Take b = 2.

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

13 is prime.

b^((n-1)/13)-1 mod n = 77, which is a unit, inverse 490.

5 is prime.

b^((n-1)/5)-1 mod n = 162, which is a unit, inverse 779.

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

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

n is prime by Pocklington's theorem.