Find the largest positive integer $n<100$, such that there exists an arithmetic progression of positive integers $a_1,a_2,...,a_n$ with the following properties.
$1)$ All numbers $a_2,a_3,...,a_{n−1}$ are powers of positive integers, that is numbers of the form $j^k$, where $j \geq 1$ and $k \geq 2$ are integers.
$2)$ The numbers $a_1$ and $a_n$ are not powers of positive integers.
Partial solution. First observe that there are arbitrary long arithmetic sequences of perfect powers. Indeed, assume for some $n\ge1$ that $a_k=mk+d$ (with $m,d\in\Bbb N$) is an arithmetic sequence where for $1\le k\le n$, $a_k$ is a perfect power (say, $a_k=b_k^{c_k}$ with $c_k\ge 2$). Then we can find an arithmetic sequence $a_k'=m'k+d'$ such that $a_k'$ is a perfect power for $1\le k\le n+1$. To do so we just let $m'=a_{n+1}^rm$ and $d'=a_{n+1}^rd$ where $r=\operatorname{lcm}(c_1,\ldots,c_n)$. That makes $a_k'=a_ka_{n+1}^r=(b_ka_{n+1}^{r/c_k})^{c_k}$ a perfect power for $1\le k\le n$, and of course also $a_{n+1}'=a_n^{r+1}$ is a perfect power. In particular, we can pick the exponents $c_k'=c_k$ for $1\le k\le n$ and $c_{n+1}'=\operatorname{lcm}(c_1,\ldots,c_n)+1$.
However, regarding the OP's question we have to watch out if $a_0'$ and/or $a_{n+2}'$ "accidentally" become perfect powers.
From a quick search, this question appears to still be open. Bounds on $n$ exist for some restricted sets of powers $k$, and it is known that the number of progressions of length $n$ for $n\geq 6$ with relatively coprime terms (if any) is finite.
See for instance:
