Take b = 7.

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

61 is prime.

b^((n-1)/61)-1 mod n = 219357, which is a unit, inverse 601026.

7 is prime.

b^((n-1)/7)-1 mod n = 963889, which is a unit, inverse 689901.

3 is prime.

b^((n-1)/3)-1 mod n = 143450, which is a unit, inverse 608055.

(3^2 * 7 * 61) divides n-1.

(3^2 * 7 * 61)^2 > n.

n is prime by Pocklington's theorem.