Take b = 2.

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

17 is prime.

b^((n-1)/17)-1 mod n = 585, which is a unit, inverse 197.

3 is prime.

b^((n-1)/3)-1 mod n = 64, which is a unit, inverse 182.

(3^2 * 17) divides n-1.

(3^2 * 17)^2 > n.

n is prime by Pocklington's theorem.