Take b = 2.

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

47 is prime.

b^((n-1)/47)-1 mod n = 4342, which is a unit, inverse 3509.

3 is prime.

b^((n-1)/3)-1 mod n = 3446, which is a unit, inverse 2235.

(3^3 * 47) divides n-1.

(3^3 * 47)^2 > n.

n is prime by Pocklington's theorem.