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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.

user26857
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Tony
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2 Answers2

1

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.

Lutz Lehmann
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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.

diffset
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    Your argument is incorrect because you are assuming the system must have a single solution$$a=1,b=0,c=-1,n=2$$Gives $(x_1,x_2)=(0,-1)$ –  Aug 02 '20 at 08:28
  • -1 because of the objections in the comments – miracle173 Aug 02 '20 at 08:40
  • @yukelid Okay your solution is correct. But can you elaborate where my error is? Does my argument at least hold for $n\geq3$ – diffset Aug 02 '20 at 08:42
  • @diffset I agree that the cyclic permutation $(1,2,…,n)$ must also be a solution, but this doesn't mean that $x_1=x_2=…=x_n$ must hold; it is one of many possibilities. As for $n≥3$, $$a=1,b=0,c=−1,n=4$$ again gives $(x_1,x_2,x_3,x_4)=(0,−1,0,-1)$. –  Aug 02 '20 at 09:01
  • @yukelid Thanks! But can you come up with a solution for odd $n>1$ where $x_i\neq x_j$ for some $i\neq j$? :-) – diffset Aug 02 '20 at 09:13
  • @miracle173 Maybe you want to reconsider after having seen my edits? – diffset Aug 02 '20 at 13:55
  • @yukelid You won't find a solution for $(a,b,c,n)=(1,0,-1,3)$. See my edit. – diffset Aug 02 '20 at 14:20
  • @diffset i agree, but what are you trying to arrive at with all the case-verification? –  Aug 02 '20 at 15:00