Take b = 3.

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

37 is prime.

b^((n-1)/37)-1 mod n = 1308, which is a unit, inverse 81.

3 is prime.

b^((n-1)/3)-1 mod n = 807, which is a unit, inverse 1729.

(3^3 * 37) divides n-1.

(3^3 * 37)^2 > n.

n is prime by Pocklington's theorem.