Take b = 2.

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

19 is prime. b^((n-1)/19)-1 mod n = 635, which is a unit, inverse 610.

5 is prime. b^((n-1)/5)-1 mod n = 66, which is a unit, inverse 565.

(5 * 19) divides n-1.

(5 * 19)^2 > n.

n is prime by Pocklington's theorem.