$x\in\left\{1,-1,2\cos\frac{2\pi}{7}-1,2\cos\frac{4\pi}{7}-1,2\cos\frac{6\pi}{7}-1,2\cos\frac{2\pi}{9},2\cos\frac{4\pi}{9},2\cos\frac{8\pi}{9}\right\}$ Where $x=-1$ is a double solution and the others are all simple solutions.

Thanks, I just now realized it also asks that $x,f(x)$ and $f(f(x))$ are all positive, that makes it trivial.

The original version of the problem is as follows: Find all positive real numbers $(x,y,z)$ such that: $$x = \frac{1}{y^2+y-1}$$$$y = \frac{1}{z^2+z-1}$$$$z = \frac{1}{x^2+x-1}$$

rchokler wrote: $x\in\left\{1,-1,2\cos\frac{2\pi}{7}-1,2\cos\frac{4\pi}{7}-1,2\cos\frac{6\pi}{7}-1,2\cos\frac{2\pi}{9},2\cos\frac{4\pi}{9},2\cos\frac{8\pi}{9}\right\}$ Where $x=-1$ is a double solution and the others are all simple solutions. can u explain?

Rewrite the given system as $y(y+1)=(x+1)/x$, $z(z+1)=(y+1)/y$, and $x(x+1)=(z+1)/z$. It is obvious that if one of the variables is equal to $-1$, they all are; similarly if one of them is equal to $1$, they all are. So $(x,y,z)=(1,1,1)$ and $(-1,-1,-1)$ are solutions. Now assume $(x-1)(x+1)\ne 0.$ Multiply the three expressions to get $xyz=\pm 1$. (a) For $xyz=-1$, the system simplifies to $y+1=-zx-z$, $z+1=-xy-x$, and $x+1=-yz-y$. Substituting, we finally get $y^5+2y^4-2y^3-5y^2-y+1=0=(y+1)^2(y^3-3y+1)$. The solutions of the cubic are easily found to be $2\cos(2\pi/9),\ 2\cos(4\pi/9),\ 2\cos(8\pi/9)$. (b) For $xyz=1$, the system simplifies to $y+1=zx+z$, $z+1=xy+x$, and $x+1=yz+y$. (Note: if $x,y,z>0$, then adding we get $xy+yz+xz=3$ and therefore the only solution in the positive numbers is $x=y=z=1$.) Substituting, we finally get $y^5+4y^4+2y^3-5y^2-3y+1=0=(y-1)(y+1)(y^3+4y^2+3y-1)$. The solutions of the cubic are found to be $-\frac{4}{3}+2\sqrt{\frac{7}{9}}\cos\left(\frac{1}{3}\text{arccos}\frac{\sqrt{7}}{14}-\frac{2k\pi}{3}\right)$, where $k=0,1,2$. These are the same as $2\cos\frac{2\pi}{7}-1,\ 2\cos\frac{4\pi}{7}-1,\ 2\cos\frac{6\pi}{7}-1$. Overall, the solutions are as given above by rchokler. Namely, $x\in\left\{1,-1,2\cos\frac{2\pi}{7}-1,2\cos\frac{4\pi}{7}-1,2\cos\frac{6\pi}{7}-1,2\cos\frac{2\pi}{9},2\cos\frac{4\pi}{9},2\cos\frac{8\pi}{9}\right\}$.

the problem comes into to solve the equation- 1/(x^2 + x - 1) = x

If $x$, $y$ and $z$ satisfy the conditions then at least one of them must be greater than or equal to 1. This is because $a<1$ implies $a^{2}+a-1<1$, thus \[ \frac{1}{a^{2}+a-a}>1. \]Supose $x$, $y$ and $z$ are different in pairs and satisfy the conditions. Choose the greatest of them, say $x$. Then $x>1$ ($x\neq1$ because this imply that $x=z=1$). Then $x^{2}+x-1>x^{2}$ so \[ z<\frac{1}{x^{2}}. \]Then \[ z^{2}+z-1<\frac{1}{x^{4}}+\frac{1}{x^{2}}-1<\frac{1}{x^{4}} \]so $y>x^{4}>x$ a contradiction. Then $x$, $y$ and $z$ cannot be different in pairs. If two of $x$, $y$, or $z$ are equal then is easy to see that $x=y=z=1$. This is the unique solution.

If $x$, $y=f(x)$ and $z=f(f(x))$ are all positive, then as stated above we see that $y+1=xy(x+1)$ and cyclic permutations, from where $xyz=\pm1$, and since again $x,y,z>0$ we have $xyz=1$. Then, $yz+z=z(y+1)=xyz(x+1)=x+1$ and cyclic permutations, which add up to $xy+yz+zx=3$. Then $xy,yz,zx$ are positive reals which have both AM and GM equal to $1$, or $xy=yz=zx=1$ and $x=y=z=1$ is the only possible solution.

Gamamal wrote: Let $f(x)=\frac{1}{x^2+x-1}$. Find all points $x$ so that $f(f(f(x)))=x$. Guess you needed to tell the domain and counterdomain of the function.

