Take b = 2.

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

109 is prime.

b^((n-1)/109)-1 mod n = 13439, which is a unit, inverse 22038.

5 is prime.

b^((n-1)/5)-1 mod n = 1292, which is a unit, inverse 22901.

(5 * 109) divides n-1.

(5 * 109)^2 > n.

n is prime by Pocklington's theorem.