Take b = 2.

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

5 is prime. b^((n-1)/5)-1 mod n = 58, which is a unit, inverse 103.

3 is prime. b^((n-1)/3)-1 mod n = 47, which is a unit, inverse 104.

(3^2 * 5) divides n-1.

(3^2 * 5)^2 > n.

n is prime by Pocklington's theorem.