A clue to solve the system of equation $$\begin{cases} x +y=\sqrt[2107]{z}\\y +z=\sqrt[2107]{x}\\z +x=\sqrt[2107]{y}\end{cases}$$ 
It is obvious that $(0,0,0)$ is an answer , but How can be sure for other solution (If other solution exists ) ?
thanks in advance.
 A: There are very few real solutions.
By enforcing the substitution $x=a^{2107},y=b^{2107},z=c^{2107}$ the problem boils down to solving
$$\left\{\begin{array}{rcl}c-b^{2107}&=&a^{2107}\\ b^{2107}+c^{2107}&=&a\\b-c^{2107}&=&a^{2107}\end{array}\right.$$
and by assuming $b\neq c$, the first and the third equation imply
$$ \frac{b^{2107}-c^{2107}}{b-c}=b^{2106}+b^{2105}c+\ldots+b c^{2105}+c^{2106} = -1 $$
which is impossible since the LHS is clearly positive. By symmetry it follows that the only solutions are given by $x=y=z=0,\,$ or $x=y=z=\pm 2^{-2107/2106}$.
A: $$\begin{cases} x +y=\sqrt[2107]{z}\\y +z=\sqrt[2107]{x}\\z +x=\sqrt[2107]{y}\end{cases}\Rightarrow \begin{cases} x -y=\sqrt[2107]{y}-\sqrt[2107]{x}\\y -z=\sqrt[2107]{z}-\sqrt[2107]{y}\\z -x=\sqrt[2107]{x}-\sqrt[2107]{z}\end{cases}$$
This implies $$\sqrt[2107]{x}+x=\sqrt[2107]{y}+y=\sqrt[2107]{z}+z$$ which is satisfied by $x=y=z$ This common value is given by the first (or second or third) equation $$2x=\sqrt[2107]{x}\Rightarrow x^{1-\frac{1}{2017}}=x^{\frac{2016}{2017}}=\dfrac 12\Rightarrow x=\sqrt[2016]{\frac{1}{2^{2017}}}$$
Uniqueness.-Suppose that x is real (see the answer of Tito Piezas III) and that $x\lt y$ so $y=x+h$ where $h\gt 0$. 
The derivative of $f(x)=\sqrt[2107]{x}$ is $f'(x)=\frac{1}{2017\sqrt[2107]{x^{2016}}}\gt 0$ then $f$ is increasing.
$$\sqrt[2107]{x}+x=\sqrt[2107]{y}+y\Rightarrow\sqrt[2107]{x}=\sqrt[2107]{x+h}+h\Rightarrow h=0\text{ because f is increasing  }$$
Contradiction with $h\gt 0$.
A: The OP didn't specify real solutions. Given the system,
$$\begin{cases} 
(x+y)^n = z\\
(y+z)^n = x\\
(x+z)^n = y\end{cases}$$ 
Assume $x = y$. Hence,
$$\begin{cases} 
(2x)^n = z\\
(x+z)^n = x\\
(x+z)^n = x\end{cases}$$ 
So we just have,
$$\big(x+(2x)^n\big)^n = x$$
By the Fundamental Theorem of Algebra (FTA), this has $n^2$ complex solutions. (And as the other answers showed, $2$ of these are real.)

P.S. For an example of a system that surprisingly turned out to have extra real solutions (from a $54$-deg poly), see this post.
