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Knowing that $p$ is prime enables us to rule out the possibility that $p+2$ and $p+4$ are both prime, except in the one trivial case that $p=3$, since at least one of $p,\ p+2,\ p+4$ is divisible by $3$. But in some cases, $p,\ p+2,\ p+6$ are all prime.

For which finite sets $0\in A\subseteq \{0,2,4,6,\ldots\}$ does there exist a prime $p$ such that every member of $p+A = \{p+a : a\in A\}$ is prime?

Are there some such sets $A$ (besides $A=\lbrace 0 \rbrace$) for which infinitely many such $p$ are known to exist? (I think the answer to that one is unknown in the case $A=\{0,2\}$.)

Later note: Above I wrote $A\subseteq \{0,2,4,6,\ldots\}$. Later I changed it to $0\in A\subseteq \{0,2,4,6,\ldots\}$. Any $A$ that doesn't contain $0$ represents the same size and shape as one that does.

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There are no sets for which infinitely many such $p$ are known to exist. See: en.wikipedia.org/wiki/De_Polignac%27s_conjecture, but I don't know about the first question –  Carl Nov 19 '11 at 21:08
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20% acceptance rate? Tsk, tsk. –  Gerry Myerson Nov 19 '11 at 23:50

2 Answers 2

If there is a number $q$ such that the set $A\cup\lbrace0\rbrace$ covers every congruence class modulo $q$ then there are obviously only finitely many primes $p$ such that every member of $p+A$ is prime, since either $p$ or at least one member of $p+A$ will be a multiple of $q$. It is believed, but not proved, that if there is no such $q$ then there will be infinitely many primes $p$ such that $p+A$ are all prime. As Carl notes, there is no non-empty set $A$ (apart from $\{0\}$ itself) for which existence has been proved.

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When you say "covers", do you mean "intersects"? –  Michael Hardy Nov 20 '11 at 1:23
    
Interesting. The conjecture is that in each such case infinitely many exist, but what if we weaken it to say that in each such case, at least one exists. Has that been proved? –  Michael Hardy Nov 20 '11 at 2:24
    
It occurs to me that I should have specified that every value of $A$ to be considered should contain $0$, since any that doesn't is really the same size and shape as one that does. –  Michael Hardy Nov 20 '11 at 2:34
    
Suppose $0\in A\subseteq\{0,2,4,6,\ldots\}$. Consider the statement that there is no integer $q$ such that $A$ intersects every congruence class mod $q$. How would one know that that statement is true, given a particular $A$? How would one make a list of all finite sets $A$ satisfying that condition (an infinite list, but one might hope there's a simple algorithm that generates the whole sequence of them if it runs forever)? Has someone done that? –  Michael Hardy Nov 20 '11 at 2:39
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@MichaelHardy: If A is a finite set, one need only check the primes q <= |A| to decide if A covers all congruence classes mod q. So one can list the finite acceptable sets. –  Charles Nov 20 '11 at 4:04

Dickson's conjecture implies that for any finite set S not containing all residue classes mod some prime q, there are infinitely many integers n such that $n+s$ is prime for all $s\in S.$ There are no $|S|>1$ for which this is known to hold.

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