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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.

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    $\begingroup$ Related : 1968 IMO Problem 3 $\endgroup$ – mathlove Nov 25 '16 at 6:16
  • $\begingroup$ @mathlove-In the last case why did it consider $\Delta\gt1$ and not $\Delta\gt0$, and also it didn't address my concern which I have stated in the last section in my post. $\endgroup$ – user350331 Nov 25 '16 at 6:35
  • $\begingroup$ I think that $\Delta\gt 1$ is a typo. I know that it does not address your concern. I just wrote "related". $\endgroup$ – mathlove Nov 25 '16 at 6:46
  • $\begingroup$ After thinking a lot on this, is it even possible to account for the solutions that I am talking about in the last section of my post, maybe some kind of a relation b/w $x_i$ and $x_j$, because the the link that you provided made me feel that it was implying that in the third case there can be only two different solutions, I don't know whether this feeling that I have is correct or not. So, please tell me if I am correct to try to account for the solutions that I have mentioned $\endgroup$ – user350331 Nov 25 '16 at 6:53
  • $\begingroup$ When $a=c=1,b=4,n=4$, the system has at least four solutions $(x_1,x_2,x_3,x_4)=(\alpha,\alpha,\alpha,\alpha),(\beta,\beta,\beta,\beta),(-2,-3,-2,-3),(-3,-2,-3,-2)$. It seems that the number of the solutions is dependent on the parity of $n$. $\endgroup$ – mathlove Nov 25 '16 at 7:03
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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.

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  • $\begingroup$ I think after reading through the Wikipedia link there are also some other points at which the function gets satisfied as I had pointed out in my post and you pretty much told me the general form. Other then that it will be a long time after which I would be able to understand all present in the link but still that was wonderful and gave me confidence that I can think ahead of the book. The bounty is definitely yours. $\endgroup$ – user350331 Nov 30 '16 at 14:01
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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.

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  • $\begingroup$ I mistakenly assumed that finding solutions to the sums of the equations on both sides, meant finding solutions to the original equations. This is not the case. However, your answer doesn't seem to answer the question posed in the OP. In case-3, do solutions exist other than $\alpha$ or $\beta$? $\endgroup$ – Jens Nov 25 '16 at 22:04

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