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orl wrote: Solve the system of equations: $ \begin{matrix} x^2 + x - 1 = y \\ y^2 + y - 1 = z \\ z^2 + z - 1 = x. \end{matrix} $ Let (x, y, z) be a solution of this system. Then, let a = x + 1, b = y + 1 and c = z + 1. Then, x = a - 1 and y = b - 1, so that the equation $ x^2+x-1=y $ becomes $ \left(a-1\right)^2+\left(a-1\right)-1=b-1 $, or, equivalently, $ a^2-a-1=b-1 $. Thus, $ a^2-a=b $, so that $ a^2=a+b $. Similarly, $ b^2=b+c $ and $ c^2=c+a $. Hence, $ \left(a-b\right)a^2=\left(a-b\right)\left(a+b\right)=a^2-b^2=\left(a+b\right)-\left(b+c\right)=a-c=-\left(c-a\right) $. Similarly, $ \left(b-c\right)b^2=-\left(a-b\right) $ and $ \left(c-a\right)c^2=-\left(b-c\right) $. Multiplying these three equations, we get $ \left(a-b\right)a^2\cdot\left(b-c\right)b^2\cdot\left(c-a\right)c^2=\left(-\left(c-a\right)\right)\cdot\left(-\left(a-b\right)\right)\cdot\left(-\left(b-c\right)\right) $. Now, if the numbers b - c, c - a and a - b are all nonzero, then we can divide this equation by (b - c) (c - a) (a - b) and obtain $ a^2b^2c^2=-1 $, what is impossible, since $ a^2b^2c^2\geq 0 $ (as squares are always nonnegative). Thus, the numbers b - c, c - a and a - b cannot all be nonzero; one of them must be zero. WLOG, assume that b - c = 0. Then, b = c, so that y = z (since b = y + 1 and c = z + 1). Hence, the equation $ y^2+y-1=z $ becomes $ y^2+y-1=y $, so that $ y^2=1 $. Thus, either y = 1 or y = -1. If y = 1, then y = z yields z = 1, and thus $ x=z^2+z-1=1^2+1-1=1 $, so we get the solution (x, y, z) = (1, 1, 1). A trivial check shows that the triple (x, y, z) = (1, 1, 1) is indeed a solution of our system of equations. If y = -1, then y = z yields z = -1, and thus $ x=z^2+z-1=\left(-1\right)^2+\left(-1\right)-1=-1 $, so we get the solution (x, y, z) = (-1, -1, -1). A trivial check shows that the triple (x, y, z) = (-1, -1, -1) is indeed a solution of our system of equations. Hence, our system of equations has the two solutions (x, y, z) = (1, 1, 1) and (x, y, z) = (-1, -1, -1). darij

I can solve this problem for all $x,y,z\ge0$ suppose that $x\ge y \Rightarrow x^2+x-1\ge y^2+y+1$ so we have $y\ge z$ and again we have $z\ge x$ hence $x\ge y\ge z\ge x \Rightarrow x=y=z$ and if we had $x\le y \Rightarrow x^2+x+1\le y^2+y+1$ so $y\le z$ and now we have $z\le x$ and again $x\le y\le z\le x \Rightarrow x=y=z$ so if we have $x,y,z\ge 0 \Rightarrow x=y=z$ and the only thing we haveto do is tosolve this equation $x^2+x-1=x \Rightarrow x^2=1 \Rightarrow x=y=z=1$ but if $x,y,z\not\ge 0$ I tried to use a similar solution of this system of equations $(x+y)^3=z$ $(y+z)^3=x$ $(z+x)^3=y$ but it didn't work for $x,y,z\not\ge 0$ : would you please help me to complete this solution. and the solution of darij was really beautiful. thanks...

darij grinberg wrote: Let (x, y, z) be a solution of this system. Then, let a = x + 1, b = y + 1 and c = z + 1. Then, x = a - 1 and y = b - 1 Nice solution! But one question - how did you think to come up with those substitutions?

