No Other Prime Must Divide It

For all nZ,n \in \mathbb{Z}, let ω(n)\omega (n) denote the number of distinct prime divisors of n.n. Find the sum of all primes pp such that ω((p1)!+1)=1.\omega ((p-1)!+1) = 1.

Details and assumptions

  • As an explicit example, since 100=22×52100= 2^2 \times 5^2 has two distinct prime divisors (22 and 55), ω(100)=2.\omega (100) = 2.
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