Take b = 2.

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

103 is prime.

b^((n-1)/103)-1 mod n = 3156376229, which is a unit, inverse 2943989098.

97 is prime.

b^((n-1)/97)-1 mod n = 3073569831, which is a unit, inverse 4442220496.

89 is prime.

b^((n-1)/89)-1 mod n = 4774961049, which is a unit, inverse 346071677.

(89 * 97 * 103) divides n-1.

(89 * 97 * 103)^2 > n.

n is prime by Pocklington's theorem.