Take b = 3.

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

13 is prime.

b^((n-1)/13)-1 mod n = 3056, which is a unit, inverse 48.

5 is prime.

b^((n-1)/5)-1 mod n = 189, which is a unit, inverse 1866.

(5 * 13) divides n-1.

(5 * 13)^2 > n.

n is prime by Pocklington's theorem.