Take b = 3.

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

19 is prime. b^((n-1)/19)-1 mod n = 214, which is a unit, inverse 126.

3 is prime. b^((n-1)/3)-1 mod n = 322, which is a unit, inverse 44.

(3 * 19) divides n-1.

(3 * 19)^2 > n.

n is prime by Pocklington's theorem.