Take b = 2.

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

73 is prime.

b^((n-1)/73)-1 mod n = 7302, which is a unit, inverse 2504.

5 is prime.

b^((n-1)/5)-1 mod n = 239, which is a unit, inverse 5022.

(5 * 73) divides n-1.

(5 * 73)^2 > n.

n is prime by Pocklington's theorem.