find all the answers of these equations

Question

find all the answers of these equations:

$x_{1} + x_{2} = x_{3} ^2$

$x_{2} + x_{3} = x_{4} ^2$

$x_{3} + x_{4} = x_{5} ^2$

$x_{4} + x_{5} = x_{1} ^2$

$x_{5} + x_{1} = x_{2} ^2$

My attempt:By summing all equations we can easily find that one answer is that all of them are zero but what about other answers?

In the hint of the book is written that first show $x_1=x_5$.

Answer 1

Not a full fledged answer but maybe you or someone else can take what I have worked on and turn it into an answer.

Given $x_1=x$ and $x_2=y$ you can determine $x_3,x_4,$ and $x_5$ by simply using three out of the 5 equations to define them namely

1. $x_3=\sqrt{x+y}$ (Using the first equation)

2. $x_5=y^2-x$ (Using the last equation)

3. $x_4=x^2+x-y^2$ (Using the fourth equation)

Given that this is true it follows that the 2nd and 3rd equation only serve as checks for your solutions. So you can rewrite them in terms of x and y, and gain two equalities.

(1). $x_2+x_3=x_4^2$ gets rewritten as $y+\sqrt{x+y}=(x^2+x-y^2)^2$

(2). $x_3+x_4=x_5^2$ gets rewritten as $\sqrt{x+y}+x^2+x-y^2=(y^2-x)^2$

Simplifying this and subtracting the two equations will give you a quadratic in terms of y.

$y+\sqrt{x+y}=x^4+x^2+y^4-2xy^2+2x^3-2x^2y^2$ (Equation 1)

$-(\sqrt{x+y}+x^2+x-y^2=y^4-2xy^2+x^2)$ (-Equation 2)

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$y-x^2-x+y^2=x^4+2x^3-2x^2y^2$(New Equation=Equation 1- Equation 2)

$(2x^2+1)y^2+y-(x^4+2x^3+x^2+x)=0$(Simplified)

Which will give you a 2 possible solutions for $y$ given an x. Problem is it is possible that both solutions are Extraneous(crops up from the use of elimination).

$\bar{y}(x)=y_{solution}=\frac{-1\pm \sqrt{1+4(2x^2+1)(x^4+2x^3+x^2+x)}}{2(2x^2+1)}$

This can be avoided by simply plugging the solutions into one of the prior equations (1) or (2) and verifying that it is also a solution there. Problem is the calculation and analysis is so messy at this point that, that verifying.

Verify that $\bar{y}(x)+\sqrt{x+\bar{y}(x)}=(x^2+x-(\bar{y}(x))^2)^2$ implies that $x=2$ or $x=0$(Pain in the ass right?)

Anyways at least it's down to one variable right?(Hell you might even be able to use some calculus on that mug.)

Answer 2

A non-trivial solution is given by $$ (x_1,\ldots ,x_5)=(2,2,2,2,2). $$

Edit: The system has $32$ complex solutions, two of them are real, even integral, namely $(0,0,0,0,0)$ and $(2,2,2,2,2)$. I used elimination and reduced the system to two equations in $x_1$ and $x_2$, but there will be more elegant solutions.