Five distinct positive integers form an arithmetic progression. Can their product be equal to $a^{2008}$ for some positive integer $a$ ?
1. Let the five distinct positive integers in the arithmetic progression be \(a, a+d, a+2d, a+3d, a+4d\), where \(a\) and \(d\) are positive integers and \(d\) is the common difference.
2. The product of these five integers is:
   \[
   P = a(a+d)(a+2d)(a+3d)(a+4d)
   \]
3. We need to determine if this product can be expressed as \(a^{2008}\) for some positive integer \(a\).
4. Consider the prime factorization of \(P\). For \(P\) to be a perfect power, specifically \(a^{2008}\), the exponents in the prime factorization of \(P\) must all be multiples of 2008.
5. Let's examine the given solution. The proposed numbers are:
   \[
   a^{804}(a+r)^{803}(a+2r)^{803}(a+3r)^{803}(a+4r)^{803}
   \]
   \[
   a^{803}(a+r)^{804}(a+2r)^{803}(a+3r)^{803}(a+4r)^{803}
   \]
   \[
   a^{803}(a+r)^{803}(a+2r)^{804}(a+3r)^{803}(a+4r)^{803}
   \]
   \[
   a^{803}(a+r)^{803}(a+2r)^{803}(a+3r)^{804}(a+4r)^{803}
   \]
   \[
   a^{803}(a+r)^{803}(a+2r)^{803}(a+3r)^{803}(a+4r)^{804}
   \]
6. The product of these five numbers is:
   \[
   P = \left(a^{804}(a+r)^{803}(a+2r)^{803}(a+3r)^{803}(a+4r)^{803}\right) \times \left(a^{803}(a+r)^{804}(a+2r)^{803}(a+3r)^{803}(a+4r)^{803}\right) \times \left(a^{803}(a+r)^{803}(a+2r)^{804}(a+3r)^{803}(a+4r)^{803}\right) \times \left(a^{803}(a+r)^{803}(a+2r)^{803}(a+3r)^{804}(a+4r)^{803}\right) \times \left(a^{803}(a+r)^{803}(a+2r)^{803}(a+3r)^{803}(a+4r)^{804}\right)
   \]
7. Simplifying the exponents, we get:
   \[
   P = a^{804+803+803+803+803} \times (a+r)^{803+804+803+803+803} \times (a+2r)^{803+803+804+803+803} \times (a+3r)^{803+803+803+804+803} \times (a+4r)^{803+803+803+803+804}
   \]
   \[
   P = a^{4016} \times (a+r)^{4016} \times (a+2r)^{4016} \times (a+3r)^{4016} \times (a+4r)^{4016}
   \]
8. Notice that each term in the product has an exponent of 4016, which is a multiple of 2008. Therefore, the product can be written as:
   \[
   P = \left(a(a+r)(a+2r)(a+3r)(a+4r)\right)^{4016/2008}
   \]
   \[
   P = \left(a(a+r)(a+2r)(a+3r)(a+4r)\right)^2
   \]
9. Since \(a(a+r)(a+2r)(a+3r)(a+4r)\) is a positive integer, its square is also a positive integer. Therefore, \(P\) can indeed be expressed as \(a^{2008}\) for some positive integer \(a\).

The final answer is \( \boxed{ a^{2008} } \)