at 20:00 comment added Gerhard Paseman OK. Lucia can give you a professional answer. I can toss one together, but you still have to do some work. I believe Chebotarev density theorem and quadratic reciprocity are the main components for the answer. Gerhard "Still Likes His Amateur Status" Paseman, 2018.11.12.
Nov 12, 2018 at 19:10 history edited Daniel Soltész CC BY-SA 4.0
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Nov 12, 2018 at 19:08 comment added Daniel Soltész Whoops, for my purposes it is enough that for every fixed $k$ there is such a sequence.
Nov 12, 2018 at 18:47 comment added Daniel Soltész Existence is fine for me. The fact that it is (more or less) obvious is consistent with my observation that no papers are bothered by this part of the argument.
Nov 12, 2018 at 18:33 comment added Gerhard Paseman In fact, there should be density results (Chebotarev?) that would say how many primes q there are larger than and close to p^k that belong to a given congruence class mod p. You should be able to show existence with such results. (Of course, it is easy for fixed k. Still thinking about the general problem.) Gerhard "Like Lucia Said, There's Plenty" Paseman, 2018.11.12.
Nov 12, 2018 at 18:18 comment added Gerhard Paseman Maybe reverse it: pick some nice primes p and look at primes q in the range p^k, 2p^k. Gerhard "Surely There Are Enough Primes" Paseman, 2018.11.12.
Nov 12, 2018 at 18:06 comment added Lucia Do you want just existence or must it be constructive? I think existence is more or less obvious since a typical prime would satisfy this.
Nov 12, 2018 at 17:57 history asked Daniel Soltész CC BY-SA 4.0