Triplet Twins

Assume that there exists three primes $p,q,r$ such that the requirements are met. Therefore we have $p= q-2, r = q+2$ and the numbers from $p$ to $r$ become $q-2, q-1, q, q+1, q+2$

Since $q-2, q, q+2$ are all prime, either $q-2 = 3 \Rightarrow \left(p,q,r\right) = \left(3,5,7\right)$ , or $q-2 > 3$ and so $3| q-1\Rightarrow 3|q+2$ or $3| q+1 \Rightarrow 3|q-2$ .

In both of the latter cases we have one of the three primes becoming a multiple of $3$ which contradicts their prime nature, and hence they can't be prime given that $p = q-2 > 3$

Thus there is only one solution, $3,5,7$

If you boil this problem down, we're simply asked how many primes are consecutive odd integers. Let's cross-examine both lists:

$1, 3, 5, 7, \boxed{9}, 11, 13, \boxed{15}, 17, 19, \boxed{21}, 23 \dots$

This is not to say that all prime numbers are simply non-multiples of $3$ , odd, integers. On the contrary, I've simply proven that given any three consecutive odd integers beginning with $n>5$ , there will always be a multiple of three present, and thus cannot be prime. We now must only consider sequences of primes starting with $n<5$ . Notice

$2, 3, 5, 7, 11, \dots$

And $3, 5, 7$ matches the conditions. Now that we have proven that no sequence of $n$ odd integers can be all prime for $n \geq 3$ . So we're finished, and the answer is thus $\boxed{1}$ .