TITLE: Solve the system of equations $x_i^2=x_{i+1}+x_{i+2}$
QUESTION [5 upvotes]: Solve the system of equations$$\left\{\begin{align*}x_1^2&=x_2+x_3\\x_2^2&=x_3+x_4\\&\ \ \vdots\\x_n^2&=x_1+x_2\end{align*}\right.$$
  where $x_i \in \mathbb R$.

I solved the case where $\forall x_i>0$. In this case $x_1 = x_2 = \cdots = x_n = 2$.(Let $a=\max {x_k}$ then $a^2\le2a$ and let $b=\min {x_k}$ then $b^2\ge2b$. Then $a=b=2$)
And $(0;0;...;0)$ another solution.
But if $x_i \in \mathbb R$ I need help.

REPLY [3 votes]: Some observations:
$\textbf{Observation 1:}$ If $x:=(x_1,...,x_n)$ is a solution to the system, then $$x\in\mathbb{S}(a,\sqrt{n}):=\{x\in\mathbb{R}^n:\|x-a\|=\sqrt{n}\}$$ where $a:=(1,1,...,1)$ and $\|\cdot\|$ is the usual Euclidean norm.
From the system, adding both sides one gets
$$x^2_1+ \cdots +x^2_n=2(x_1+\cdots+x_n)\Rightarrow (x_1-1)^2+\cdots+(x_n-1)^2=n$$ hence the claim.
$\textbf{Observation 2:}$ We can write the system as 
$$A(x)x=Bx$$
where $A(x)$ is a diagonal matrix with entries $a_{ii}=x_i$ for  $i=1,\dots,n$ and $B$ is a matrix whose entries are $0$ and $1$ in a specified order (which you get from your equations on the right side). Therefore
$$(A(x)-B)x=0$$
is a well defined equation for each given $x$. Clearly $x=0$ is a trivial solution. If $x\neq 0$ then this implies that $\det(A(x)-B)=0$. For example in the case $n=2$ this determinant equation gives $$x_1x_2=x_1+x_2$$
which combined with the original equations $x^2_1=x^2_2=x_1+x_2$ yields the nontrivial solution $x_1=x_2=2$. You can go for higher values of $n$ though computing the determinant becomes more laborious.