Take b = 2.

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

199 is prime.

b^((n-1)/199)-1 mod n = 7685141, which is a unit, inverse 16164448.

83 is prime.

b^((n-1)/83)-1 mod n = 13722120, which is a unit, inverse 22679122.

(83 * 199) divides n-1.

(83 * 199)^2 > n.

n is prime by Pocklington's theorem.