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Is there an elementary(or not) proof that there are at least two consecutive primes which have difference $2n$ for every natural number $n$? i remind that Polignac's conjecture states that there should be infinite such pairs for every $n$ so it is a much more easy question to ask and of course should be valid.

In addition, for every $n\in\mathbb N$, is there any $m_n\in\mathbb N$ such that there are two natural numbers $k_1,k_2$ so that $k_2-k_1=2n$, $k_1, k_2$ are not divisible by $p_1,\dots p_{m_n}$, where $p_i$ is the $i$-th prime, and for every $k\in\{k_1+1,\dots k_2-1\}$ there exists a $p\in\{p_1,\dots p_{m_n}\}$ such that $p|k$? (elementary proof)

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I think your title is a little misleading, as Polignac's conjecture is only related to your question and not your actual question. Would you mind changing it to something that more accurately represents what you're after? –  Mike Spivey Mar 17 '11 at 17:14
    
do you have something better to propose, if so feel free to edit the title –  minasteris Mar 17 '11 at 17:39
    
@minasteris: How is this? (Feel free to edit or change back; this is your question, after all.) –  Mike Spivey Mar 17 '11 at 17:42
    
The answer to the new question is no. Take $m=1$ and $n=2$. –  Jonas Meyer Mar 18 '11 at 18:10
    
sorry wrong edit –  minasteris Mar 18 '11 at 19:25

2 Answers 2

up vote 8 down vote accepted

I don't know the state of the art on this question, but some searches indicate that it is not known whether every even number is a difference of two primes, let alone a difference of consecutive primes. Perhaps a little old, but this is mentioned in The new book of prime number records by Paulo Ribenboim, 1996, on page 250. (This problem is called the "Goldbach Variation" in Hofstadter's Gödel, Escher, Bach.)

The following source seems more recent, and mentions that Chen's work related to the Goldbach conjecture showed that every even number is a difference of a prime and a product of at most two primes: http://primes.utm.edu/notes/conjectures/

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where can i find chen's article? –  minasteris Mar 18 '11 at 6:38
    
I'm not sure, but it might be this one: books.google.com/… –  Jonas Meyer Mar 18 '11 at 7:05
    
@minasteris: Perhaps what you had in mind was related to the comments on Chris Caldwell's Prime Pages just below the item about Chen's work related to the Goldbach conjecture (see Jonas's last link in Answer). That next item concerns Polignac's conjecture and says (in part): "It is easy to show that for every positive integer m there is an even number 2n such that there are more than m pairs of consecutive primes with difference 2n." Note that n is not free here, but has an existence conditioned on m. –  hardmath Mar 18 '11 at 19:48
    
@hardmath: my question is different –  minasteris Mar 18 '11 at 20:41

The reworked addition to the second question has a positive answer. For any positive integer $n$ we can find an initial segment of the primes $p_1,p_2,...,p_m$ (where $m > 0$) and a positive number $k$ such that neither $k$ nor $k+2n$ is divisible by any of those primes, but each integer strictly between $k$ and $k+2n$ is divisible by at least one of them.

For motivation consider the special case in which $2n+1$ happens to be a prime. Then take $k=1$ and the primes to be those less than $2n+1$. Clearly $k$ and $k+2n$ are free from those small divisors, while every integer in between must be divisible by at least one of them.

Edited: Changed to make larger primes those above $2n$ (rather than above $n$)

For the general case we will use the CRT. It is simply a matter of choosing remainders for $k$ modulo the primes less than or equal to $2n$ so that neither $k$ nor $k + 2n$ is divisible, and then deploying the larger primes as needed to knock out any in-between integers not already divided by one of the smaller primes.

Obviously we want $k$ to be odd, so that $k+2n$ is odd as well. For odd primes $p_i$ up to (and if necessary) including $n$, there will always be at least one remainder for $k$ mod $p_i$ such that neither $k$ nor $k+2n$ is divisible by $p_i$. If $p_i$ divides $2n$, then this is really just a matter of choosing a nonzero remainder at $k$. If not, the fact that $p_i$ > 3 means there are only two "bad" choices for the remainder at $k$, so there's always at least one good choice.

Now consider any integers strictly between $k$ and $k+2n$ that are not already "sieved out" by those smaller primes. Then use the successive primes above $2n$ to get one-by-one a divisibility condition on those intervening numbers, which of course specifies a nonzero remainder for $k$ modulo each such prime. A crude estimate is that we need no more than $n$ primes greater than $2n$ to accomplish knockouts of all the entries unsieved by 2 (and hence by the collective set of smaller primes).

Finally determine a $k > 0$ having the specified remainders modulo the primes $p_1,...,p_m$ and we are done. None of them divide $k$ or $k+2n$, but at least one divides each of integers strictly in between.

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