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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: http://en.wikipedia.org/wiki/De_Polignac%27s_conjecture, but I don't know about the first question2011-11-19

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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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    @GerryMyerson : I've accepted answers that well and fully answered the questions. Are there particular ones that I have not accepted that you think I should have accepted?2011-11-22
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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.