Take b = 2.

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

79 is prime.

b^((n-1)/79)-1 mod n = 5754, which is a unit, inverse 92.

5 is prime.

b^((n-1)/5)-1 mod n = 4052, which is a unit, inverse 4943.

(5^2 * 79) divides n-1.

(5^2 * 79)^2 > n.

n is prime by Pocklington's theorem.