Take b = 2.

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

11 is prime.

b^((n-1)/11)-1 mod n = 10518, which is a unit, inverse 14426.

5 is prime.

b^((n-1)/5)-1 mod n = 7926, which is a unit, inverse 9599.

(5^2 * 11) divides n-1.

(5^2 * 11)^2 > n.

n is prime by Pocklington's theorem.