Take b = 2.

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

1019 is prime.

b^((n-1)/1019)-1 mod n = 1064388, which is a unit, inverse 1785403.

37 is prime.

b^((n-1)/37)-1 mod n = 2595732, which is a unit, inverse 72503.

(37 * 1019) divides n-1.

(37 * 1019)^2 > n.

n is prime by Pocklington's theorem.