Take b = 2.

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

3 is prime. b^((n-1)/3)-1 mod n = 2, which is a unit, inverse 7.

2 is prime. b^((n-1)/2)-1 mod n = 11, which is a unit, inverse 6.

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

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

n is prime by Pocklington's theorem.