Solution :
Let , , and be integers satisfying , and . Then
divides 3 – 2 = 1 and,
divides 2 – 1 = 1.
Therefore, . Since , , and are different integers, they are three consecutive integers satisfying .
Suppose that and a is an integer. Then
divides 5 – 1 = 4
divides 5 – 2 = 3
divides 5 – 3 = 2
Therefore, , , and . It can be easily seen that there is a unique integer a satisfying these conditions. Thus, the equation has at most one integer solution.