Take b = 2.

b^(n-1) mod n = 1.

13 is prime.

b^((n-1)/13)-1 mod n = 22508, which is a unit, inverse 18902.

11 is prime.

b^((n-1)/11)-1 mod n = 102, which is a unit, inverse 15882.

3 is prime.

b^((n-1)/3)-1 mod n = 3587, which is a unit, inverse 17108.

(3 * 11 * 13) divides n-1.

(3 * 11 * 13)^2 > n.

n is prime by Pocklington's theorem.