Solution. We show that two such primes exist: 3 and 7.
p=3 satisfies the conditions since 1+1.2=3 is a prime.
p=5 does not satisfies the condition since 1+2.4=9 is composite.
p=7 satisfies the conditions since 1+1.6=7, 1+2.6=13 and 1+3.6=19 are all primes.
p=11 does not satisfies the condition since 1+2.10=21 is composite.
p=13 does not satisfies the condition since 1+2.12=25 is composite.
In the rest of the solution we assume p17.
Suppose that some prime does not divide (p-1). Then q and p-1 are coprime, so there is such a , for which q|1+k(p-1). By qp-1<1+k(p-1), the number q is a proper divisor of (1+k(p-1)), contradicting the assumption that 1+k(p-1) is a prime.
Hence, p-1 is divisible by all prime numbers less than p/2. Let r be the greatest prime which is less than p/2. By Chebysev's theorem we have . The numbers 2,3,r are distinct prime divisors of p-1, so 6r|p-1. But this is not possible because p-14r+2<6r.