Take b = 2.

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

131 is prime.

b^((n-1)/131)-1 mod n = 230861, which is a unit, inverse 30204.

23 is prime.

b^((n-1)/23)-1 mod n = 200722, which is a unit, inverse 243261.

(23 * 131) divides n-1.

(23 * 131)^2 > n.

n is prime by Pocklington's theorem.