Take b = 2.

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

29 is prime.

b^((n-1)/29)-1 mod n = 12013, which is a unit, inverse 4750.

3 is prime.

b^((n-1)/3)-1 mod n = 1982, which is a unit, inverse 19059.

2 is prime.

b^((n-1)/2)-1 mod n = 29579, which is a unit, inverse 14790.

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

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

n is prime by Pocklington's theorem.