Solve a system of equation with absolute value and rotation Solve
$\begin{cases} \left|x_1-x_2\right|=\left|x_2-x_3\right|=...=\left|x_{2018}-x_1\right|,\\ x_1+x_2+...+x_{2018}=2018. \end{cases}$
I think there must be such a way to solve systems of equation with the form of rotation and absolute value like this.
I have difficulty in solving the first equation. Since to me, there are quite a lot of cases to consider, for example, $\left|x_1-x_2\right|=\left|x_2-x_3\right|$ leads to $x_1-x_2=x_2-x_3$ and $x_1-x_2=x_3-x_2$ and so on.

Here is the answer to verify my comment.
Let $|x_{1} - x_{2}| = C \geq 0$ and note that
\begin{align}
|x_{1} - x_{2}| = C  \implies x_{1} = \pm C + x_{2}
\end{align}
Similarly, we have
\begin{align}
 x_{1} &= \pm C + x_{2} \\
 x_{2} &= \pm C + x_{3} \\
 & \; \; \vdots \\
 x_{2017} &= \pm C + x_{2018} \\
 x_{2018} &= \pm C + x_{1} \\
\end{align}
Keep substituting, we obtain
\begin{align}
x_{1} &= \pm C + x_{2} \\
      &= \pm C \pm C + x_{3} \\
      & \; \; \vdots \\
      &= \pm C \pm C \pm \cdots \pm C + x_{1}
\end{align}
Thus, we have
\begin{align}
\pm 2018C = 0 \implies C = 0
\end{align}
It means that $x_{1} = x_{2} \cdots = x_{2018}$ and let them be $A$.
So now, we use the second equation
\begin{align}
x_{1} + x_{2} + \cdots + x_{2018} &= 2018 \\
2018A &= 2018 \\
A &= 1
\end{align}
Conclusion: $x_{1} = x_{2} \cdots = x_{2018} = 1$
