Take b = 3.

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

79 is prime.

b^((n-1)/79)-1 mod n = 62063, which is a unit, inverse 15871.

2 is prime.

b^((n-1)/2)-1 mod n = 80895, which is a unit, inverse 40448.

(2^10 * 79) divides n-1.

(2^10 * 79)^2 > n.

n is prime by Pocklington's theorem.