Take b = 2.

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

23 is prime.

b^((n-1)/23)-1 mod n = 8016, which is a unit, inverse 6960.

11 is prime.

b^((n-1)/11)-1 mod n = 1560, which is a unit, inverse 12642.

(11^2 * 23) divides n-1.

(11^2 * 23)^2 > n.

n is prime by Pocklington's theorem.