Take b = 2.

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

5 is prime.

b^((n-1)/5)-1 mod n = 1617, which is a unit, inverse 576.

3 is prime.

b^((n-1)/3)-1 mod n = 1566, which is a unit, inverse 1638.

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

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

n is prime by Pocklington's theorem.