There are $n\geq2$ numbers $x_1, x_2, \ldots, x_n$ such that $x^2_i=1 (1\leq i\leq n)$ and $$x_1x_2+x_2x_3+\dots+x_{n-1}x_n+x_nx_1=0.$$Prove that $n$ is divisible with $4$.
Source: Moldova EGMO TST 2022
Tags: number theory
There are $n\geq2$ numbers $x_1, x_2, \ldots, x_n$ such that $x^2_i=1 (1\leq i\leq n)$ and $$x_1x_2+x_2x_3+\dots+x_{n-1}x_n+x_nx_1=0.$$Prove that $n$ is divisible with $4$.