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so my question is if this has been considered

imagine a big prime $ q $ then my uqestion is if we can find an infinite group of primes so

$ P_{n}=P_{n+1}+k $ with $ P_{0}=q $ and k a positive integer

this would mean that we can find an infinite sequence of primes from a given prime

that's it $ q , q +k , q+2k, q+3k ... $ are always primes for a given q and k integer

Jose Garcia
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  • Your question is not clear. Are you asking if there is an infinite arithmetic progression consisting only of primes? The answer to that is no: if $P_0=q$ then we'd get $P_q=q+q\times k$ which is divisible by $q$. I'm guessing that you meant to write $P_{n+1}=P_n+k$ here. – lulu Feb 20 '20 at 14:43
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    The $q$th term, $P_q=q+qk$, is clearly not prime. – Barry Cipra Feb 20 '20 at 14:48
  • On the plus side: Dirichlet tells us that, with obvious exceptions, every arithmetic progression contains infinitely many primes, and it even gives us the density for such primes. Doesn't give us an explicit construction, of course. – lulu Feb 20 '20 at 14:48
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    To add to lulu's comment, the Green-Tao theorem guarantees the existence of arbitrarily (but still finitely) long arithmetic progressions of primes. See https://en.wikipedia.org/wiki/Primes_in_arithmetic_progression – Barry Cipra Feb 20 '20 at 14:50
  • It comes even worse, there is not even a non-constant univariate polynomial $\ f(x)\ $ with integer coefficients, such that $\ f(n)\ $ is prime for every natural number $\ n\ $. – Peter Feb 20 '20 at 15:50

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