Take b = 7.

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

3 is prime.

b^((n-1)/3)-1 mod n = 496799, which is a unit, inverse 497952.

2 is prime.

b^((n-1)/2)-1 mod n = 995327, which is a unit, inverse 497664.

(2^12 * 3^5) divides n-1.

(2^12 * 3^5)^2 > n.

n is prime by Pocklington's theorem.