Take b = 5.

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

3 is prime. b^((n-1)/3)-1 mod n = 83, which is a unit, inverse 100.

2 is prime. b^((n-1)/2)-1 mod n = 191, which is a unit, inverse 96.

(2^6 * 3) divides n-1.

(2^6 * 3)^2 > n.

n is prime by Pocklington's theorem.