Take b = 5.

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

7 is prime.

b^((n-1)/7)-1 mod n = 619, which is a unit, inverse 162.

3 is prime.

b^((n-1)/3)-1 mod n = 254, which is a unit, inverse 363.

2 is prime.

b^((n-1)/2)-1 mod n = 671, which is a unit, inverse 336.

(2^5 * 3 * 7) divides n-1.

(2^5 * 3 * 7)^2 > n.

n is prime by Pocklington's theorem.