Take b = 5.

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

17 is prime.

b^((n-1)/17)-1 mod n = 216811, which is a unit, inverse 219983.

7 is prime.

b^((n-1)/7)-1 mod n = 2008, which is a unit, inverse 111282.

3 is prime.

b^((n-1)/3)-1 mod n = 135452, which is a unit, inverse 137633.

(3^2 * 7 * 17) divides n-1.

(3^2 * 7 * 17)^2 > n.

n is prime by Pocklington's theorem.