Given that $x,y,z\in\mathbb R$, solve $$\begin{cases}6x^2-12x=y^3\\6y^2-12y=z^3\\6z^2-12z=x^3\end{cases}$$

I've tried adding the equalities but to no avail. I'd add what I've tried, but it'd be useless at best. I've tried using some cube identities, etc. but I seem to lack the general idea of a solution this problem could have.

P.S.: no calculus, logs or anything like that can be used. Just basic algebraic manipulations. Thanks.


Let $f(x)=\sqrt[3]{6x^2-12x}$. Then solutions to this system of equations correspond to $f^3(x)=x$, that is, order-3 fixpoints of $f$. (Note that $f^3=f\circ f\circ f$ here and below.)

Numerically, I count no less than $7$ real solutions:


These are in the cycles $-0.889\to2.488\to1.940$ and $-0.527\to2.0000001\to0.0123$, as well as the order 1 fixpoint $f(0)=0$.

I found these numbers by numerically finding the roots of $f^3(x)-x$ (which is an very complicated function). But we can bound the roots of this function, thus justifying the brute force solution of just looking for roots in that region.

First, note that any cycle must contain a nonpositive entry, because $f(x)<x$ for all $x>0$, Now $f(x)<x$ iff $6x^2-12x<x^3$ iff $x(x^2-6x+12)>0$, and $x^2-6x+12>0$ for all $x$ since the discriminant $\Delta=6^2-4\cdot12=-12$ is negative, so if the smallest element of the cycle is positive, $f(x)<x$ is a contradiction.

Since $6x^2-12x$ takes a minimum at $x=1$, $f(x)\ge f(1)=-\sqrt[3]6$, so this is our lower bound on any cycle in $f$. The upper bound is $f(-\sqrt[3]6)=3.46,$ because $f$ is decreasing for $x\le0$. Thus by inspecting the solutions to $f^3(x)=x$ in this region, be can be sure to find all solutions.

The final way of whittling down the solution space is to note that there is a bound on how wiggly $f^3(x)$ can be, given $f(x)$. Since $f$ is decreasing on $(-\infty,1]$ and increasing on $[1,\infty)$, $f^2$ will have $4$ regions of increasing/decreasing, and $f^3$ will have exactly $8$. (This happens because the minimum of $f$, $-\sqrt[3]6$, is less than the $x$ value for the minimum, that is, $1$.) Each one of these regions has a crossing with the line $y=x$ except the last one, where $f^3(x)<f^2(x)<f(x)<x$, as we have already observed. Thus, the $7$ solutions here are all of them.

I know of no closed form for any of them except for $0$. (Edit: Peter Sheldrick's solution gives these numbers as roots of a certain degree $24$ polynomial, which passes as a "closed form" in some circles. I'll let you be the judge.)

Here's a cobweb graph of the two cycles:

  • $\begingroup$ This is not very elementary as I'm not familiar with "order-$k$ fixpoints" and don't know what that is. $\endgroup$ – user26486 Apr 10 '14 at 21:43
  • $\begingroup$ A fixpoint of the function $f$ is a point such that $f(x)=x$. An order 3 fixpoint is a point such that $f(f(f(x)))=x$ (which I write $f^3(x)=x$ above). I'm afraid that the equation itself is not very elementary, so there is not much I can do. Even using calc and logs and stuff won't give you a closed form for these numbers. $\endgroup$ – Mario Carneiro Apr 10 '14 at 23:21
  • $\begingroup$ The reason it is called that is the dynamical systems viewpoint, where you "drop a point" $x$ in the function and watch where it goes, which is to say, you examine the sequence $x,f(x),f(f(x)),\dots$. If this sequence is periodic of length $n$, you have a fixpoint of order $n$. $\endgroup$ – Mario Carneiro Apr 10 '14 at 23:24
  • $\begingroup$ Even if the math goes over your head, you should be able to type the numbers in your calculator and verify that they do indeed solve the system of equations (which is all that really matters in the end). Note that your $x,y,z$ correspond to the three values of the cycle in order, that is, $-.889,2.49,1.94$ or cyclic permutations thereof, and similar for the other cycle. $\endgroup$ – Mario Carneiro Apr 11 '14 at 0:05

Symmetry dictates $x=y=z$ is a solution. Hence, we get $$x^3 - 6x^2+12x = 0\implies x(x^2-6x+12) = 0 \implies x = 0, x = 3 \pm i\sqrt3$$

  • 1
    $\begingroup$ wow, how can you so simply claim it's because of symmetry and leave it at that? Can a proof really consist of such statement? Even if you're right, I don't think leaving the proof as strict as you've made it to be is reasonable. I don't have a lot of experience, so tell me more about the way of proving all this using your given strategy. Thanks. $\endgroup$ – user26486 Apr 10 '14 at 21:01
  • $\begingroup$ @mathh I am not claiming all the solutions are given by this. All I am saying is the following: If $x=f(y,z)$, $y=f(z,x)$ and $z= f(x,y)$, then if we find $a$ such that $a = f(a,a)$, then $x=y=z=a$ is one solution. $\endgroup$ – user141421 Apr 10 '14 at 21:02
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    $\begingroup$ @Amzoti How can I figure out the amount of solutions nonlinear systems of equations contain? $\endgroup$ – user26486 Apr 10 '14 at 21:10
  • $\begingroup$ Note that $x, y, z$ are supposed to be real numbers, and $x=y=z=0$ is easily seen to be a solution. $\endgroup$ – Nicky Hekster Apr 10 '14 at 21:11
  • $\begingroup$ I'm not convinced you've provided the full set of solutions the system has. And if you haven't, we must find the rest of the solutions as we're asked to actually solve the system of equations, not just find a few solutions. $\endgroup$ – user26486 Apr 10 '14 at 21:19

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