Primality proof for n = 127:
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.