11.2. Find all triples of real numbers such that each of these numbers is equal to the square of the difference of the other two.
Answer. $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$.

Solution. Write the condition as a system of equations in the unknowns $x, y, z$ as follows: $x=(y-z)^{2}, y=(x-z)^{2}, z=(x-y)^{2}$. Subtract the second equation from the first, the third from the first, and the third from the second, to obtain three equations - consequences:

$x-y=(x-y)(2 z-x-y), x-z=(x-z)(2 y-x-z), y-z=(y-z)(2 x-y-z)$. If the variables are not equal in pairs, then by canceling the first parentheses, we get: $2 z-x-y=2 y-x-z=2 x-y-z=1$. Adding the three obtained equalities, we have 0=3 - a contradiction. Therefore, among the variables there are equal ones. Let, for example, $x=y$. Then from the third equation of the original system $z=0$, and from the first $x=x^{2}$, from which $x=y=0$ or $x=y=1$. Similarly, the equality of the other variables is considered. Verification shows that the found triples of numbers are solutions of the system.

Evaluation. Just the answer with verification: 0 points. Loss of each solution minus 2 points. Unjustified cancellation of parentheses - minus 3 points.