Take b = 2.

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

19 is prime.

b^((n-1)/19)-1 mod n = 195104, which is a unit, inverse 82906.

3 is prime.

b^((n-1)/3)-1 mod n = 111667, which is a unit, inverse 28441.

(3^4 * 19) divides n-1.

(3^4 * 19)^2 > n.

n is prime by Pocklington's theorem.