Take b = 5.

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

13 is prime.

b^((n-1)/13)-1 mod n = 13172, which is a unit, inverse 1862.

11 is prime.

b^((n-1)/11)-1 mod n = 20039, which is a unit, inverse 1301.

3 is prime.

b^((n-1)/3)-1 mod n = 142, which is a unit, inverse 6816.

(3^2 * 11 * 13) divides n-1.

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

n is prime by Pocklington's theorem.