Solve system of $n$ equations of the form $2x_k^3+4=x_k^2(x_{k+1}+3)$ 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:
$$x_1=x_2=x_3=\cdots=x_n=-1$$
$$x_1=x_2=x_3=\cdots=x_n=2$$

$$2x_n^3+4=x_n^2(x_{n+1}+3)$$
$$2x_n+\frac{4}{x_n^2}=x_{n+1}+3$$
$$x_{n+1}=2x_n+\frac{4}{x_n^2}-3$$
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


*

*If $x_1<x_2$, then $x_1<x_2<\cdots<x_n<x_1\rightarrow$ contradicts.

*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 $x_1=x_2=\cdots=x_n=2$
