Take b = 3.

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

97 is prime.

b^((n-1)/97)-1 mod n = 7360, which is a unit, inverse 6061.

59 is prime.

b^((n-1)/59)-1 mod n = 422, which is a unit, inverse 5615.

(59 * 97) divides n-1.

(59 * 97)^2 > n.

n is prime by Pocklington's theorem.