
1. In the domain of real numbers solve the following system of equations

$$
\begin{aligned}
& y+3 x=4 x^{3}, \\
& x+3 y=4 y^{3} .
\end{aligned}
$$

(Pavel Calábek)


Solution. The sum and the difference of the equations gives

$$
\begin{aligned}
& 4(x+y)=4\left(x^{3}+y^{3}\right), \\
& 2(x-y)=4\left(x^{3}-y^{3}\right),
\end{aligned}
$$

or

$$
\begin{aligned}
x+y & =(x+y)\left(x^{2}-x y+y^{2}\right), \\
\frac{1}{2}(x-y) & =(x-y)\left(x^{2}+x y+y^{2}\right) .
\end{aligned}
$$

If $x+y=0$, then the substitution $y=-x$ into the original first equation gives $2 x=4 x^{3}$, or $x\left(2 x^{2}-1\right)=0$. This equation has three roots $x=0$ and $x= \pm \frac{1}{2} \sqrt{2}$, thus we get three solutions $(0,0),\left(\frac{1}{2} \sqrt{2},-\frac{1}{2} \sqrt{2}\right)$, and $\left(-\frac{1}{2} \sqrt{2}, \frac{1}{2} \sqrt{2}\right)$.

If $x-y=0$, then substituting $y=x$ we get $4 x=4 x^{3}$, or $x\left(x^{2}-1\right)=0$ and we get two new solutions $(1,1)$ and $(-1,-1)$.

Now let $x+y$ and $x-y$ be both non-zero. Then

$$
\begin{aligned}
& 1=x^{2}-x y+y^{2}, \\
& \frac{1}{2}=x^{2}+x y+y^{2} .
\end{aligned}
$$

The sum and the difference of the equations is $x^{2}+y^{2}=\frac{3}{4}$ and $x y=-\frac{1}{4}$, thus

$$
(x+y)^{2}=x^{2}+y^{2}+2 x y=\frac{3}{4}-\frac{1}{2}=\frac{1}{4},
$$

and $x+y=\frac{1}{2}$ or $x+y=-\frac{1}{2}$, it means $x, y$ are then the solutions of the quadratic equations

$$
t^{2}-\frac{1}{2} t-\frac{1}{4}=0, \quad \text { resp. } \quad t^{2}+\frac{1}{2} t-\frac{1}{4}=0
$$

with roots $t_{1,2}=\frac{1}{4} \pm \frac{1}{4} \sqrt{5}$, resp. $t_{3,4}=-\frac{1}{4} \pm \frac{1}{4} \sqrt{5}$, and we get four other solutions: $\left(t_{1}, t_{2}\right),\left(t_{2}, t_{1}\right),\left(t_{3}, t_{4}\right),\left(t_{4}, t_{3}\right)$.

Conclusion. The solutions of the problem are

$$
\begin{gathered}
(0,0),\left(\frac{1}{2} \sqrt{2},-\frac{1}{2} \sqrt{2}\right),\left(-\frac{1}{2} \sqrt{2}, \frac{1}{2} \sqrt{2}\right),(1,1),(-1,-1), \\
\left(\frac{1}{4}+\frac{1}{4} \sqrt{5}, \frac{1}{4}-\frac{1}{4} \sqrt{5}\right),\left(\frac{1}{4}-\frac{1}{4} \sqrt{5}, \frac{1}{4}+\frac{1}{4} \sqrt{5}\right) \\
\left(-\frac{1}{4}+\frac{1}{4} \sqrt{5},-\frac{1}{4}-\frac{1}{4} \sqrt{5}\right),\left(-\frac{1}{4}-\frac{1}{4} \sqrt{5},-\frac{1}{4}+\frac{1}{4} \sqrt{5}\right) .
\end{gathered}
$$
