TITLE: 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
QUESTION [13 upvotes]: 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 $x_4^2\ge x_1^2$.

REPLY [8 votes]: 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 $x^2-2x=0$.