Problem

Source: Mexico 2023/2

Tags: combinatorics



The numbers from $1$ to $2000$ are placed on the vertices of a regular polygon with $2000$ sides, one on each vertex, so that the following is true: If four integers $A, B, C, D$ satisfy that $1 \leq A<B<C<D \leq 2000$, then the segment that joins the vertices of the numbers $A$ and $B$ and the segment that joins the vertices of $C$ and $D$ do not intersect inside the polygon. Prove that there exists a perfect square such that the number diametrically opposite to it is not a perfect square.