Take b = 2.

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

29 is prime.

b^((n-1)/29)-1 mod n = 346, which is a unit, inverse 1321.

5 is prime.

b^((n-1)/5)-1 mod n = 711, which is a unit, inverse 100.

(5^2 * 29) divides n-1.

(5^2 * 29)^2 > n.

n is prime by Pocklington's theorem.