Take b = 3.

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

7 is prime. b^((n-1)/7)-1 mod n = 3, which is a unit, inverse 85.

3 is prime. b^((n-1)/3)-1 mod n = 106, which is a unit, inverse 6.

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

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

n is prime by Pocklington's theorem.