Take b = 2.

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

67 is prime.

b^((n-1)/67)-1 mod n = 26615, which is a unit, inverse 5766.

29 is prime.

b^((n-1)/29)-1 mod n = 15932, which is a unit, inverse 30780.

(29 * 67) divides n-1.

(29 * 67)^2 > n.

n is prime by Pocklington's theorem.