# Solving $x_1+x_2=x_3^2, x_2+x_3=x_4^2, x_3+x_4=x_5^2,x_4+x_5=x_1^2, x_5+x_1=x_2^2$ in reals

find answers of this system of equations in real numbers$$\left\{ \begin{array}{c} x_1+x_2=x_3^2 \\ x_2+x_3=x_4^2 \\ x_3+x_4=x_5^2 \\ x_4+x_5=x_1^2 \\ x_5+x_1=x_2^2 \end{array} \right.$$

Things I have done: by observation I was able to see that $x_i=0$ and $x=2$ are answers.

but for solving it in a better this was my idea: W.L.O.G i assumed that $x_1^2 \ge x_2^2$. So $$x_1^2\ge x_2^2 \rightarrow x_4+x_5\ge x_5+x_1 \rightarrow x_4\ge x_1$$ but I ran in to problem because $x_1$ could be negative so I can't conclude from $x_4\ge x_1$ that $x_4^2\ge x_1^2$.

There's probably a more clever solution.

Case 1: Suppose one of the $x_i$ is $0$: say $x_1$ is $0$. Then, $$x_2-x_4=(x_2+x_3)-(x_3+x_4)=x_4^2-x_5^2=(x_4-x_5)(x_4+x_5)=(x_4-x_5)x_1^2=0$$ so that $x_2=x_4$. This implies $$-x_2=x_1-x_2=x_1-x_4=(x_1+x_5)-(x_4+x_5)=x_2^2-x_1^2=(x_2-x_1)(x_2+x_1)=x_2^2.$$ This means $x_2=0$ or $x_2=-1$. But $x_3^2=x_1+x_2=x_2$ so $x_2\geq 0$ therefore we infer that $x_2=x_4=0$. From here, it's easy to see that $x_3=x_5=0$ as well.

Case 2: each $x_i$ is nonzero. Label the given equations as (1)-(5). From (1) and (2), $$x_1-x_3=(x_3-x_4)x_5^2.$$ Suppose $x_1>x_3$, then because $x_5^2>0$, we have $x_3>x_4$. In particular, we have $x_1>x_4$, which implies $x_2>x_1$ thanks to equations (4) and (5). So we have $x_2>x_1>x_3>x_4$. Equations (2) and (3) now imply that $x_4>x_5$. But then equations (3) and (4) imply $x_5>x_1$. So you have the impossibility $x_2>x_1>x_3>x_4>x_5>x_1$.

Supposing $x_1<x_3$ will lead to a similar contradiction as in the previous paragraph.

It must be that $x_1=x_3$ so that $x_3=x_4$. But then $x_1=x_4$ implies $x_1=x_2$. In turn, $x_2=x_4$ implies $x_4=x_5$. So $x_1=x_2=x_3=x_4=x_5=x$ for some nonzero $x$. You can now easily see that $x=2$.

Conclusion: 2 solutions $x_i=0$ $\forall i$ and $x_i=2$ $\forall$ $i$.

• thanks for posting your solution. do you know any good books about techniques in solving system of equations? all of my searches on Google leads to physics related books that mostly covered differential equations. Commented Sep 7, 2014 at 17:06
• Sorry, I don't. My solution above was ad-hoc, based on an intuition that I needed to use the cyclical nature of the problem somehow. Commented Sep 7, 2014 at 17:09

Assume, $x_1$ is the biggest of $x_1, x_2, x_3, x_4, x_5$. Then $x_1+x_2 = x_3^2 \ge 0$ and $x_1 \ge x_2$, therefore $x_1\ge 0$ and $x_1^2\ge x_2^2$.

Hence, $x_4+x_5 \ge x_5 + x_1$, therefore $x_4 \ge x_1$, or $x_4=x_1$. We obtained that $x_4$ is also the biggest of those numbers. In the same way you prove, that $x_4=x_2$, then $x_2 = x_5%$ and $x_5=x_1$.

So all numbers are equal to each other and they are solutions of $x^2-2x=0$.