Task 4. Does there exist an arithmetic progression of five natural numbers whose product is exactly the 2008th power of a natural number?
Solution: Let's choose an arbitrary arithmetic progression of five natural numbers; for example, $1,2,3,4$ and 5. Their product is 120. If we multiply each term by $120^{n}$, we will get a new arithmetic progression with a product of $120^{5 n+1}$. The product is an exact 2008th power when $5 n+1$ is divisible by 2008. Since $5 n+1=2 \cdot 2008$ when $n=403$, one possible progression is

$$
120^{403}, 2 \cdot 120^{403}, 3 \cdot 120^{403}, 4 \cdot 120^{403}, 5 \cdot 120^{403}
$$