10.4. Find all sets of four real numbers such that each number, when added to the product of the other three, equals 2.
Answer. $\{1,1,1,1\},\{-1,-1,-1,3\},\{-1,-1,3,-1\},\{-1,3,-1,-1\}$ and $\{3,-1,-1,-1\}$.

Solution. Let the numbers of the sought quadruple be $a, b, c, d$. By the condition, $a+bcd=b+acd=2$, from the first equation we have $(a-b)(1-cd)=0$, and if $a \neq b, cd=1$ and from the second equation $a+b=2$. Therefore, for any pair of numbers in our quadruple, either these numbers are equal, or their sum is 2 and the product of the remaining pair of numbers is 1. In particular, among the numbers of our quadruple, there are at most two different numbers.

1) All four numbers are equal to each other, then $a^3 + a = 2$. Since the function $f(x) = x^3 + x$ is the sum of two monotonically increasing functions $f(x) = x^3$ and $f(x) = x$, the solution will be unique: $a=1$. We obtain the first sought quadruple $\{1,1,1,1\}$.
2) Some of the numbers (at least two) are equal to $a$, and the rest (at least one) are equal to $2-a$. If the numbers equal to $a$ are exactly two, then $a(2-a)=1$, from which $a=1$ and we actually have case 1). If the numbers equal to $a$ are exactly three, then $a^2=1$ and either $a=1$ and we again get case 1), or $a=-1$ and we get a new sought quadruple $\{-1,-1,-1,3\}$. Adding the solutions obtained by permuting the variables, we get three new quadruples.

Grading criteria. Only the case where all numbers are equal is discussed with proof: 2 points. Losing the case of equal numbers: minus 2 points. Losing symmetric solutions: minus 1 point.