Find all positive solutions of the system of equations

Question

Find all positive solutions of the system of equations

$x_1+x_2=(x_3)^2$ , $x_2+x_3=(x_4)^2$ , $x_3+x_4=(x_5)^2$ , $x_4+x_5=(x_1)^2$ , $x_5+x_1=(x_2)^2$

What i have done :

$(x_1+x_2)+(x_2+x_3)+(x_3+x_4)+(x_4+x_5)+(x_5+x_1)=(x_3)^2+(x_4)^2+(x_5)+(x_1)^2+(x_2)^2$

Thus $ 2(x_1+x_2+x_3+x_4+x_5)=(x_1)^2+(x_2)^2+(x_3)^2+(x_4)^2+(x_5)^2$

Let $ x = (x_1,x_2,x_3,x_4,x_5) $ . I am currently attempting to find $ x$ such that $||x||$ is at maximum.By symmetry $||x||$ attains maximum when $ x_1 = x_2 =x_3 =x_4 = x_ 5 $. Thus $ 2(5x_1) =5(x_1)^2 $. Hence $x_1=x_2 =x_3=x_4=x_5=2 $.I am unsure how to tackle this problem from this points onward.

Any insight is greatly appreciated !

Thanks !

Answer 1

Let $x_s$ be the smallest of the five variables and $x_b$ the biggest. Looking at the biggest one, $$ x_b^2 = x_{b-2} + x_{b-1} \le x_b+x_b = 2x_b \implies x_b \le 2 $$ (since we're assuming $x_b$ is positive). Here we are considering the indices modulo $5$, so that $x_{-1}=x_4$ and $x_0=x_5$. Similarly, looking at the smallest one, $$ x_s^2 = x_{s-1} + x_{s-2} \ge x_s + x_s = 2x_s \implies x_s \ge 2 $$ (again since $x_s$ is positive). Therefore $2\le x_s\le x_b\le 2$, and so the solution $x_1=x_2=x_3=x_4=x_5=2$ is the unique solution.