Take b = 2.

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

13 is prime. b^((n-1)/13)-1 mod n = 102, which is a unit, inverse 89.

3 is prime. b^((n-1)/3)-1 mod n = 213, which is a unit, inverse 241.

(3 * 13) divides n-1.

(3 * 13)^2 > n.

n is prime by Pocklington's theorem.