most recent 30 from http://mathoverflow.net 2013-06-19T23:43:48Z http://mathoverflow.net/feeds/question/17597 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17597/prime-numbers-with-given-difference Alexei Fedotov 2010-03-09T11:16:55Z 2010-03-16T03:38:33Z

<p>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. </p>

http://mathoverflow.net/questions/17597/prime-numbers-with-given-difference/17606#17606 François G. Dorais 2010-03-09T14:43:02Z 2010-03-16T03:38:33Z

<p>(This question has been killed in the comments, but it is still lacking the useful pointers.)</p> <p>This is a weak form of the <a href="http://mathworld.wolfram.com/k-TupleConjecture.html" rel="nofollow">Hardy-Littlewood Conjecture</a> which moreover predicts an asymptotic density for the number of such <a href="http://en.wikipedia.org/wiki/Prime_k-tuple" rel="nofollow">prime k-tuplets</a>. Special cases of this include: <a href="http://en.wikipedia.org/wiki/Twin_prime" rel="nofollow">twin primes</a>, <a href="http://en.wikipedia.org/wiki/Cousin_prime" rel="nofollow">cousin primes</a>, <a href="http://en.wikipedia.org/wiki/Sexy_prime" rel="nofollow">sexy primes</a>, <a href="http://en.wikipedia.org/wiki/Prime_quadruplet" rel="nofollow">prime quadruplets, quintuplets, and sextuplets</a>. 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.</p> <p>As far as I know, the infinitude of prime k-tuplets is an open problem for all fixed admissible patterns with k &ge; 2. Note that the <a href="http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem" rel="nofollow">Green-Tao Theorem</a> 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.)</p>