Take b = 2.

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

23 is prime.

b^((n-1)/23)-1 mod n = 199, which is a unit, inverse 937.

5 is prime.

b^((n-1)/5)-1 mod n = 1059, which is a unit, inverse 588.

(5^2 * 23) divides n-1.

(5^2 * 23)^2 > n.

n is prime by Pocklington's theorem.