mersenneforum.org Primes of the Form Mod(p,q) = Mod(x,q)
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 2018-12-10, 06:27 #1 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 41·59 Posts Primes of the Form Mod(p,q) = Mod(x,q) Hi all, Is there a proof that primes p such that Mod (p,q) = Mod (x,q) For any given prime q and integer x such that 0 < x < q Are Infinity many? For example for q = 5 and x = 3 Is it provable that the set of primes p: 13, 23, 43, ... Is infinity large. Thank you for any reference you may provide.
 2018-12-10, 09:19 #2 Nick     Dec 2012 The Netherlands 2·3·307 Posts This may help: https://primes.utm.edu/notes/Dirichlet.html I think it is also worth experimenting with specific numbers to get a feel for this type of problem (and not just with q prime). For example: Prove that there are infinitely many prime numbers. Prove that there are infinitely many primes which are 3 mod 4. Prove that there are infinitely many primes that are 1 mod 4.
 2018-12-10, 12:51 #3 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 97316 Posts Much obliged Nick, For the record the Wikipedia entry: https://en.m.wikipedia.org/wiki/Diri...c_progressions
 2018-12-10, 15:37 #4 Batalov     "Serge" Mar 2008 San Diego, Calif. 2×3×1,733 Posts
 2018-12-11, 02:31 #5 science_man_88     "Forget I exist" Jul 2009 Dartmouth NS 5×13×131 Posts http://mathworld.wolfram.com/Modular...gFunction.html is indirectly related.
2018-12-11, 02:51   #6
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Quote:
 Originally Posted by Nick 3. Prove that there are infinitely many primes that are 1 mod 4.
4. Prove that there are infinitely many primes that are 6 mod 7.

Oh... wait... that was the other guy...

 2018-12-11, 03:34 #7 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 97316 Posts To much meat on those links. Thank you Gents both. Currently I am exhausted from my day job. Digestion will have to wait for another day if at all possible. The straight forward proof seems way too advanced. But Google seems to indicate there has been searches for an "elementary" proof. Will see if that's easier on the mind. Thank you for all the replies.

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