Take b = 3.

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

29 is prime.

b^((n-1)/29)-1 mod n = 31, which is a unit, inverse 1213.

3 is prime.

b^((n-1)/3)-1 mod n = 825, which is a unit, inverse 1813.

(3^2 * 29) divides n-1.

(3^2 * 29)^2 > n.

n is prime by Pocklington's theorem.