Solve the system : $ x^2 + x - 1 = y $, $ y^2 + y - 1 = z $, $ z^2 + z - 1 = x $ .

Question

Solve the system of equations in x,y,z : $ x^2 + x - 1 = y $ , $ y^2 + y - 1 = z $, $ z^2 + z - 1 = x $ .

I'd tried analysing various manipulations but can't figure out .

What I've tried: $$ x^2 + x - 1 = y \ ...(1)$$ $$ y^2 + y - 1 = z \ ...(2)$$ $$ z^2 + z - 1 = x \ ...(3)$$

From (1), (2) and (3) , $ x^2 + y^2 + z^2 = 3 $

From (1), $ x(x+1) = y+1 $ & $ (x+1)(x-1) = y-x $

From (2), $ y(y+1) = z+1 $ & $ (y+1)(y-1) = z-y $

From (3), $ z(z+1) = x+1 $ & $ (z+1)(z-1) = x-z $

Any way it's quite obvious to see that $ x = y = z = 1 $ is one of the possible solutions. But no idea how to use these informations to get all the possible values of x,y and z .

Answer 1

We know $$ x^2 + x - 1 = y \implies x(1+x) = 1 +y $$ Similarly $$ y^2 + y - 1 = z \implies y(1+y) = 1 +z $$ $$ z^2 + z - 1 = x \implies z(1+z) = 1 +x $$ Substituting this value of $1+x$ in the 1st equation, we get $$ xz(1+z) = 1 + y$$ Again substituting this in the second equation, we get, $$ xyz(1+z) = 1 + z$$ This basically implies either $z = -1$ or $xyz = 1$ Assuming $z = -1$ we get $$x = (-1)^2 -1 -1 = -1$$ Similarly, $y = -1$ Thus, clearly $(-1,-1,-1)$ is a solution. Now in the second case, we have $xyz = 1$ Also, as you have correctly pointed out, $x^2 + y^2 + z^2 = 3$. Thus we can say, $$x^2 + y^2 + z^2 = 3\times 1 = 3xyz$$ $$\implies (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 0$$ This is possible if and only if $x=y=z$ or if $x+y+z=0$. Considering $x=y=z$, we get, $x^3 = 1$. Hence for real values of x, $x = y = z = 1$. If you include complex values, $\omega$ and $\omega^2$ also satisfy the condition $xyz = 1$. Hence these are also valid solutions. Now, let's consider the other case where $x + y + z = 0$.

Then, $$x + y + \frac{1}{xy} = 0 \implies y = \frac{-x^2 \pm \sqrt{x^4 - 4x}}{2x}$$ This expression can be used to solve for x, and hence for y and z.

Answer 2

Set $f(t) = t^2+t-1$. Then $x, y, z$ are solutions of the equation $$ 0 = f(f(f(t))) - t = (t-1)(t+1)(t^6+4t^5+5t^4+2t^3+1) \, . $$ The last factor $g(t) = t^6+4t^5+5t^4+2t^3+1$ has no real zeros, $g$ attains a positive global minimum at $t = (\sqrt{13}-7)/6$.

It follows that $x=y=z=1$ and $x=y=z=-1$ are the only real solutions.