Take b = 2.

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

37 is prime.

b^((n-1)/37)-1 mod n = 743, which is a unit, inverse 1970.

13 is prime.

b^((n-1)/13)-1 mod n = 1303, which is a unit, inverse 1500.

(13 * 37) divides n-1.

(13 * 37)^2 > n.

n is prime by Pocklington's theorem.