## Task 11/89

We are looking for all pairs $(n ; k)$ of natural numbers $n$ and $k$ for which the following holds: The sum of the squares of $n$ and its $k$ immediate predecessors is equal to the sum of the squares of the $k$ immediate successors of $n$.
According to the text of the task, it should hold that

$$
\begin{gathered}
\sum_{i=0}^{k}(n-i)^{2} \sum_{i=1}^{k}(n+i)^{2} \\
\sum_{i=0}^{k} n^{2}-2 n \sum_{i=0}^{k} i+\sum_{i=0}^{k} i^{2}=\sum_{i=1}^{k} n^{2}+2 n \sum_{i=1}^{k} i+\sum_{i=1}^{k} i^{2}
\end{gathered}
$$

due to $\sum_{i=0}^{k} n^{2}=n^{2}+\sum_{i=1}^{k} n^{2}, \sum_{i=0}^{k} i=\sum_{i=1}^{k} i$ and $\sum_{i=0}^{k} i^{2}=\sum_{i=1}^{k} i^{2}$, it follows from this by equivalent transformation that

$$
n^{2}=4 n \sum_{i=1}^{k} i=4 n \frac{k(k+1)}{2}=2 n k(k+1)
$$

Now, it is certain that $n \neq 0$, as otherwise the immediate predecessors would not be natural numbers. Thus, it follows that $n=2 k(k+1)$. Therefore, there exist infinitely many pairs $(n ; k)$ with $(n ; k)=[2 k(k+1) ; k]$ where $k \in \mathbb{N}, k \geq 1$.