Solve the system of equations $x_i^2=x_{i+1}+x_{i+2}$ 
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.
 A: 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.  
A: Define $x_{k + n} = x_k$ for any $k \in \mathbb{N}_+$ and $m = \min\limits_{1 \leqslant k \leqslant n} x_k$, $M = \max\limits_{1 \leqslant k \leqslant n} x_k$. Since for some $k$,$$
M^2 = x_k + x_{k + 1} \leqslant 2M,
$$
then $M \leqslant 2$. For some other $k$,$$
m^2 = x_k + x_{k + 1} \leqslant 2M \leqslant 4,
$$
then $m \geqslant -2$.

Now it will be proved that $m \geqslant 0$. Suppose $m < 0$, without loss of generality, assume that $x_2 = m$. Note that$$
-2 \leqslant m < 0 \Longrightarrow (m + 2)m \leqslant 0\\\Longrightarrow m < m^2 + m \leqslant -m \Longrightarrow |m^2 + m| \leqslant |m| = -m.
$$
It will be proved by induction that$$
x_{2k} \leqslant m^2 + m \leqslant -m,\ x_{2k + 1} \geqslant -m > 0. \quad \forall k \geqslant 1 \tag{1}
$$
For $k = 1$, $x_2 = m \leqslant m^2 + m \leqslant -m$ and $x_3 = x_1^2 - x_2 \geqslant -x_2 = -m$. Assume that (1) holds for $k$. Because $|x_{2k}| \leqslant |m|$ and $x_{2k + 1} \geqslant -m > 0$, then$$
x_{2k + 2} = x_{2k}^2 - x_{2k + 1} \leqslant m^2 - (-m) = m^2 + m \leqslant -m,\\
x_{2k + 3} = x_{2k + 1}^2 - x_{2k + 2} \geqslant m^2 - (m^2 + m) = -m.
$$
End of induction.
If $n$ is odd, suppose $n = 2l + 1$, then$$
0 > -m = x_2 = x_{2l + 3} \geqslant -m > 0,
$$
a contradiction. If $n$ is even, suppose $n = 2l$, then$$
x_1 = x_{2l + 1} \geqslant -m > 0 \Longrightarrow x_3 = x_1^2 - x_2 = x_1^2 - m \geqslant m^2 - m,\\
x_4 = x_2^2 - x_3 = m^2 - x_3 \leqslant m^2 - (m^2 - m) = m.
$$
Note that $x_4 \geqslant m$, thus in fact, $x_4 = m$, which implies all the inequalities that have appeared in this part so far, i.e. from (1) to $x_4 \leqslant m$, are also equalities. Particularly,$$
m^2 - m = x_3 = -m \Longrightarrow m = 0,
$$
a contradiction. Therefore, $m \geqslant 0$.

Now, because $0 \leqslant x_k \leqslant 2$ for each $k$, then$$
\sum_{k = 1}^n (x_k - 1)^2 \leqslant \sum_{k = 1}^n 1 = n.
$$
However, this inequality is, in fact, an equality as is pointed out by @Adrian, which implies $x_k = 0$ or $2$ for each $k$. If there exists $k_0$ such that $x_{k_0} = 0$, then $x_{k_0 + 1} + x_{k_0 + 2} = 0$ implies $x_{k_0 + 1} = x_{k_0 + 2} = 0$, and by induction, all $x_k$ equals $0$. If there exists $k_0$ such that $x_{k_0} = 2$, analogous deduction shows that all $x_k$ equals $2$.
Therefore, all solutions are $(0, \cdots, 0)$ and $(2, \cdots, 2)$.
A: We can eliminate $x_3=x_1^2-x_2$, $x_4=x_2^2-x_3,\ldots ,x_{n}=x_{n-2}^2-x_{n-1}$. Then we are left with just two equations in the variables $x_1$ and $x_2$. Taking resultants we obtain a polynomial in one variable. For example, for $n=5$ we obtain a quadratic equation in $x_1$, namely
$$
x_1^2(2x_2^2+1)+x_1-x_2(x_2^3 + 2x_2^2 + x_2 + 1)=0,
$$
which we can solve, and then substitute in the other equation. In this case the resultant has just two real solutions. So we only obtain $(x_1,x_2)=(0,0),(2,2)$.
I suppose one can prove this in general.
However, we obtain a curve $f(x,y)=0$, which is difficult to solve algebraically.
A: This is an example of a non-linear difference equation due to the quadratic term in $x_i^2$ 
These are more difficult to solve analytically than linear difference equations. Usually they are examined via numerical methods.
See http://de2de.synechism.org/c1/sec15.pdf - page 2 for a similar example.
