Take b = 2.

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

19 is prime.

b^((n-1)/19)-1 mod n = 2823, which is a unit, inverse 1120.

5 is prime.

b^((n-1)/5)-1 mod n = 2743, which is a unit, inverse 2719.

(5^2 * 19) divides n-1.

(5^2 * 19)^2 > n.

n is prime by Pocklington's theorem.