9.2. Find all such triples of numbers such that each number is equal to the square of the sum of the other two.
Answer: $(0,0,0) ;\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$.

Solution. Let's write the condition algebraically: $a=(b+c)^{2}, b=(a+c)^{2}, c=(a+b)^{2}$. Suppose among the numbers there are different ones, for example, $a \neq b$. Subtracting the second equation from the first, we get $a-b=(b+c)^{2}-(a+c)^{2}=(b-a)(b+a+2 c)$. Since $a \neq b$, it follows from the obtained equation that $a+b+2 c=-1$. But this is impossible, since the numbers $a$, $b$, and $c$ are squares, so their sum is non-negative.

Thus, $a=b=c$. Then each equation has the form $a=(2 a)^{2}$, so $a=b=c=0$ or $a=b=c=\frac{1}{4}$.

Grading criteria.

«+» A complete and justified solution is provided

«Ғ» It is proven that the three numbers must be equal, but an error is made in the answer (for example, only one triplet is found)

«Ғ» The correct answer is obtained based on the assumption that the three numbers are equal, but this is not proven

«Ғ» Only the correct answer is provided

«-» Only one of the triplets is guessed

«-» The problem is not solved or is solved incorrectly