Problem Statement:-
If $a,b,c\in\Bbb{R}$ and $a\neq0$, solve the following system of equations in $n$ unknowns $x_1,x_2,x_3,\ldots,x_n$ $$a{x_1}^{2}+bx_1+c=x_2\\ a{x_2}^{2}+bx_2+c=x_3\\ \ldots\ldots\ldots\ldots\ldots\ldots..\\ \ldots\ldots\ldots\ldots\ldots\ldots..\\ a{x_{n-1}}^{2}+bx_{n-1}+c=x_n\\ a{x_n}^{2}+bx_n+c=x_1$$ when
$\text{(i)}~~(b-1)^2\lt4ac\\ \text{(ii)}~~(b-1)^2=4ac\\ \text{(iii)}~~(b-1)^2\gt4ac$
My Solution:-
Let $f(x_i)=a{x_i}^2+(b-1)x_i+c$
On summing all the given equations, we get $$\sum_{i=1}^{n}{\left(a{x_i}^2+(b-1)x_i+c\right)}=0\implies \sum_{i=1}^{n}{f(x_i)}=0$$
Consider the following quadratic equation $$ax^2+(b-1)x+c=0\tag{1}$$
Also, consider the following cases:-
Case-1:-$\qquad(b-1)^2\lt4ac$
In this case the eq. $(1)$ has no real solutions and has the same sign as that of $a$. So $f(x_i)\gt 0$, hence the system of equation does not have any solution.
Case-2:-$\qquad(b-1)^2=4ac$
In this case eq. $(1)$ has repeated roots as $D=0$. So, $f(x_i)=0$ only at $x_i=\dfrac{1-b}{2a}$
Hence, in this case the system of equation has the solution $x_i=\dfrac{1-b}{2a}$, where $i\in\left\{1,2,3,\ldots,x_n\right\}$
Case-3:-$\qquad(b-1)^2\gt4ac$
In this case the eq $(1)$ has two distinct real roots, $\because D\gt0$.
The roots are given by $x=\dfrac{1-b\pm\sqrt{\left(b-1\right)^2-4ac}}{2a}$
So, in this case $f(x_i)=0$, when $$x_i=\alpha=\dfrac{1-b-\sqrt{\left(b-1\right)^2-4ac}}{2a}$$ or $$x_i=\beta=\dfrac{1-b+\sqrt{\left(b-1\right)^2-4ac}}{2a}$$
My deal with the problem:-
This was the approach that I had taken while solving the question in the first go, and so did the book that I am solving from, except in the third case it also showed what would happen if $x_i\in(\alpha,\beta)$ or if $x_i\in\Bbb{R}-(\alpha,\beta)$.
But, on analysing my combined with the book's solution I thought that the I didn't handle Case-3, well enough.
There can also be a condition such that $$f(x_i)+f(x_j)=0$$ where $x_i$ and $x_j$ are such value of $x$ which aren't the roots of the equation $(1)$.
Which implies either $x_i\in(\alpha,\beta)$ and $x_j\in\Bbb{R}-(\alpha,\beta)$ or vice versa.
So how to account for these solutions.