Take b = 2.

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

131 is prime.

b^((n-1)/131)-1 mod n = 234602, which is a unit, inverse 430556.

19 is prime.

b^((n-1)/19)-1 mod n = 415397, which is a unit, inverse 133995.

(19 * 131) divides n-1.

(19 * 131)^2 > n.

n is prime by Pocklington's theorem.