If $a,b,c \in \mathbb R$ and $a \neq 0$ then for the system of quadratic equations in n variables $x_1, x_2 ... x_n$ 
If $a,b,c  \in \mathbb R$ and $a \neq 0$ then for the system of quadratic equations in n variables $x_1, x_2 ... x_n$ 
$ax_1^2+bx_1+c=x_2$ 
$ax_2^2+bx_2+c=x_3$
. 
. 
. 
$ax_n^2+bx_n+c=x_1$ 
$1)$ show that that the equations have no solutions if $(b-1)^2<4ac$ 
$2)$ find the solutions if $(b-1)^2>4ac$

My Attempt:
I was able to do the first part by adding all the equations and converting them into summations of quadratic equations,
$$ax_1^2+x_1(b-1)+c + ax_2^2+x_2(b-1)+c ... ax_n^2+x_n(b-1) + c=0$$
for this to have no solution, $(b-1)^2<4ac$
But, how do you find the solutions if there are any (for the condition $(b-1)^2>4ac$)
One thing that can be done is just to substitute and then solve the resulting $f(x^{2n})$ polynomial. But that would be too much work. Is there any other easier way to find the solutions?
Any help would be appreciated.
 A: You can normalize the system by setting $y_i=ax_i+\frac{b}2$ so that
$$
y_{(i+1)\bmod n}=ax_{(i+1)\bmod n}+\frac{b}2
=a^2x_i^2+abx_i+ac+\frac{b}2
=y_i^2+ac+\frac{b}2-\frac{b^2}4
=y_i^2+\tilde c
$$
so that in the end there is only one free parameter determining the dynamic of the map. This map is well-known, you are looking for period-$n$ cycles of the Mandelbrot map on the real line. This can also be transformed to cycles of the Feigenbaum/logistic map.
A: Edit
I think one trick is to realize that the cyclic permutation $(x_1,x_2,\ldots,x_n)$ yields the same system. Therefore there is a solution with $x_1=x_2=\ldots{}=x_n$. Then you end up with just one equation $ax^2+(b-1)x+c=0$ which is easily solvable.
Edit 2
One can write the system as follows: $x_{i+i}=ax_i+bx+c$ and take the subscripts$\mod n$. This leads to a quadratic map. According to that website there are (at least for certain $a,b,c$) not only period 2 (as already given in the comments) but also period 3 fixed point, which means that there are solutions for $n=3$ where all $x_i$ are distinct. (In fact, there are such solutions for every $n$.) I tried to find a solution for $(a,b,c)=(1,0,-1)$ and $n=3$. This leads to the map $x_{i+i}=f(x_i)=x_i^2-1$. Its period 3 fixed point $x$ satisfies $f(f(f(x)))-x=0$ which leads to finding the roots of an eighth degree polynomial. Dividing the polynomial $f(x)-x=x^2-x-1$ which is satisfied by the (period one) fixed point the polynomial $x^6+x^5-2x^4-x^3+x^2+1$ remains. (The Wolfram page says $\ldots -x^2 \ldots$ which is wrong). However, this polynomial only has non-real roots. Therefore, for the parameters given there is no solution except for the one given in the first part.
