Take b = 2.

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

31 is prime.

b^((n-1)/31)-1 mod n = 8886, which is a unit, inverse 2872.

3 is prime.

b^((n-1)/3)-1 mod n = 10972, which is a unit, inverse 1736.

(3^2 * 31) divides n-1.

(3^2 * 31)^2 > n.

n is prime by Pocklington's theorem.