Take b = 11.

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

7 is prime.

b^((n-1)/7)-1 mod n = 10661, which is a unit, inverse 1552.

5 is prime.

b^((n-1)/5)-1 mod n = 8732, which is a unit, inverse 1156.

3 is prime.

b^((n-1)/3)-1 mod n = 644, which is a unit, inverse 8745.

2 is prime.

b^((n-1)/2)-1 mod n = 13439, which is a unit, inverse 6720.

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

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

n is prime by Pocklington's theorem.