I got it through a slightly different method but yeah darij your solution really is beautiful Ashwath

Orl when all problems will be solved, could you post the solutions at PDF like the problems? Thanks

pkothari13 wrote: darij grinberg wrote: Let (x, y, z) be a solution of this system. Then, let a = x + 1, b = y + 1 and c = z + 1. Then, x = a - 1 and y = b - 1 Nice solution! But one question - how did you think to come up with those substitutions? Yeah how?Any trick or just practice?

pkothari13 wrote: darij grinberg wrote: Let (x, y, z) be a solution of this system. Then, let a = x + 1, b = y + 1 and c = z + 1. Then, x = a - 1 and y = b - 1 Nice solution! But one question - how did you think to come up with those substitutions? I just looked for a substitution of the kind a = x + k, b = y + k, c = z + k which would make the equations $\begin{matrix} x^2 + x - 1 = y \\ y^2 + y - 1 = z \\ z^2 + z - 1 = x\end{matrix}$ not contain the nasty 1's, but just contain the variables a, b, c. And found that the right k is 1. Darij

I also used the substitution $a=x+1$, but not in the same way. Let $a=x+1, b=y+1, c=z+1$. We then have the system $\begin{matrix}xa=b \\ yb=c\\ zc=a\\ \end{matrix}$ which gives us (assuming $a,b,c\not =0$, I'll come back to that one) $xyz=1$ In a trivial way (add the three eequations to each other), we also find that the system $\begin{matrix} x^2 + x - 1 = y \\ y^2 + y - 1 = z \\ z^2 + z - 1 = x. \end{matrix}$ gives us $x^2+y^2+z^2=3$. Since $xyz=\frac{x^2+y^2+z^2}{3}=1$, we have by the AMGM-inequality that $x=y=z=1$ and (if we include negative numbers, which we do) $x=y=-1, z=1$ and permutaions of that. But since we here assumed that $a,b,c\not=0$ which is equivalent to $x,y,x=-1$, we can throw away that solution (it is false). Now assume one of $a,b,c$ equals $0$. We then have (assuming because of symmetry that $a=0$, or $x=-1$) $\begin{matrix} - 1 = y \\ y^2 + y - 1 = z \\ z^2 + z = 0. \end{matrix}$ and it is obvious that the only solution to this equation is that $x=y=z=-1$ Therefore the two solutions are $x=y=z=1$ and $x=y=z=-1$ Hopefully I haven't made too many errors... Edit: Fixed one minor I found.

I only noticed that (-1;-1;-1) works. My solution is based upon the cartesian representation and the growing and the functional equation f^3(x)=x with f(x)=x^2+x-1 and the roots of the polinomial and derivatives...

Add the equations to get $x^2+y^2+z^2=3$ Rewrite the equations as $x(x+1)=y+1$ $y(y+1)=z+1$ $z(z+1)=x+1$ If $x,y,z\neq -1$, multiplying the equations we get $xyz=1 \Rightarrow x^2y^2z^2=1$ You can easily check none of $x, y, z$ is 0, thus $x^2,y^2,z^2 \in R^+$ And by AM-MG $1=\frac{3}{3}=\frac{x^2+y^2+z^2}{3}\geq \sqrt[3]{x^2y^2z^2}=\sqrt[3]{1}=1$ We have the equality, thus $x=y=z=1$ If some of $x, y, z$ is $-1$, it´s easy to see $x=y=z=-1$ Thus, the solutions are $x=y=z=t \in$ {$1, -1$}

Found 8 Solutions. Real: $\pm(1,1,1)$ Complex: Cyclic permutations of $(A_3,A_2,A_1)$ its conjugate. $A_k=\frac{-2-i}{3}+\frac{1}{6}e^{2\pi ki/3}\sqrt[3]{4(5+14i)+12(2+3i)\sqrt{3}}+\frac{1}{6}e^{-2\pi ki/3}\sqrt[3]{4(5+14i)-12(2+3i)\sqrt{3}}$

A particular case of OMM 2011 P3