Take b = 2.

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

23 is prime.

b^((n-1)/23)-1 mod n = 68, which is a unit, inverse 573.

3 is prime.

b^((n-1)/3)-1 mod n = 124, which is a unit, inverse 234.

(3^2 * 23) divides n-1.

(3^2 * 23)^2 > n.

n is prime by Pocklington's theorem.