Take b = 3.

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

787 is prime.

b^((n-1)/787)-1 mod n = 2240226, which is a unit, inverse 1632100.

73 is prime.

b^((n-1)/73)-1 mod n = 65535, which is a unit, inverse 1138904.

(73 * 787) divides n-1.

(73 * 787)^2 > n.

n is prime by Pocklington's theorem.