$x_i \geq 0 $ Let $S=x_1+...+x_n$ $x_i=(S-x_i)^{2018}$ If $x_i \geq x_j \to S-x_i \geq S-x_j \to x_i=x_j \to (nx)^{2018}=x \to x=\frac{1}{n\sqrt[2017]{n}},0$

RagvaloD wrote: $x_i \geq 0 $ Let $S=x_1+...+x_n$ $x_i=(S-x_i)^{2018}$ If $x_i \geq x_j \to S-x_i \geq S-x_j \to x_i=x_j \to (nx)^{2018}=x \to x=\frac{1}{n\sqrt[2017]{n}},0$ Flipped inequality sign I guess?

RagvaloD wrote: $x_i \geq 0 $ Let $S=x_1+...+x_n$ $x_i=(S-x_i)^{2018}$ If $x_i \geq x_j \to S-x_i \geq S-x_j \to x_i=x_j \to (nx)^{2018}=x \to x=\frac{1}{n\sqrt[2017]{n}},0$ 1. The problem says "integer solutions". Your solution $\frac{1}{n\sqrt[2017]{n}}$ does not exactly look very integer. 2. But if you don't assume that $x_i$ are all integer (as you seem not to assume based on your answer), then you can't go from $(S-x_i)^{2018} \geq (S-x_j)^{2018}$ to $S-x_i \geq S-x_j$. 3. $(nx)^{2018}$ should be really $((n-1)x)^{2018}$. The correct answer is: for $n=2$ there are two solutions: $x_1=x_2=0$ and $x_1=x_2=1$; and for $n \geq 3$ there is only one solution: $x_i=0$ for $1 \leq i \leq n$.

$ x_1 = (x_2 + x_3 + x_4 +.. + x_n)^{n} x_2 = (x_1 + x_3 + x_4 +.. + x_n)^{n} . . x_n = (x_1 + x_2 + x_3 +.. + x_{n - 1})^{n} $ Find all solution

$x_i=0$ for all $i$ is clearly a solution. Now suppose, some $x_i \neq 0$. Note all $x_i$ are non-negative, so consider the maximum of them, say $x_1$ is the maximum, then $x_i \geq x_1^{2018}\geq x_1$ for $i>1$ implies $x_i=1=x_1$ for all $i>1$. Then the only possibility is $n=2$.

From the equations, all $x_i \ge 0$. Take the minimum $x_1$ and suppose $x_1 = c^{2018}$ where $c$ is a positive integer. Then $c^{2018} = (S-c^{2018})^{2018} \ge ((n-1)c^{2018})^{2018}$ which is equivalent to $c \ge (n-1)c^{2018}$. If $n \ge 3$ then we must have $c = 0$ in which case $S =0$ gives all $x_i = 0$. If $n=2$ then $x_1 = 1, x_2 = 1$ or $x_1 = 0, x_2 = 0$.

Consider $S=x_1+x_2...+x_n$ then clearly we have $x_n = (S-x_n)^{2018}$, we know $(S-x_i)^{2018} = ((S-x_1)^{2018}+...(S-x_n)^{2018})^{2018}$ (if i it's different of 1 and n of course) then we have $S-x_i = (S-x_1)^{2018}...+(S-x_n)^{2018}$ assume $WLOG$ that $S-x_j=n, n\geq2$ (with i different of j) then $(S-x_i)^{2018} \geq n^{2018}$ but $(S-x_j) \geq (S-x_i)^{2018}$ since every $(S-x_k)^{2018}\geq 0$ this implies $n\geq n^{2018}$ absurd then we have $1\geq (S-x_k)$ for every k. Assume $(S-x_i) = 1$ then we necessarily have only another $(S-x_j)=1$ other way we obtain a absurd again. if we have also a $S-x_k = 0$ we have a absurd too because we'll have $0\geq1$ that means in the case $S-x_i=1$ we have only n=2, then how $x_i=x_j$ and $(S-x_i)=1$ we must have $x_1=x_2=1$. Finally if we have only $S-x_k=0$ we have that every $x_j$ it's equal, and how $x_1+x_2...+x_i=0$ then every $x_k=0$ Note that since every $x_k$ it's a perfect square, then every $x_k$ it's non negative this also implies $S-x_k$ non negative too

look for the smallest number between these numbers.then at n> 2 it will all be zero.