Let's find all the quadruples of real numbers $x_{1}, x_{2}, x_{3}$, $x_{4}$, such that by adding to any of its elements the product of the other three, the sum is always 2.
I. solution. None of the $x_{i}$ can be 0, because it is easily seen that then the others could only be 2, but then $0+2 \cdot 2 \cdot 2 \neq 2$. Let us denote the product $x_{1} x_{2} x_{3} x_{4}$ by $y$, then we have to solve the system of equations

$$
x_{i}+\frac{y}{x_{i}}=2 \quad(i=1,2,3,4)
$$

From this,

$$
x_{i}^{2}-2 x_{i}+y=0, x_{i}=1 \pm \sqrt{1-y}, \text { otherwise }\left|x_{i}-1\right|=\sqrt{1-y}
$$

According to this, by reducing the unknowns by 1, we can get at most two different numbers. This can happen in three ways:

$\alpha$) all four $x_{i}-1$ differences are equal,

$\beta$) three are equal, the fourth is -1 times as large,

$\gamma$) two are equal, the other two are -1 times as large.

In the $\alpha$ case, $x_{1}=x_{2}=x_{3}=x_{4}$, so from (1)

$$
x_{1}+x_{1}^{3}=2, \quad x_{1}^{3}+x_{1}-2=\left(x_{1}-1\right)\left(x_{1}^{2}+x_{1}+2\right)=0
$$

and since the second factor is positive for all real $x_{1}$ values:

$$
x_{1}^{2}+x_{1}+2=\left(x_{1}+\frac{1}{2}\right)^{2}+\frac{9}{4} \geqq \frac{9}{4}>0
$$

only $x_{1}=1$ is possible. The number quadruple $1,1,1,1$ indeed satisfies the requirements.

In the $\beta$ case, for example, $x_{1}-1=x_{2}-1=x_{3}-1$, and thus $x_{1}=x_{2}=x_{3}$, and $x_{4}-1=1-x_{1}$. So $x_{4}=2-x_{1}$, and from (1) for $i=4$

$$
2-x_{1}+x_{1}^{3}=2, \quad x_{1}\left(x_{1}^{2}-1\right)=0
$$

and since $x_{1} \neq 0, x_{1}$ can be 1 or -1. The first leads to the previous solution, the latter to the number quadruple $-1,-1,-1,3$, which is also indeed a solution.

In the $\gamma$ case, for example, $x_{1}-1=x_{2}-1=1-x_{3}=1-x_{4}=x_{4}$, so $x_{1}=x_{2}$, and $x_{3}=x_{4}=2-x_{1}$, and from (1) for $i=3$

$$
2-x_{1}+x_{1}^{2}\left(2-x_{1}\right)=2, \quad x_{1}\left(x_{1}-1\right)^{2}=0
$$

we only get the solution $1,1,1,1$.

According to all this, two real number quadruples satisfy the requirement: $1,1,1,1$ and $-1,-1,-1,3$.

János Lörincz (Sárospatak, Rákóczi F. g. IV. o. t.)

II. solution. There is one solution where $x_{1}=x_{2}=x_{3}=x_{4}$, as we saw in the previous solution: $x_{1}=x_{2}=x_{3}=x_{4}=1$. If there are different numbers among the numbers, then there is one that differs from at least two others. Choose the numbering so that $x_{1} \neq x_{2}, x_{1} \neq x_{3}$ holds. Write down the requirement starting from $x_{1}$ and $x_{2}$, then form the difference of the two equalities:

$$
\begin{gathered}
x_{1}+x_{2} x_{3} x_{4}=2, \quad x_{2}+x_{1} x_{3} x_{4}=2 \\
x_{1}-x_{2}+x_{3} x_{4}\left(x_{2}-x_{1}\right)=\left(x_{1}-x_{2}\right)\left(1-x_{3} x_{4}\right)=0
\end{gathered}
$$

Similarly

$$
\left(x_{1}-x_{3}\right)\left(1-x_{2} x_{4}\right)=0
$$

Then from (3) and (4) $x_{3} x_{4}=1$, and $x_{2} x_{4}=1$, so from (2) $x_{1}+x_{2}=2$, and $x_{1}+x_{3}=2$, so $x_{2}=x_{3}$.

From the equation $x_{4}+x_{1} x_{2} x_{3}=2$ starting from $x_{4}$, $x_{1}$, $x_{3}$, $x_{4}$ can be easily eliminated if we multiply by $x_{2}$:

$$
\begin{gathered}
x_{2} x_{4}+x_{1} x_{2}^{2} x_{3}=1+\left(2-x_{2}\right) x_{2}^{3}=2 x_{2}, \quad x_{2}^{4}-2 x_{2}^{3}+2 x_{2}-1=0 \\
x_{2}^{4}-1-2 x_{2}\left(x_{2}^{2}-1\right)=\left(x_{2}^{2}-1\right)\left(x_{2}^{2}+1-2 x_{2}\right)=\left(x_{2}-1\right)^{3}\left(x_{2}+1\right)=0
\end{gathered}
$$

From this, either $x_{2}=x_{3}=1, x_{1}=2-x_{2}=1, x_{4}=1 / x_{2}=1$, which does not contain different values, or $x_{2}=x_{3}=-1$, $x_{4}=1 / x_{2}=-1, x_{1}=2-x_{2}=3$, and this indeed satisfies the requirements.