Solving $x_1+x_2=x_3^2, x_2+x_3=x_4^2, x_3+x_4=x_5^2,x_4+x_5=x_1^2, x_5+x_1=x_2^2$ in reals 
find answers of this system of equations in real numbers$$
\left\{ 
\begin{array}{c}
x_1+x_2=x_3^2 \\ 
x_2+x_3=x_4^2  \\
x_3+x_4=x_5^2   \\
x_4+x_5=x_1^2    \\
x_5+x_1=x_2^2  
\end{array}
\right. 
$$

Things I have done: by observation  I was able to see that $x_i=0$ and $x=2$ are answers.
but for solving it in a better this was my idea: W.L.O.G i assumed that $x_1^2 \ge x_2^2$. So $$x_1^2\ge x_2^2 \rightarrow x_4+x_5\ge x_5+x_1 \rightarrow x_4\ge x_1$$ but I ran in to problem because $x_1$ could be negative so I can't conclude from $x_4\ge x_1$ that $x_4^2\ge x_1^2$.
 A: Assume, $x_1$ is the biggest of $x_1, x_2, x_3, x_4, x_5$.
Then $x_1+x_2 = x_3^2 \ge 0$ and $x_1 \ge x_2$, therefore $x_1\ge 0$ and $x_1^2\ge x_2^2$.
Hence, $x_4+x_5 \ge x_5 + x_1$, therefore $x_4 \ge x_1$, or $x_4=x_1$. We obtained that $x_4$ is also the biggest of those numbers.
In the same way you prove, that $x_4=x_2$, then $x_2 = x_5%$ and $x_5=x_1$.
So all numbers are equal to each other and they are solutions of $x^2-2x=0$.
A: There's probably a more clever solution.
Case 1: Suppose one of the $x_i$ is $0$: say $x_1$ is $0$. Then,
$$
x_2-x_4=(x_2+x_3)-(x_3+x_4)=x_4^2-x_5^2=(x_4-x_5)(x_4+x_5)=(x_4-x_5)x_1^2=0
$$
so that $x_2=x_4$. This implies
$$
-x_2=x_1-x_2=x_1-x_4=(x_1+x_5)-(x_4+x_5)=x_2^2-x_1^2=(x_2-x_1)(x_2+x_1)=x_2^2.
$$
This means $x_2=0$ or $x_2=-1$. But $x_3^2=x_1+x_2=x_2$ so $x_2\geq 0$ therefore we infer that $x_2=x_4=0$. From here, it's easy to see that $x_3=x_5=0$ as well.
Case 2: each $x_i$ is nonzero. Label the given equations as (1)-(5). From (1) and (2),
$$
x_1-x_3=(x_3-x_4)x_5^2.
$$ 
Suppose $x_1>x_3$, then because $x_5^2>0$, we have $x_3>x_4$. In particular, we have $x_1>x_4$, which implies $x_2>x_1$ thanks to equations (4) and (5). So we have $x_2>x_1>x_3>x_4$. Equations (2) and (3) now imply that $x_4>x_5$. But then equations (3) and (4) imply $x_5>x_1$. So you have the impossibility $x_2>x_1>x_3>x_4>x_5>x_1$.
Supposing $x_1<x_3$ will lead to a similar contradiction as in the previous paragraph.
It must be that $x_1=x_3$ so that $x_3=x_4$. But then $x_1=x_4$ implies $x_1=x_2$. In turn, $x_2=x_4$ implies $x_4=x_5$. So $x_1=x_2=x_3=x_4=x_5=x$ for some nonzero $x$. You can now easily see that $x=2$.
Conclusion: 2 solutions $x_i=0$ $\forall i$ and $x_i=2$ $\forall$ $i$.
