Take b = 3.

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

13 is prime.

b^((n-1)/13)-1 mod n = 2941, which is a unit, inverse 2842.

5 is prime.

b^((n-1)/5)-1 mod n = 2013, which is a unit, inverse 2928.

(5^3 * 13) divides n-1.

(5^3 * 13)^2 > n.

n is prime by Pocklington's theorem.