Take b = 2.

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

61 is prime.

b^((n-1)/61)-1 mod n = 1524, which is a unit, inverse 3921.

5 is prime.

b^((n-1)/5)-1 mod n = 2004, which is a unit, inverse 516.

(5^2 * 61) divides n-1.

(5^2 * 61)^2 > n.

n is prime by Pocklington's theorem.