Take b = 2.

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

103 is prime.

b^((n-1)/103)-1 mod n = 5010448, which is a unit, inverse 270428.

23 is prime.

b^((n-1)/23)-1 mod n = 4901119, which is a unit, inverse 803080.

7 is prime.

b^((n-1)/7)-1 mod n = 2957565, which is a unit, inverse 1102302.

(7^3 * 23 * 103) divides n-1.

(7^3 * 23 * 103)^2 > n.

n is prime by Pocklington's theorem.