Problem. For each integer $n\ge 2$, find all integer solutions of the following system of equations: $$x_1=(x_2+x_3+x_4+\dots +x_n{)}^{2018},$$ $$x_2=(x_1+x_3+x_4+\dots +x_n{)}^{2018},$$ $$⋮$$ $$x_n=(x_1+x_2+x_3+\cdots +x_{n-1}{)}^{2018}.$$

Origin: Iberoamerican Mathematical Olympiad 2018, problem 1.

Solution. First, notice that $y^{2018}\ge 0$ for any $y$, thus the right sides of all equations are non-negative. Therefore, so are the left sides, and so all $x_i\ge 0$, $1\le i\le n$.

Now, let's consider $x_{i_0}$ which is the smallest among $x_1$, \ldots, $x_n$. Then we have $$x_{i_0}={\left(\sum _{i\ne i_0}{x}_i\right)}^{2018}\ge {((n-1)\cdot x_{i_0})}^{2018}=(n-1{)}^{2018}\cdot x_{i_0}^{2018}.$$ This can only be true either when $x_{i_0}=0$, or when $x_{i_0}=1$ and $n-1=1$. Consider these two cases separately.

• $x_{i_0}=0$. Then ${\left(\sum _{i\ne i_0}{x}_i\right)}^{2018}=0$. Then $\sum _{i\ne i_0}{x}_i=0$. Then all $x_i=0$. Finally, we can check directly that this result satisfies the original equations.
• $x_{i_0}=1$ and $n-1=1$. Then $n=2$. Without loss of generality, let's assume that $i_0=1$. Then we have $x_1=1$ and the two equations are $1=x_2^{2018}$ and $x_2=1^{2018}$. Therefore, $x_2=1$. Finally, we can check directly that this result satisfies the original equations.

To summarize, for $n=2$, there are two solutions: $x_1=x_2=0$ and $x_1=x_2=1$; for $n\ge 3$, there is only one solution: $x_i=0$ for $1\le i\le n$.