Take b = 3.

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

13 is prime.

b^((n-1)/13)-1 mod n = 1023, which is a unit, inverse 606.

7 is prime.

b^((n-1)/7)-1 mod n = 1237, which is a unit, inverse 967.

(7 * 13) divides n-1.

(7 * 13)^2 > n.

n is prime by Pocklington's theorem.