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.