62nd Ukrainian National Mathematical Olympiad

Suppose that such numbers exist. Since they are equal to some squares, each of these numbers is nonnegative. Let's denote the sum of all these 10 numbers by $S$. Let's pick one of these numbers and denote it by $a$, then $S\ge a$, and also $$a=(S-a{)}^2\Rightarrow a^2-(2S+1)a+S^2=0.$$ If there are two different $a$, satisfying this condition, then by Vieta's theorem their product is equal to $S^2$. However, this means that one of them is greater than $S$, which contradicts what we proved above. Thus, there is only one possible $a$, and it follows that all numbers are equal.