Take b = 2.

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

13 is prime.

b^((n-1)/13)-1 mod n = 125, which is a unit, inverse 942.

2 is prime.

b^((n-1)/2)-1 mod n = 1611, which is a unit, inverse 806.

(2^2 * 13) divides n-1.

(2^2 * 13)^2 > n.

n is prime by Pocklington's theorem.