Take b = 2.

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

107 is prime.

b^((n-1)/107)-1 mod n = 83987, which is a unit, inverse 561110.

19 is prime.

b^((n-1)/19)-1 mod n = 247037, which is a unit, inverse 868572.

(19 * 107) divides n-1.

(19 * 107)^2 > n.

n is prime by Pocklington's theorem.