### Iberoamerican integer system of equations, 2018

June 30, 2019

Problem. For each integer $n\ge 2$$n \geq 2$, find all integer solutions of the following system of equations: ${x}_{1}=\left({x}_{2}+{x}_{3}+{x}_{4}+\dots +{x}_{n}{\right)}^{2018},$ ${x}_{2}=\left({x}_{1}+{x}_{3}+{x}_{4}+\dots +{x}_{n}{\right)}^{2018},$ $⋮$ ${x}_{n}=\left({x}_{1}+{x}_{2}+{x}_{3}+\cdots +{x}_{n-1}{\right)}^{2018}.$

Origin: Iberoamerican Mathematical Olympiad 2018, problem 1.

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

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

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

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