Argentine National Olympiad 2016

For each $n\ge 3$ there exists an arithmetic progression of length $n$ with these properties. The solution uses the remark that if $l\in \mathbb{N}$ is divisible by a prime $p$ but not by $p^2$—in which case we say that $l$ is exactly divisible by $p$—then $l$ is not a perfect power.

Start a construction by choosing two primes $p$ and $q$ such that $n<p<q$. Consider the arithmetic progression $P$ with first term $p$, common difference $q-p$ and length $n$; its terms are $a_k=p+k(q-p)$, $k=0,1,\dots ,n-1$. Observe that $a_0=p$ is the only term divisible by $p$. Indeed, if $p$ divides $a_k$ with $k\ge 1$ then $p$ divides $k$ or $q-p$. Both are impossible: Since $p$ is a prime, it divides neither $k$ (as $0<k<n<p$) nor $q-p$ (as $q$ is a prime greater than $p$). Similarly, $a_1=q$ is the only term divisible by $q$. If $q$ divides $a_k$ with $k\ge 2$ then $q$ divides $k-1$ or $p$. Neither one is possible as $q$ is prime and $0<k-1<n<q$, $p<q$. In addition, $q$ does not divide $a_0=p$ since $p<q$.

Next, let $A=a_0{a}_1\dots a_{n-1}$ be the product of the terms of $P$. By the above, $A$ is exactly divisible by $p$ and $q$. Multiply each $a_k$ by $A$ to obtain a new progression $a_{0}A,a_{1}A,\dots ,a_{n-1}A$ of length $n$. The product of its terms $a_0{a}_1\dots a_{n-1}{A}^{\prime \prime }=A^{n+1}$ is a perfect $(n+1)$-st power. Because $p$ divides $a_k$ only for $k=0$ and $A$ is exactly divisible by $p$, it follows that each of the terms $a_{1}A,\dots ,a_{n-1}A$ is also exactly divisible by $p$, hence not a perfect power. Similarly, since $a_0=p$ is not divisible by $q$ and $A$ is exactly divisible by $q$, the first term $a_{0}A$ of the new progression is exactly divisible by $q$. So $a_{0}A$ is not a perfect power, which completes the justification that the progression $a_{0}A,a_{1}A,\dots ,a_{n-1}A$ satisfies the given conditions.