253. Find all possible systems of four real numbers such that the sum of each of them with the product of the others is 2.
253. Let $x, y, z, t$ be the desired numbers and $x y z t = A$; note that $A \neq 0$, since if, for example, $x=0$, then the conditions of the problem lead to contradictory equalities $y=z=t=2$ and $y z t=2$. Further, the equation $x + y z t = 2$ can be rewritten as:

$$
x + \frac{A}{x} = 2, \text{ or } x^2 - 2x + A = 0
$$

Similarly, we obtain

$$
y^2 - 2y + A = 0, \quad z^2 - 2z + A = 0, \quad t^2 - 2t + A = 0
$$

But the equation $x^2 - 2x + A = 0$ can have (for a given $A$) only two different roots; therefore, among the numbers $x, y, z, t$ there are no more than two different ones. We will consider several possible cases separately.

$1^{\circ}$. If $x = y = z = t$, then our equations give

$x + x^3 = 2$, or $x^3 + x - 2 = 0$,

or, finally, $(x-1)(x^2 + x + 2) = 0$,

from which $x_1 = 1, x_{2,3} = \frac{-1 \pm \sqrt{\overline{ }}}{2}$. Thus, here we have the only solution to the system in real numbers: $x = y = z = t = 1$.

$2^{\circ}$. If $x = y = z$, and $t$ may be different, then the conditions of the problem give

$$
x + x^2 t = 2 \text{ and } t + x^3 = 2
$$

Subtracting one of the equations (*) from the other, we get:

$$
x^3 - x^2 t - x + t = 0, \text{ or } (x - t)(x^2 - 1) = 0
$$

therefore, either $x = t$ (this case we have already considered above), or $x = \pm 1$. But if $x = 1$, the first of the equations (*) immediately gives $t = 1$ - and we again arrive at the solution obtained above; if $x = -1$, then the same equation gives $t = 3$.

$3^{\circ}$. If $x = y$ and $z = t$, then the system reduces to the following:

$$
x + x z^2 = 2, \quad z + x^2 z = 2
$$

Subtracting these equations term by term, we get:

$$
x - z + x z^2 - x^2 z = 0, \quad \text{or} \quad (x - z)(1 - x z) = 0
$$

But the equality $x = z$ immediately leads us to case $1^{\circ}$, and if $x z = 1$ and (due to the first of the equations (**)) $x + z = 2$, then we again find the already known solution $x = z = 1$.

Answer: either all 4 numbers are equal to 1, or three of them are equal to -1 and one is equal to 3.