## Task A-2.1.

Determine all pairs $(n, k)$ of natural numbers for which

$$
n^{2}+n=k^{2}+2 k-9
$$
## First solution.

Let's write the given equality as

$$
n^{2}+n+10=(k+1)^{2}
$$

Notice that the expression on the right side is a perfect square, so the expression on the left side must also be a perfect square. Furthermore, observe that

$$
n^{2}<n^{2}+n+10<n^{2}+8 n+16=(n+4)^{2}
$$

for any natural number $n$.

Therefore, the only possibilities are as follows:

- $n^{2}+n+10=(n+1)^{2}$, or $n^{2}+n+10=n^{2}+2 n+1$, or $n=9$, which gives us the solution $(n, k)=(9,9)$.
- $n^{2}+n+10=(n+2)^{2}=n^{2}+4 n+4$, i.e., $n=2$, which gives us the solution $(n, k)=(2,3)$.
- $n^{2}+n+10=(n+3)^{2}=n^{2}+6 n+9$, i.e., $5 n=1$ and we see that there is no solution in this case.

Finally, the solutions we are looking for are $(n, k)=(9,9)$ and $(n, k)=(2,3)$.