Are there infinitely many quintuples of type $p, p + 2, p + 14, p + 26, p + 38$? Are there infinitely many quintuples of type $p, p + 2, p + 14, p + 26, p + 38$? I think there are not... but I don't know exactly why this isn't true.
My homework isn't requiring that I formally prove it, they just want a yes/no answer. So i was just wondering on the logic behind it.
 A: Hint: if $p$ is not divisible by $5$, then one of the remaining numbers is divisible by $5$. Use this to infer that $p\leq 5$. Now you have just 3 cases to check.
A: There are not, and you can see this easily enough by considering the primes modulo $10$, and the fact that there is only one prime ending in $5$ in base $10$.


*

*If $p \equiv 1 \pmod{10}$, then $p + 14 \equiv 5 \pmod{10}$.

*If $p \equiv 3 \pmod{10}$, then $p + 2 \equiv 5 \pmod{10}$.

*If $p \equiv 7 \pmod{10}$, then $p + 38 \equiv 5 \pmod{10}$.

*If $p \equiv 9 \pmod{10}$, then $p + 26 \equiv 5 \pmod{10}$.

A: If $p$ is a prime, then the question of whether there are infinitely many pairs of primes $p$ and $p+2$ is called the twin prime conjecture, to which the answer is unknown. I doubt the answer is yes, as that would essentially involve proving the twin prime conjecture. So the answer is either "no" or "unknown."
A: Or, the first two are twin primes, which modulo 30 would be 11|13, 17|19, and 29|31.  11+14, 17+38, and 29+26 would all be 25 (mod 30) and divisible by 5, ruling out each of the three twin prime candidates.  Of course, since the answer has to be yes or no, and since the twin prime conjecture is still conjecture, one could embrace no as the only choice of an answer.
