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 :


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 !


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.

  • 1
    $\begingroup$ Given the chain-like nature of the equation set, in my opinion to assume the $x_i$ are in nondecreasing order is not the most general case. (Rearranging them affects what gets equated to what in the system.) $\endgroup$ – coffeemath Nov 20 '14 at 13:30
  • $\begingroup$ Maple gives all 0's, also all 2's, and a third case involving Root_Of with polynomials of degree 30. The latter may mean there are other positive solutions (with the $x_i$ not in ascending or descending order). $\endgroup$ – coffeemath Nov 20 '14 at 13:38
  • $\begingroup$ @coffeemath: You're right, the way I wrote it was sloppy. Fixed. $\endgroup$ – Greg Martin Nov 20 '14 at 20:24
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    $\begingroup$ Just to be picky: OK to choose $x_1$ as the smallest, but is it then OK to assume $x_5$ is the largest? That is, in a sequence of five cyclically arranged numbers, does it follow that the largest and smallest are adjacent around the cycle? The equations themselves admit only the renaming scheme obtained on replacing each $x_k$ by $x_{k+1}$ [subscripts mod 5] so given an arrangement like 1,4,5,2,3 we can't shuffle them around either way and get the lowest next to the highest. $\endgroup$ – coffeemath Nov 21 '14 at 5:04
  • 2
    $\begingroup$ hehe, picky is good. Edited again.... $\endgroup$ – Greg Martin Nov 21 '14 at 5:06

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