# No Other Prime Must Divide It

For all $$n \in \mathbb{Z},$$ let $$\omega (n)$$ denote the number of distinct prime divisors of $$n.$$ Find the sum of all primes $$p$$ such that $$\omega ((p-1)!+1) = 1.$$

Details and assumptions

• As an explicit example, since $$100= 2^2 \times 5^2$$ has two distinct prime divisors ($$2$$ and $$5$$), $$\omega (100) = 2.$$
×