Take b = 2.

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

103 is prime.

b^((n-1)/103)-1 mod n = 8786, which is a unit, inverse 1634.

29 is prime.

b^((n-1)/29)-1 mod n = 5957, which is a unit, inverse 3102.

(29 * 103) divides n-1.

(29 * 103)^2 > n.

n is prime by Pocklington's theorem.