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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.