Solved with Elliott Liu, Groovy (\help), Jeffrey Chen, Raymond Feng. Let \(b=f(a)\), \(c=f(b)\), \(a=f(c)\), which are given to be positive in #3. Observe that we have \begin{align*} b(a^2+a-1)&=1\\ \implies b+1&=ab(a+1).\end{align*}Analogously, \begin{align*} c+1&=bc(b+1)\\ a+1&=ca(c+1). \end{align*}Multiplying these together, we obtain \(abc=1\). It follows that \begin{align*} a^2+a=\frac1b+1&=ca+1,\end{align*}and analogously \begin{align*} b^2+b&=ab+1\\ c^2+c&=bc+1. \end{align*} Summing, we have \[\left(a^2+b^2+c^2\right)+(a+b+c)=(ab+bc+ca)+3.\]However \(a^2+b^2+c^2\ge ab+bc+ca\) and \(a+b+c\ge3\), so equality must hold, implying \(a=b=c=1\).

Solved with Titu Andreescu ('s methods). We claim that the only solution is $(x, y, z) = (1, 1, 1)$. This works. First note that $f(t) = t^2 + t - 1 = (t + \phi)\left(t - \frac1\phi\right)$, so if $x < \frac1\phi$, then $z < 0$, impossible (and if $x=\frac1\phi$, $z$ is undefined). Similarly for $y, z$, so $x, y, z > \frac1\phi$. Then for $t > \frac1\phi$, $f(t)$ is strictly increasing, so $\frac1{f(t)}$ is strictly decreasing. Now we claim that $x=y=z$. Assume for the sake of contradiction that two of $x, y, z$ are not equal. WLOG assume that it is $x, y$. Note that $f^3 (x) = x$ and $f^3(y) = y$ Then if $x < y$, \[ x < y \implies f(x) > f(y) \implies f^2(x) < f^2(y) \implies f^3(x) > f^3(y) \implies x > y,\]contradiction, and similarly for $x > y$. We conclude that $x=y=z$. Now it suffices to find all solutions to $x = \frac1{x^2+x-1}$. This implies that $x^3+x^2-x-1=0$ or $(x+1)^2(x-1)=0$. Since $x>0$, $x=1$, and the only solution is $(x, y, z) = (1, 1, 1)$.

Why is everyone's solution so complicated? We claim the answer is $x=y=z=1$ Let $f(x)=\dfrac{1}{x^2+x-1}$. Consider the following observation, if $a<1$ then $f(a)>1$, if $a=1$ then $f(a)=1$, if $a>1$ then $f(a)<1$. Assume $x \not = 1$. If $x>1$ then \[ \Rightarrow f(x) <1 \Rightarrow f(f(x))>1 \Rightarrow f(f(f(x)))=x<1 \]which is a contradiction. A similar argument follows for $x<1$. Thus, we have $x=1$ and $y=z=1$ trivially follows.

Si x=y=z entonces (x,y,z)=(1,1,1) Supongamos que x<=y<=z (E1) Entonces x²+x-1 <= y²+y-1 <= z²+z-1 1/z <= 1/x <= 1/y Donde y <= x pero por E1 teniamos que x <= y entonces x=y Sabemos que x = 1/(y²+y-1) Pero z = 1/(x²+x-1) y como x=y entonces x=z Por lo tanto x=y=z=1

tk1 wrote: Rewrite the given system as $y(y+1)=(x+1)/x$, $z(z+1)=(y+1)/y$, and $x(x+1)=(z+1)/z$. It is obvious that if one of the variables is equal to $-1$, they all are; similarly if one of them is equal to $1$, they all are. So $(x,y,z)=(1,1,1)$ and $(-1,-1,-1)$ are solutions. Now assume $(x-1)(x+1)\ne 0.$ Multiply the three expressions to get $xyz=\pm 1$. (a) For $xyz=-1$, the system simplifies to $y+1=-zx-z$, $z+1=-xy-x$, and $x+1=-yz-y$. Substituting, we finally get $y^5+2y^4-2y^3-5y^2-y+1=0=(y+1)^2(y^3-3y+1)$. The solutions of the cubic are easily found to be $2\cos(2\pi/9),\ 2\cos(4\pi/9),\ 2\cos(8\pi/9)$. (b) For $xyz=1$, the system simplifies to $y+1=zx+z$, $z+1=xy+x$, and $x+1=yz+y$. (Note: if $x,y,z>0$, then adding we get $xy+yz+xz=3$ and therefore the only solution in the positive numbers is $x=y=z=1$.) Substituting, we finally get $y^5+4y^4+2y^3-5y^2-3y+1=0=(y-1)(y+1)(y^3+4y^2+3y-1)$. The solutions of the cubic are found to be $-\frac{4}{3}+2\sqrt{\frac{7}{9}}\cos\left(\frac{1}{3}\text{arccos}\frac{\sqrt{7}}{14}-\frac{2k\pi}{3}\right)$, where $k=0,1,2$. These are the same as $2\cos\frac{2\pi}{7}-1,\ 2\cos\frac{4\pi}{7}-1,\ 2\cos\frac{6\pi}{7}-1$. Overall, the solutions are as given above by rchokler. Namely, $x\in\left\{1,-1,2\cos\frac{2\pi}{7}-1,2\cos\frac{4\pi}{7}-1,2\cos\frac{6\pi}{7}-1,2\cos\frac{2\pi}{9},2\cos\frac{4\pi}{9},2\cos\frac{8\pi}{9}\right\}$. You didn't realize $x,y,z$ are positive.