Take b = 2.

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

7 is prime.

b^((n-1)/7)-1 mod n = 4099, which is a unit, inverse 1222.

5 is prime.

b^((n-1)/5)-1 mod n = 9779, which is a unit, inverse 1149.

(5^3 * 7) divides n-1.

(5^3 * 7)^2 > n.

n is prime by Pocklington's theorem.