Let be given natural numbers N_1,N_2,N_3,...,N_k such that for every prime p less or equal k set N_1,N_2,N_3,...,N_k does not contain all reminders modulo p. Is it right that there exists number X such that all X+N_1,X+N_2,X+N_3,...,X+N_k are prime? I think it must follow from some theorems about prime numbers in arithmetical progression.

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    $\begingroup$ If you could solve the case N_1=1, N_2=3 I'd be very interested ... $\endgroup$ – gowers Mar 9 '10 at 11:21
  • $\begingroup$ I don't think the OP is asking for infinitely many. I can settle your case by exhibiting X=2, can't I? $\endgroup$ – TonyK Mar 9 '10 at 11:25
  • $\begingroup$ Even this variant is not going to work easily: simply look for pairs of consecutive primes arbitrarily far apart from each other to find different twin primes. $\endgroup$ – damiano Mar 9 '10 at 11:29
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    $\begingroup$ Yes---perhaps one should instead do $N_1=1$, $N_2=3$, $N_3=N+1$ and $N_4=N+3$ for some super-large $N$ congruent to 0 mod 6. This definitely forces infinitely many twin primes. $\endgroup$ – Kevin Buzzard Mar 9 '10 at 11:36

(This question has been killed in the comments, but it is still lacking the useful pointers.)

This is a weak form of the Hardy-Littlewood Conjecture which moreover predicts an asymptotic density for the number of such prime k-tuplets. Special cases of this include: twin primes, cousin primes, sexy primes, prime quadruplets, quintuplets, and sextuplets. While your conjecture is much weaker than Hardy-Littlewood, Kevin Buzzard's trick in the comments shows that it globally implies the infinitude of prime k-tuplets for any admissible pattern.

As far as I know, the infinitude of prime k-tuplets is an open problem for all fixed admissible patterns with k ≥ 2. Note that the Green-Tao Theorem falls short of proving any instance of this since the step size of the arithmetic progressions is not fixed. (Even the Tao-Ziegler Theorem falls short since the polynomials are required to have vanishing constant term.)

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    $\begingroup$ It has been conjectured that, for any $k>0$ and $n>2$, there are at least as many primes between 2 and $n$ (inclusive) as there are between $k+2$ and $k+n$. About 35 years ago, Hensley and Richards proved that this conjecture is incompatible with the prime $k$-tuples conjecture. It's the only example that comes to my mind of two plausible conjectures being proved incompatible. It's generally believed now that prime $k$-tuples is true and the other conjecture is false. $\endgroup$ – Gerry Myerson Mar 9 '10 at 23:24

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