Take b = 3.

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

1663 is prime.

b^((n-1)/1663)-1 mod n = 1701215345, which is a unit, inverse 1534051064.

347 is prime.

b^((n-1)/347)-1 mod n = 1380226035, which is a unit, inverse 1288538855.

(347 * 1663) divides n-1.

(347 * 1663)^2 > n.

n is prime by Pocklington's theorem.