Take b = 2.

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

109 is prime.

b^((n-1)/109)-1 mod n = 17445, which is a unit, inverse 2228.

2 is prime.

b^((n-1)/2)-1 mod n = 17657, which is a unit, inverse 8829.

(2 * 109) divides n-1.

(2 * 109)^2 > n.

n is prime by Pocklington's theorem.