Take b = 2.

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

23 is prime.

b^((n-1)/23)-1 mod n = 363871, which is a unit, inverse 227161.

19 is prime.

b^((n-1)/19)-1 mod n = 126065, which is a unit, inverse 189376.

13 is prime.

b^((n-1)/13)-1 mod n = 460982, which is a unit, inverse 337377.

(13 * 19 * 23) divides n-1.

(13 * 19 * 23)^2 > n.

n is prime by Pocklington's theorem.