[ Systems of nonlinear algebraic equations ]

Solve the systems:

a) $\left\{\begin{array}{l}x+3 y=4 y^{3} \\ y+3 z=4 z^{3} \\ z+3 x=4 x^{3}\end{array}\right.$

b) $\left\{\begin{array}{l}2 x+x^{2} y=y \\ 2 y+y^{2} z=z \\ 2 z+z^{2} x=x\end{array}\right.$

c) $\left\{\begin{array}{l}3\left(x+\frac{1}{x}\right)=4\left(y+\frac{1}{y}\right)=5\left(z+\frac{1}{z}\right) \\ x y+y z+x z=1 ;\end{array}\right.$

d) $\left\{\begin{array}{c}\frac{1-x^{2}}{1+x^{2}}=\frac{2 y}{1+y^{2}}=\frac{1-z^{2}}{1+z^{2}} \\ x y+y z+x z\end{array}\right.$
a) If one of the unknowns (for example, $x$) is greater than 1, then $z=4 x^{3}-3 x=x^{3}+3\left(x^{3}-x\right)>1$; similarly, $y>1$. But then

$x^{3}+y^{3}+z^{3}>x+y+z$, which contradicts the equality obtained by adding all the equations. Therefore, each of the unknowns does not exceed 1. Similarly, it is proved that each of them is not less than -1.

Therefore, we can make the substitution $x=\cos \varphi$. Then we get $z=\cos 3 \varphi, y=\cos 9 \varphi, x=\cos 27 \varphi$. Thus, $\cos 27 \varphi=$ $\cos \varphi$, that is,

$27 \varphi= \pm \varphi+2 k \pi$, from which $\varphi=k \pi / 14$ or $k \pi / 13$.

b) It is easy to check that none of the unknowns can equal 1. Therefore, the first equation can be written as $y=\frac{2 x}{1-x^{2}}$. After the substitution $x=\operatorname{tg} \varphi$ we get $y=\operatorname{tg} 2 \varphi, z=\operatorname{tg} 4 \varphi, x=\operatorname{tg} 8 \varphi$. Hence $\operatorname{tg} 8 \varphi=\operatorname{tg} \varphi$, that is, $8 \varphi=\varphi+k \pi$.

c) It is clear that all variables have the same sign, so we will assume that they are all positive.

Let $x=\operatorname{tg} \alpha, y=\operatorname{tg} \beta, z=\operatorname{tg} \gamma(0<\alpha, \beta, \gamma<\pi / 2)$. Then $\operatorname{tg} \gamma(\operatorname{tg} \alpha+\operatorname{tg} \beta)=1-\operatorname{tg} \alpha \operatorname{tg} \beta, \operatorname{tg} \gamma=\operatorname{ctg}(\alpha+\beta)$, that is, $\alpha+\beta+\gamma=\pi / 2$.

On the other hand, $x+\frac{1}{x}=\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha}=\frac{2}{\sin 2 \alpha}$, so $\frac{3}{\sin 2 \alpha}=\frac{4}{\sin 2 \beta}=\frac{5}{\sin 2 \gamma}$.

Therefore, $2 \alpha, 2 \beta, 2 \gamma$ are the angles of a right triangle with sides $3,4,5$.

From this, $z=\operatorname{tg} \pi / 4=1$. To find $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$, we can use either the half-angle formulas or the bisector property: the bisector of angle $2 \alpha$ divides the leg of length 3 into segments of length $15 / 9$ and $12 / 9$, so $\operatorname{tg} \alpha=12 / 9: 4=1 / 3$; similarly, $\operatorname{tg} \beta=1 / 2$.

d) Let $x=\operatorname{tg} \alpha / 2, y=\operatorname{tg} \beta / 2, z=\operatorname{tg} \gamma / 2(-\pi<\alpha, \beta, \gamma<\pi)$. Then the first two equations will be written as $\cos$ $\alpha=\sin \beta=\cos \gamma$, and the last one, under our constraints, similarly to c), will give the condition $\alpha+\beta+\gamma= \pm \pi$. From

the equality $\cos \alpha=\cos \gamma$ we get $\gamma= \pm \alpha$. But if $\gamma=-\alpha$, then $\beta= \pm \pi$, which is impossible.

Thus, $\gamma=\alpha, \beta= \pm \pi-2 \alpha, \sin 2 \alpha=\sin \beta=\cos \alpha$, from which $\cos \alpha=0$ or $\sin \alpha=1 / 2$.

In the first case, $\alpha= \pm \pi / 2, \beta=0$. In the second case, $\alpha=\pi / 6, \beta=2 \pi / 3$ or $\alpha=5 \pi / 6, \beta=-2 \pi / 3$.

Recall that $\operatorname{tg} \frac{\pi}{12}=2-\sqrt{3}$ (see problem $\underline{61169}$ ).

## Answer

a) $(0,0,0), \pm(1,1,1), \pm(\cos \pi / 14,-\cos 5 \pi / 14, \cos 3 \pi / 14), \pm(\cos \pi / 7,-\cos 2 \pi / 7, \cos 3 \pi / 7), \pm(\cos \pi / 13,-\cos 4 \pi / 13, \cos$ $3 \pi / 13)$

$\pm\left(\cos 2 \pi / 13,-\cos 5 \pi / 13, \cos { }^{6 \pi} / 13\right)$ and all sets obtained from the specified ones by cyclic permutations.

b) $(0,0,0), \pm(\operatorname{tg} \pi / 7, \operatorname{tg} 2 \pi / 7,-\operatorname{tg} 3 \pi / 7)$ and all sets obtained from the specified ones by cyclic permutations.

c) $\left( \pm \frac{1}{3}, \pm \frac{1}{2}, \pm 1\right)$.

d) $( \pm 1,0, \pm 1),(2 \pm \sqrt{3}, \mp \sqrt{3}, 2 \pm \sqrt{3})$.