Take b = 3.

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

7 is prime.

b^((n-1)/7)-1 mod n = 2662, which is a unit, inverse 317.

2 is prime.

b^((n-1)/2)-1 mod n = 3135, which is a unit, inverse 1568.

(2^6 * 7^2) divides n-1.

(2^6 * 7^2)^2 > n.

n is prime by Pocklington's theorem.