Take b = 2.

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

19 is prime.

b^((n-1)/19)-1 mod n = 100235, which is a unit, inverse 45420.

17 is prime.

b^((n-1)/17)-1 mod n = 139111, which is a unit, inverse 130038.

3 is prime.

b^((n-1)/3)-1 mod n = 87669, which is a unit, inverse 110313.

(3^3 * 17 * 19) divides n-1.

(3^3 * 17 * 19)^2 > n.

n is prime by Pocklington's theorem.