Take b = 2.

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

101 is prime.

b^((n-1)/101)-1 mod n = 16333, which is a unit, inverse 1114.

3 is prime.

b^((n-1)/3)-1 mod n = 12853, which is a unit, inverse 10259.

(3^3 * 101) divides n-1.

(3^3 * 101)^2 > n.

n is prime by Pocklington's theorem.