Solve system of $n$ equations of the form $2x_k^3+4=x_k^2(x_{k+1}+3)$

Question

Solve the system of $n$ equations, $n\geq2$: $$ \begin{cases} 2x_1^3+4=x_1^2(x_2+3)\\ 2x_2^3+4=x_2^2(x_3+3)\\ \qquad \vdots\\ 2x_{n-1}^3+4=x_{n-1}^2(x_n+3)\\ 2x_n^3+4=x_n^2(x_1+3)\\ \end{cases} $$

I think that there are only two solutions, but I don't know how to prove it:

$$x_1=x_2=x_3=\cdots=x_n=-1$$ $$x_1=x_2=x_3=\cdots=x_n=2$$

Answer 1

$$2x_n^3+4=x_n^2(x_{n+1}+3)$$ $$2x_n+\frac{4}{x_n^2}=x_{n+1}+3$$ $$x_{n+1}=2x_n+\frac{4}{x_n^2}-3$$ Define $f(x)$: $$f(x)=2x+\frac{4}{x^2}-3=x+\frac{x^3-3x^2+4}{x^2}=x+\frac{(x-2)^2(x+1)}{x^2}$$ \begin{align} &f(x)=x\quad(x=-1, \space 2)\\ &f(x)>x\quad(x>-1,\space x\ne2)\\ &f(x)<x\quad(x<-1)\\ \end{align} If $x_1\ne-1,2$ then

1. If $x_1<x_2$, then $x_1<x_2<\cdots<x_n<x_1\rightarrow$ contradicts.
2. If $x_1>x_2$, then $x_1>x_2>\cdots>x_n>x_1\rightarrow$ contradicts.

Therefore, $x_1=x_2=\cdots=x_n=-1$ or $x_1=x_2=\cdots=x_n=2$