Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I would like to revisit a closed question of asterios in a more MO kind of way, because it cuts quite close to a related question about sieving that might be of general interest.

The original question: http://mathoverflow.net/questions/49669/finite-or-infinite-again-closed amounts to the following. Given a positive integer $a \ge 3$, do there exist infinitely many integers $k$ such that (simultaneously):

  1. $ak+1$ does not have any non-trivial factors of the form $\pm 1 \mod a$.
  2. $ak-1$ does not have any non-trivial factors of the form $\pm 1 \mod a$.

Presumably the answer is yes, because one expects there to be infinitely many twin primes $p$, $p+2$ satisfying any reasonable congruence condition. If one wants to prove something unconditionally, however, one could look towards generalizing the result of Chen. Drifting away from the original problem slightly, and making things more explicit, one could make the following conjecture:

(*?) There exist infinitely many primes $p$ such that $p+2$ is either prime or a product $qr$ of two primes $q$ and $r$, where $q$ and $r$ are of the form $1 \mod 4$.

The question is: Is (*?) amenable to known sieving techniques, or, in the other extreme, does the imposition of congruence conditions on $q$ and $r$ create a difficulty similar to the parity problem?

(Of course, one can easily modify the conjecture in various ways, imposing congruence conditions on $p$ and different congruence conditions on $q$ and $r$ and then ask the analogous conjecture, providing that the congruence conditions don't combine in unpleasant ways. This is slightly tricky: one would not want to insist that $p \equiv 1 \mod 4$ and that $p+2$ was either prime or had two prime factors, both of the form $-1 \mod 4$, not because the resulting conjecture is false, but because it would then be equivalent to the twin prime conjecture, and would fall prey to the parity problem.)

(*?) is almost an amalgam of two (non-trivial!) sieving problems. Drop the congruence conditions on $q$ and $r$ and one gets Chen's theorem. Simply requiring that every prime divisor of $p+2$ is of the form $1 \mod 4$ on the other hand is close (in fact, slightly stronger) to asking that $p$ be represented by the quadratic form $a^2+b^2-2$, and the question of counting primes represented by quadratic forms was answered by Iwaniec in '74.

share|improve this question
add comment

2 Answers

Here is a partial answer to your question. Basically, the methods of Goldston, Graham, Pintz, and Yildirim (see here) have some bearing on your question.

Look at the series

$$\sum_{N < n < 2N} \bigg( \chi_1(n) + \chi_1(n + 2) + \chi_2(n + 2) \bigg) \bigg( \sum_d \lambda_d \bigg)^2$$

where the $\lambda_d$ are the usual Selberg sieve coefficients or some variant, and $\chi_1$ and $\chi_2$ are the characteristic functions of products of exactly one or two primes. Then I worked out a back of a napkin calculation using Theorems 7-9 in GGPY (possibly I made some mistake), and on the Elliott-Halberstam conjecture (a big assumption!) the above is positive, which gives another proof of Chen's theorem (this one conditional).

One nice feature of the GGPY sieve is that it is compatible with the kind of conditions you describe. For example, let $\chi_2$ be the characteristic function of $E_2$ numbers whose prime factors are both 1 mod 4, or split completely in your favorite Galois extension, or some other appropriate condition. (See the GGPY paper as well as a followup which I wrote.) Then the machinery still works. You just multiply the contribution from the $E_2$ numbers by a fourth. In fact you can do better and only sieve on integers $n$ congruent to 1 mod 4, in which case you only have to divide by a half.

According to my napkin calculation, this is numerically not enough, the series is asymptotically slightly negative, and so this argument fails to prove your assertion, even on EH. So perhaps your question really is "difficult" in the way suggested by the parity problem. But this at least shows you how you can hope to prove statements related to what you asked for.

share|improve this answer
add comment

one should add this :There are infinitely many twin primes if and only if there are infinitely many natural numbers that are not of the form 6nm+/-n +/-m. Proof: Every number that is not a multiply of 2 or 3 is of the form 6N+/-1. So the only couples that are not divisible by 2 or 3 are (6N-1,6N+1) for any N.Now are there infinitely many such prime couples (twin primes)?. If the number 6N-1 is prime it should not be written as a product of some numbers 6n+1 ,6m-1 for any n,m. So (6n+1)(6m-1)=6(6nm-n+m)-1 which means that N should not be of the form 6nm-n+m for any n,m. Similarly if 6N+1 is a prime it should not be a product of some numbers (6n-1)(6m-1) =6(6nm-n-m)+1 , or (6n+1)(6m+1) =6(6nm+n+m)+1 .Which means that we have a prime couple of the form (6N-1,6N+1) if and only if N is not of the form 6nm+/-n+/-m for any n,m. Is this a (well known) open problem?(infinitness and more on $anm \pm n\pm m$ )

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.