Take b = 3.

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

47 is prime.

b^((n-1)/47)-1 mod n = 2689, which is a unit, inverse 1463.

5 is prime.

b^((n-1)/5)-1 mod n = 2743, which is a unit, inverse 1947.

(5 * 47) divides n-1.

(5 * 47)^2 > n.

n is prime by Pocklington's theorem.