Take b = 3.

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

89 is prime.

b^((n-1)/89)-1 mod n = 9674, which is a unit, inverse 9239.

2 is prime.

b^((n-1)/2)-1 mod n = 11391, which is a unit, inverse 5696.

(2^7 * 89) divides n-1.

(2^7 * 89)^2 > n.

n is prime by Pocklington's theorem.