Take b = 3.

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

17 is prime.

b^((n-1)/17)-1 mod n = 2804013, which is a unit, inverse 37564.

13 is prime.

b^((n-1)/13)-1 mod n = 2844423, which is a unit, inverse 3844608.

(13^2 * 17) divides n-1.

(13^2 * 17)^2 > n.

n is prime by Pocklington's theorem.