Take b = 2.

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

107 is prime.

b^((n-1)/107)-1 mod n = 275749, which is a unit, inverse 230230.

23 is prime.

b^((n-1)/23)-1 mod n = 267571, which is a unit, inverse 186914.

(23 * 107) divides n-1.

(23 * 107)^2 > n.

n is prime by Pocklington's theorem.