System of equations and quadratic equation

Question

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

$\text{(i)}~~(b-1)^2\lt4ac\\ \text{(ii)}~~(b-1)^2=4ac\\ \text{(iii)}~~(b-1)^2\gt4ac$

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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 $$x_i=\beta=\dfrac{1-b+\sqrt{\left(b-1\right)^2-4ac}}{2a}$$

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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.

Answer 1

Your problem can be rephrased as given the polynomial $g(x) = ax^2 + bx + c$ with real coeficients and $a \neq 0$, find all periodic orbits of $g$ of length $n \in \mathbb{N}$, that is, all values of $x$ such that iterating $n$ times function $g$ you get $g^{(n)}(x) = x$. Your solution only checks whether $g(x) = x$ has solutions.

Not sure if there is a closed answer for this. Looking at wikipedia you can see some info on this problem.

Answer 2

For $n=2$, by substitution

$$ax_1^2+bx_1+c=x_2,\\ax_2^2+bx_2+c=x_1$$

yields a quartic equation

$$a(ax_1^2+bx_1+c)^2+b(ax_1^2+bx_1+c)+c=x_1$$

which has four solutions (in $\mathbb C$). As there are three independent coefficients, I don't think that there can be any simplification (and the expressions can be terrible).

For $n>2$, by similar substitutions you get equations of degree $2^n$, hence $2^n$ solutions.