Let $P(x)=x^{2000}-x^{1000}+1$. Do there exist distinct positive integers $a_1,\dots,a_{2001}$ such that $a_ia_j|P(a_i)P(a_j)$ for all $i\neq j$?
Proposed by A. Baranov
Answer is no. Assume the contrary and let $a_1<\cdots<a_{2001}$. Noting that $(a_i,P(a_i))=1$ for any $i$, we get that $a_i\mid P(a_j)$ for every $i\ne j$. It is evident that $(a_i,a_j)=1$ for every $i\ne j$, so we find that $a_2\cdots a_{2001}\mid P(a_1)$. This is, however, impossible as $P(a_1)=a_1^{2000}-a_1^{1000}+1\le a_1^{2000}<\textstyle \prod_{2\le i\le 2001}a_i$.