Take b = 2.

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

11 is prime.

b^((n-1)/11)-1 mod n = 15828, which is a unit, inverse 32387.

3 is prime.

b^((n-1)/3)-1 mod n = 2345, which is a unit, inverse 45682.

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

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

n is prime by Pocklington's theorem.