Take b = 2.

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

7 is prime. b^((n-1)/7)-1 mod n = 18, which is a unit, inverse 39.

5 is prime. b^((n-1)/5)-1 mod n = 209, which is a unit, inverse 161.

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

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

n is prime by Pocklington's theorem.