Take b = 2.

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

101 is prime.

b^((n-1)/101)-1 mod n = 94230, which is a unit, inverse 177344.

19 is prime.

b^((n-1)/19)-1 mod n = 112540, which is a unit, inverse 298625.

(19 * 101) divides n-1.

(19 * 101)^2 > n.

n is prime by Pocklington's theorem.