Are there an infinite number of prime quadruples of the form $10n + 1$, $10n + 3$, $10n + 7$, $10n + 9$?

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.

• Would proving a series help here? Or would you still need to prove it's a prime? – Cole Johnson Aug 3 '13 at 22:56
• @Cole"Cole9"Johnson Could you elaborate one what you mean by "proving a series"? – Strants Aug 3 '13 at 23:04
• I mean that thing where you add another thing on the left and then you replace k with k+1 on the right. I think it's called induction. – Cole Johnson Aug 3 '13 at 23:07
• @Cole"Cole9"Johnson I doubt induction would work. If I suspected a pattern existed in the $n$, and if I had some trick for proving a number was prime, I could use induction to prove the pattern worked for all $n$ (which would also prove that an infinite number of such $n$ existed). However, I don't think there's any (obvious) pattern, and I doubt there'd be any trick for proving the numbers given by the pattern were prime. – Strants Aug 3 '13 at 23:27
• For lists of examples, see OEIS A007811 and A173037 etc. – Jeppe Stig Nielsen Jul 23 '18 at 22:06