Take b = 3.

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

13 is prime.

b^((n-1)/13)-1 mod n = 48873, which is a unit, inverse 3336.

11 is prime.

b^((n-1)/11)-1 mod n = 31686, which is a unit, inverse 34566.

5 is prime.

b^((n-1)/5)-1 mod n = 47886, which is a unit, inverse 6888.

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

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

n is prime by Pocklington's theorem.