9. Stubborn Squares (from 7th grade. 2 points). Given 100 numbers. 2 was added to each of them. The sum of the squares of the numbers did not change. 2 was added to each of the resulting numbers again. How did the sum of the squares change now?
Solution. Let the given numbers be $x_{1}, x_{2}, \ldots, x_{100}$. By adding 2 to each number, we get the set $y_{j}=x_{j}+2$. Adding 2 again, we get the set $z_{j}=y_{j}+2=x_{j}+4$. However, the variances of all three sets are equal, that is,

$$
\overline{x^{2}}-\bar{x}^{2}=\overline{y^{2}}-\bar{y}^{2}=\overline{z^{2}}-\bar{z}^{2}
$$

By the condition, $\overline{x^{2}}=\overline{y^{2}}$ and, obviously, $\bar{y}=\bar{x}+2$. Therefore, $\bar{x}^{2}=(\bar{x}+2)^{2}$, from which $\bar{x}=-1$. Then $\bar{z}=-1+4=3$. Therefore,

$$
\overline{z^{2}}=\overline{x^{2}}+\bar{z}^{2}-\bar{x}^{2}=\overline{x^{2}}+9-1=\overline{x^{2}}+8
$$

Since there are 100 numbers in total, the sum of the squares has increased by $8 \cdot 100=800$.

Answer: the sum of the squares increased by 800.