$\underline{\text { Folklore }}$

Can one find ten such consecutive natural numbers that the sum of their squares equals the sum of the squares of the next nine consecutive natural numbers?

#
Let's denote 10 consecutive numbers as $n-9, \ldots, n-1, n$. According to the condition,

$$
(n-9)^{2}+\ldots+(n-1)^{2}+n^{2}=(n+1)^{2}+\ldots+(n+9)^{2}
$$

$$
n^{2}=(n+9)^{2}-(n-9)^{2}+\ldots+(n+1)^{2}-(n-1)^{2}=2 n \cdot(18+16+\ldots+2)=180 n .
$$

From this, $n=180$, that is, $171^{2}+\ldots+180^{2}=181^{2}+\ldots+189^{2}$.

## Answer

It is possible.