Take b = 2.

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

23 is prime.

b^((n-1)/23)-1 mod n = 252, which is a unit, inverse 607.

2 is prime.

b^((n-1)/2)-1 mod n = 1011, which is a unit, inverse 506.

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

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

n is prime by Pocklington's theorem.