Sum all $2016$ equations together. We get $x_1^2+x_2^2+...+x_{2016}^2=2016$ if $x_1>1$ we'll have $x_2>1$ and so on.But it's impossible since $x_1^2+x_2^2+...+x_{2016}^2>2016$ Apply the same method again,we get that it's impossible to have $x_1<-1$ Therefore $x_i\in[-1,1]$ Notice that $\forall x\in[-1,1];x^2+x-1\leq x$ Therefore, $x_2\leq x_1$ , $x_3\leq x_2$ and so on.Because it's cyclic,we'll have $x_1=x_2=...=x_{2016}=1$ or $x_1=x_2=...=x_{2016}=-1$

See http://www.artofproblemsolving.com/community/c6h594720p3528050

jeneva I d'ont understand $x_1< -1$ that not implicates $x_2< -1$. For example: If $x_1=-3$ than $x_2=5$

Similar solution. By adding all equalities we get $x_1^2+x_2^2+...+x_{2016}^2=2016$ If any $x_i=-1$ then all of $x_i=-1$ so $\boxed{(-1,-1,...,-1)}$ is a solution. By multiplying the equalities $x_i(x_i+1)=(x_{i+1}+1)$ we get $x_1x_2...x_{2016}=1$ By AM-GM we get $2016=x_1^2+x_2^2+...+x_{2016}^2 \geq 2016\sqrt[2016]{(x_1...x_{2016})^2}=2016$ It's the equality case so $x_1^2=x_2^2=...=x_{2016}^2$ Let $x_k$ be the largest. \[x_k(x_k+1)=x_{k+1}+1 \leq x_k+1\]If $x_k+1<0$ then $x_i<-1$ but $x_1...x_{2016}=1 $ Contradiction $x_k+1 $ is positive. Let's divide both sides by $x_k+1$ We get $x_k \leq 1$ So $ -1\leq x_i \leq 1$ Also we assume that $x_i \neq -1$ So \[1 \geq x_i>-1\]And $x_1x_2...x_{2016}=1 \implies $ $x_i=1, \boxed{(1,1,...,1)}$

Different but similar solution $x_{i} ^2+x_{i}-1=x_{i+1} $ By adding every side $x_{1} ^2+x_{2}^2....+x_{2016}^2=2016 $ Let's say $x_{j}$ be max in this sequence Then $x_{j} ^2+x_{j}-1=x_{j+1} $(....1) In the (.... 1) $ x_{j} >= x_{j} ^2+x_{j}-1 $ Then $1 >= x_{j} >= smth... $ but we get $2016 >=x_{1} ^2+x_{2}^2....+x_{2016}^2$ so all of the $x_{smth} $ are equal to each other So $2016x_{j}^2=2016$ Then we have $(1,......1);(-1,.....-1)$ for the solutions