Task 4. Given are three different natural numbers, one of which is equal to the half-sum of the other two. Is it possible for the product of these three numbers to be exactly the 2008th power of a natural number?
Solution: Let's choose three different natural numbers, one of which is equal to the half-sum of the other two; for example, 1, 2, and 3. Their product is 6. If we multiply each of the numbers 1, 2, and 3 by $6^{n}$, we will get new three numbers $6^{n}, 2 \cdot 6^{n}$, and $3 \cdot 6^{n}$, which satisfy the condition that one of them is equal to the half-sum of the other two. These numbers have a product of $6^{3n+1}$. It remains to choose $n$ such that $3n+1$ is divisible by 2008. Since $3n+1=2008$ when $n=669$, one of the possible solutions is

$$
6^{669}, 2 \cdot 6^{669}, 3 \cdot 6^{669}
$$