Task 2. Find all quadruples $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ of real numbers such that the following six equalities hold:

$$
\begin{aligned}
& x_{1}+x_{2}=x_{3}^{2}+x_{4}^{2}+6 x_{3} x_{4}, \\
& x_{1}+x_{3}=x_{2}^{2}+x_{4}^{2}+6 x_{2} x_{4}, \\
& x_{1}+x_{4}=x_{2}^{2}+x_{3}^{2}+6 x_{2} x_{3}, \\
& x_{2}+x_{3}=x_{1}^{2}+x_{4}^{2}+6 x_{1} x_{4}, \\
& x_{2}+x_{4}=x_{1}^{2}+x_{3}^{2}+6 x_{1} x_{3}, \\
& x_{3}+x_{4}=x_{1}^{2}+x_{2}^{2}+6 x_{1} x_{2} .
\end{aligned}
$$
Solution I. Subtracting the first two equations from each other gives $x_{2}-x_{3}=x_{3}^{2}-x_{2}^{2}+6 x_{4}(x_{3}-x_{2})$, which we can factor as $0=(x_{3}-x_{2})(x_{3}+x_{2}+1+6 x_{4})$. We see that $x_{2}=x_{3}$ or $x_{2}+x_{3}+1+6 x_{4}=0$. Similarly, it also holds that $x_{2}=x_{3}$ or $x_{2}+x_{3}+1+6 x_{1}=0$. Therefore, if $x_{2} \neq x_{3}$, then in both cases the second equality holds; subtracting these gives $x_{1}=x_{4}$. We conclude that $x_{2}=x_{3}$ or $x_{1}=x_{4}$. Similarly, for any permutation $(i, j, k, l)$ of $(1,2,3,4)$, it holds that $x_{i}=x_{j}$ or $x_{k}=x_{l}$.

We now prove that at least three of the $x_{i}$ are equal. If they are all equal, this is obviously true. Otherwise, there are two that are unequal, say without loss of generality $x_{1} \neq x_{2}$. Then $x_{3}=x_{4}$. If now $x_{1}=x_{3}$, then there are three elements equal. Otherwise, $x_{1} \neq x_{3}$, so $x_{2}=x_{4}$ and also then there are three elements equal. The quadruple $(x_{1}, x_{2}, x_{3}, x_{4})$ is thus, up to order, of the form $(x, x, x, y)$, where it may be that $x=y$.

Substituting this into the given equations gives $x+y=8 x^{2}$ and $2 x=x^{2}+y^{2}+6 x y$. Adding these two equations: $3 x+y=9 x^{2}+y^{2}+6 x y$. The right-hand side can be factored as $(3 x+y)^{2}$. With $s=3 x+y$ we thus have $s=s^{2}$, from which it follows that $s=0$ or $s=1$. It holds that $s=3 x+y=2 x+(x+y)=2 x+8 x^{2}$. Therefore, $8 x^{2}+2 x=0$ or $8 x^{2}+2 x=1$.

In the first case, $x=0$ or $x=-\frac{1}{4}$. We find $y=0-3 x=0$ respectively $y=0-3 x=\frac{3}{4}$. In the second case, we can factor the equation as $(4 x-1)(2 x+1)=0$, so $x=\frac{1}{4}$ or $x=-\frac{1}{2}$. We find $y=1-3 x=\frac{1}{4}$ respectively $y=1-3 x=\frac{5}{2}$.

All together, we have found the following quadruples: $(0,0,0,0),\left(-\frac{1}{4},-\frac{1}{4},-\frac{1}{4}, \frac{3}{4}\right)$ and the permutations thereof, $\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$ and $\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \frac{5}{2}\right)$ and the permutations thereof. Checking shows that all these quadruples satisfy the equations.