Take b = 5.

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

19 is prime.

b^((n-1)/19)-1 mod n = 6036, which is a unit, inverse 1013.

3 is prime.

b^((n-1)/3)-1 mod n = 6682, which is a unit, inverse 2636.

2 is prime.

b^((n-1)/2)-1 mod n = 14591, which is a unit, inverse 7296.

(2^8 * 3 * 19) divides n-1.

(2^8 * 3 * 19)^2 > n.

n is prime by Pocklington's theorem.