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Find answers of this system of equations in reals$$ \left\{ \begin{array}{c} x+3y=4y^3 \\ y+3z=4z^3 \\ z+3x=4x^3 \end{array} \right. $$

Things O have done: summing these 3 equations give $$4(x+y+z)=4(x^3+y^3+z^3)$$ $$x+y+z=x^3+y^3+z^3$$

I've also tried to show that $x,y,z$ should be between $1$ and $-1$(As $(-1,-1,-1)$ and $(1,1,1)$ are answers of this system of equations) but I was not successful.

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  • $\begingroup$ It reminds me of $$\cos3A$$ formula $\endgroup$ Sep 9, 2014 at 6:24
  • $\begingroup$ Is this a homework problem? $\endgroup$ Sep 9, 2014 at 6:28
  • $\begingroup$ @MhenniBenghorbal No.i self studying algebra and this was a question from book that i was not able to solve it. $\endgroup$ Sep 9, 2014 at 6:29
  • $\begingroup$ You can see that $(0,0,0),(1,1,1), (-1,-1,-1)$ are solutions. $\endgroup$ Sep 9, 2014 at 6:49
  • $\begingroup$ @MhenniBenghorbal,thanks.but is there any better way beside guessing? I wish i knew about functions so i could understand the answer is posted here. $\endgroup$ Sep 9, 2014 at 6:54

2 Answers 2

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Suppose that we had $x>1$. Then, since $4x^3-3x > x$, we have $z>x$, and similarly $y>z$, $x>y$, contradiction. By symmetry, we conclude that $x,y,z\in [-1,1]$.

So there exist $\alpha,\beta,\gamma \in [0,\pi]$ with $x=\cos \alpha$, $y=\cos \beta$, $z=\cos \gamma$. By the formula for $\cos 3\alpha$, we can rewrite the system of equations as:

$$ \left\{ \begin{array}{c} \alpha \equiv \pm 3 \beta\ (\operatorname{mod}\ 2\pi) \\ \beta = \pm 3 \gamma\ (\operatorname{mod}\ 2\pi) \\ \gamma = \pm 3\alpha\ (\operatorname{mod}\ 2\pi) \end{array} \right. $$

So we have $\pm 27\alpha \equiv \alpha$, so either $26\alpha \equiv 0$ or $28\alpha \equiv 0$. We find that $\alpha = \pi k / 13$ or $\alpha = \pi k / 14$ for some nonnegative integer $k$.

This gives $27$ solutions,* $x=\cos{\frac{\pi k}{13}}$ for $0\leq k\leq 13$, and $x=\cos{\frac{\pi k}{14}}$ for $1\leq k\leq 13$.

For example, one solution is $(\cos\frac{\pi}{14},\cos\frac{9\pi}{14},\cos\frac{3\pi}{14})$.

* This is exactly the number we expect, up to multiplicity, when intersecting three cubic surfaces; in particular there are no more solutions over $\mathbb{C}$.

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    $\begingroup$ Well thought and well written; the techniques used are well within the precalculus realm. $\endgroup$
    – Mirlan
    Oct 18, 2014 at 23:29
  • $\begingroup$ I think this solution is incorrect because if X lies between -1 and 0 also we'll get z>x, y>z and x>y. Similarly if X is not 1, 0 and -1 we get other way round. Which means x should be either 0 or 1 or -1. And these are the only real solutions. $\endgroup$ Apr 18, 2020 at 11:40
  • $\begingroup$ @saisaandeep I think you're mistaken. I found a small error—I mixed up y and z in the example solution—but you can verify computationally that $(x,y,z) = (\cos\frac{\pi}{14},\cos\frac{9\pi}{14},\cos\frac{3\pi}{14})$ is a solution. $\endgroup$
    – Slade
    Apr 21, 2020 at 12:15
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Consider $f(x) = 4x^3 - 3x$. We have: $f'(x) = 12x^2 - 3 = 0 \iff x = \pm \dfrac{1}{2}$. Thus:

If $x \in \left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ then $f'(x) < 0$, and $f$ decreases. This means:

If $x > y > z$, then: $x = f(y) < f(z) = y$, a contradiction. And we can obtain a contradiction for any other inequalities of $x,y,z$ using the same argument. The same argument works for $x \leq -\dfrac{1}{2}$ or $x \geq \dfrac{1}{2}$ since then $f$ increases.

For if $x > y = z$, then: $y = f(z) > f(x) = z$, contradiction again. Thus we must have:

$x = y = z$, and we have: $x = y = z = \pm 1$ and $0$

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  • $\begingroup$ How about $x=y=z=0$? $\endgroup$ Sep 9, 2014 at 6:41
  • $\begingroup$ The tags on the question, and op's comment, indicate that this is to be done with precalculus techniques. So the derivative can't be used. $\endgroup$ Sep 9, 2014 at 6:55
  • $\begingroup$ This assumes that all of $x,y,z$ lie in the same region. But there are solutions where they don't. $\endgroup$
    – Slade
    Sep 9, 2014 at 7:04

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