Take b = 3.

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

5 is prime.

b^((n-1)/5)-1 mod n = 31, which is a unit, inverse 581.

3 is prime.

b^((n-1)/3)-1 mod n = 1726, which is a unit, inverse 24.

(3^2 * 5^2) divides n-1.

(3^2 * 5^2)^2 > n.

n is prime by Pocklington's theorem.