Take b = 3.

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

13 is prime.

b^((n-1)/13)-1 mod n = 2969, which is a unit, inverse 823.

2 is prime.

b^((n-1)/2)-1 mod n = 3327, which is a unit, inverse 1664.

(2^8 * 13) divides n-1.

(2^8 * 13)^2 > n.

n is prime by Pocklington's theorem.