Take b = 2.

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

67 is prime.

b^((n-1)/67)-1 mod n = 12682, which is a unit, inverse 21642.

3 is prime.

b^((n-1)/3)-1 mod n = 7857, which is a unit, inverse 19491.

(3 * 67) divides n-1.

(3 * 67)^2 > n.

n is prime by Pocklington's theorem.