Take b = 3.

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

11 is prime.

b^((n-1)/11)-1 mod n = 991, which is a unit, inverse 300.

2 is prime.

b^((n-1)/2)-1 mod n = 1407, which is a unit, inverse 704.

(2^7 * 11) divides n-1.

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

n is prime by Pocklington's theorem.