Take b = 2.

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

7 is prime.

b^((n-1)/7)-1 mod n = 3693, which is a unit, inverse 8403.

5 is prime.

b^((n-1)/5)-1 mod n = 4911, which is a unit, inverse 5349.

(5 * 7^2) divides n-1.

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

n is prime by Pocklington's theorem.