David Ali

2022-09-20

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

Nathalie Rivers

Suppose that we had $x>1$. Then, since $4{x}^{3}-3x>x$, we have $z>x$, and similarly $y>z$, $x>y$, contradiction. By symmetry, we conclude that $x,y,z\in \left[-1,1\right]$.
So there exist $\alpha ,\beta ,\gamma \in \left[0,\pi \right]$ with $x=\mathrm{cos}\alpha$, $y=\mathrm{cos}\beta$, $z=\mathrm{cos}\gamma$. By the formula for $\mathrm{cos}3\alpha$, we can rewrite the system of equations as:

So we have $±27\alpha \equiv \alpha$, 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, $x=\mathrm{cos}\frac{\pi k}{13}$ for $0\le k\le 13$, and $x=\mathrm{cos}\frac{\pi k}{14}$ for $1\le k\le 13$.
For example, one solution is ($\left(\mathrm{cos}\frac{\pi }{14},\mathrm{cos}\frac{9\pi }{14},\mathrm{cos}\frac{3\pi }{14}\right)$).

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