amparvardi wrote: Given three distinct positive integers such that one of them is the average of the two others. Can the product of these three integers be the perfect 2008th power of a positive integer? Yes, choose for example the three numbers : $a^{669}(a-b)^{670}(a+b)^{669}$, $a^{670}(a-b)^{669}(a+b)^{669}$, $a^{669}(a-b)^{669}(a+b)^{670}$, where $a>b$ are any two positive integers.

pco's idea is very nice! I have an easy solution. Let three distinct positive integers are x,2x,3x.Product of these three integers is 2・3・x^3.So if x is 2^669・3^669, the product is the perfect 2008th power of a positive integer. 2^669・3^669, 2^700・3^669, 2^669・3^700 are satisfied the condition.