Take b = 3.

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

11 is prime.

b^((n-1)/11)-1 mod n = 9117, which is a unit, inverse 13638.

7 is prime.

b^((n-1)/7)-1 mod n = 3701, which is a unit, inverse 7582.

3 is prime.

b^((n-1)/3)-1 mod n = 16509, which is a unit, inverse 24065.

(3 * 7 * 11) divides n-1.

(3 * 7 * 11)^2 > n.

n is prime by Pocklington's theorem.