Take b = 2.

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

13 is prime.

b^((n-1)/13)-1 mod n = 1349, which is a unit, inverse 546.

5 is prime.

b^((n-1)/5)-1 mod n = 3594, which is a unit, inverse 4252.

(5^2 * 13) divides n-1.

(5^2 * 13)^2 > n.

n is prime by Pocklington's theorem.