Take b = 3.

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

13 is prime.

b^((n-1)/13)-1 mod n = 475196, which is a unit, inverse 814521.

11 is prime.

b^((n-1)/11)-1 mod n = 157914, which is a unit, inverse 339622.

5 is prime.

b^((n-1)/5)-1 mod n = 421969, which is a unit, inverse 847141.

(5^3 * 11 * 13) divides n-1.

(5^3 * 11 * 13)^2 > n.

n is prime by Pocklington's theorem.