Balkan Mathematical Olympiad Shortlist

Letting $n=p-1$, we have $x_i^{p-1}\equiv 0$ or $1(\mathrm{mod}p)$. Therefore, the congruence $x_1^n+x_2^n+\cdots +x_p^n\equiv 0(\mathrm{mod}p)$ is true when either all $x_i$ are divisible by $p$ or no $x_i$ is divisible by $p$.

On the other hand, if no $x_i$ is divisible by $p$ we have $$\sum _{i=1}^p(x_i-x_1{)}^n=\sum _{j=0}^n\left(\sum _{i=1}^p{x}_i^j\right)(\genfrac{}{}{0ex}{}{n}{j})(-x_1{)}^{n-j}\equiv 0(\mathrm{mod}p).$$ Hence, $0,x_2-x_1,x_3-x_1,\dots ,x_p-x_1$ satisfy the condition, so, all $x_i$ are congruent modulo $p$.