Let $a$ be a real number. Find the real numbers $x_1$, $x_2$, ..., $x_n$ knowing that $$x^2_1+ax_1+(\frac{a-1}{2})^2=x_2$$ $$x^2_2+ax_2+(\frac{a-1}{2})^2=x_3$$ $$...$$ $$x^2_n+ax_n+(\frac{a-1}{2})^2=x_1.$$
It's easy to see that $x_1=x_2$ is a solution as it implies $x_1=x_2=...=x_n$. From here, I split the problem into the two cases: $x_1>x_2$ and $x_1<x_2$, trying to show that both these cases have no solutions. It is intuitively easy to see that $x_1>x_2$ implies $x_1>x_2>...>x_n$ and the other case shows something similar. However, I'm stuck trying to prove these rigorously.
Suppose that $x_{k-1}<x_k$. Therefore $$x_{k-1}^2+ax_{k-1}+(\frac{a-1}{2})^2=x_{k}$$ $$x_{k-1}=\frac{x_{k}-a\pm \sqrt{(a-x_{k})^2-(a-1)^2}}{2}<x_k.$$ Screwing with the above equation along with $(a-x_{k})^2-(a-1)^2\geq0$ hasn't gotten me anywhere so far. Any hints for proving either case?
If you sum all equations, you will get $\sum_{k=1}^n x_k^2+(a-1)x_k+(\frac{a-1}{2})^2=\sum_{k=1}^n (x_k+\frac{a-1}{2})^2=0$, from where we get $x_k=\frac{1-a}{2}$ for $k=1,\dots n$.