amparvardi wrote: Five distinct positive integers form an arithmetic progression. Can their product be equal to $a^{2008}$ for some positive integer $a$ ? Yes, choose for example the five numbers : $a^{804}(a+r)^{803}(a+2r)^{803}(a+3r)^{803}(a+4r)^{803}$, $a^{803}(a+r)^{804}(a+2r)^{803}(a+3r)^{803}(a+4r)^{803}$, $a^{803}(a+r)^{803}(a+2r)^{804}(a+3r)^{803}(a+4r)^{803}$, $a^{803}(a+r)^{803}(a+2r)^{803}(a+3r)^{804}(a+4r)^{803}$, $a^{803}(a+r)^{803}(a+2r)^{803}(a+3r)^{803}(a+4r)^{804}$, Where $a,r$ are any two positive integers.

Let $x=2^{401}3^{803}5^{803}$. We choose the numbers $x, 2x, 3x, 4x, 5x$.

RaleD wrote: Let $x=2^{401}3^{803}5^{803}$. We choose the numbers $x, 2x, 3x, 4x, 5x$. Simillar solution work for next problem: Let $n,k \in \mathbb{N}, gcd(n,k)=1$ prove that exist arithmetic progression with $n$ natural terms such that product of it's terms eqals $a^k$ for any natural $a$.