Take b = 2.

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

27109 is prime.

b^((n-1)/27109)-1 mod n = 407331351, which is a unit, inverse 614376904.

67 is prime.

b^((n-1)/67)-1 mod n = 26234084, which is a unit, inverse 259062632.

(67 * 27109) divides n-1.

(67 * 27109)^2 > n.

n is prime by Pocklington's theorem.