11 positive numbers each equal to the sum of the squares of the other 10. Determine the numbers.
Solution. Let $a_{1}, a_{2}, \ldots, a_{11}$ be the 11 numbers, and denote their sum of squares by $Q$. According to the condition,

$$
a_{i}=Q-a_{i}^{2} \quad (i=1,2, \ldots, 11)
$$

which means for all $i$,

$$
a_{i}+a_{i}^{2}=Q
$$

However, the function $f(x)=x+x^{2}$ is strictly monotonic for non-negative $x$, so if $a_{1}+a_{1}^{2}=\ldots=a_{11}+a_{11}^{2}$, then necessarily $a_{1}=a_{2}=\ldots=a_{11}$.

Let the common value of the $a_{i}$ numbers be $a$, then according to the condition $a=10 a^{2}$, hence $a=\frac{1}{10}$. Therefore, all 11 numbers are $\frac{1}{10}$.

Remarks. 1. If $n+1$ positive numbers each equal the sum of the squares of the other $n$ numbers, then similarly we find that each number is $\frac{1}{n}$.

2. In the problem, it is not really essential to require that each number be positive, since the sum of squares of real numbers is at least non-negative. In this case, either each number is $\frac{1}{10}$, or all are 0.