Take b = 2.

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

107 is prime.

b^((n-1)/107)-1 mod n = 3503, which is a unit, inverse 15373.

13 is prime.

b^((n-1)/13)-1 mod n = 4955, which is a unit, inverse 2840.

(13 * 107) divides n-1.

(13 * 107)^2 > n.

n is prime by Pocklington's theorem.