Take b = 2.

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

47 is prime.

b^((n-1)/47)-1 mod n = 8557480, which is a unit, inverse 10629675.

29 is prime.

b^((n-1)/29)-1 mod n = 12453323, which is a unit, inverse 9887646.

5 is prime.

b^((n-1)/5)-1 mod n = 5784941, which is a unit, inverse 1191016.

(5^2 * 29 * 47) divides n-1.

(5^2 * 29 * 47)^2 > n.

n is prime by Pocklington's theorem.