# Prime numbers with given difference

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.

• If you could solve the case N_1=1, N_2=3 I'd be very interested ... – gowers Mar 9 '10 at 11:21
• I don't think the OP is asking for infinitely many. I can settle your case by exhibiting X=2, can't I? – TonyK Mar 9 '10 at 11:25
• 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. – damiano Mar 9 '10 at 11:29
• 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. – Kevin Buzzard Mar 9 '10 at 11:36

• 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. – Gerry Myerson Mar 9 '10 at 23:24