TITLE: Are there an infinite number of prime quadruples of the form $10n + 1$, $10n + 3$, $10n + 7$, $10n + 9$?
QUESTION [16 upvotes]: In base 10, any prime number greater than 5 must end with the digits $1$, $3$, $7$, or $9$.  For some $n$, $10n + 1$,  $10n + 3$, $10n + 7$, $10n + 9$ are all prime: for example, when $n=1$, we have that $11$, $13$, $17$, and $19$ are all prime.  My question is, can anyone disprove the claim that there are an infinite number of such primes. (If you can prove it, that is fine, too, but seeing as it is a stronger form of the twin primes conjecture, I'd imagine that would be difficult).
The only progress I've been able to make is to show that $n$ must be of the form $3k + 1$ by considering the system of modular inequalities
$$p \not\equiv 0 \mod{2}$$
$$p \not\equiv 0 \mod{3}$$
$$p \not\equiv 0 \mod{5}.$$
Beyond that, I don't know where to go.

REPLY [16 votes]: I suppose, you will not find a proof for neither the positive nor the negative result here. 

The positive result would obviously imply the (unproven and presumably difficult) twin prime conjecture.
The negative result would disprove the first Hardy-Littlewood conjecture about the density of prime sets with a given pattern, which (among other things) conjectures a (positive) density for prime quadruples.