Problem 2. Find the integers n, such that $\sqrt{24 n+1}$ is a natural number.
## Solution. 1p ex officio

The condition for the existence of the root is $24 n+1 \geq 0 \Rightarrow n \geq-\frac{1}{24}$ and $n \in \mathbb{Z} \Rightarrow n \in \mathbb{N}$. 0.5p

$24 n +1$ is an odd number and since $\sqrt{24 n+1} \in \mathbb{N} \Rightarrow 24 n+1=(2 k+1)^{2},(\forall) k \in \mathbb{N}$..... 1p

We obtain $24 n+1=4 k^{2}+4 k+1 \Rightarrow 24 n=4 k(k+1) \Rightarrow n=\frac{k(k+1)}{6}$ 1p

Since $n \in \mathbb{N} \Rightarrow k(k+1) \vdots 6$ 0.5p

But $k$ and $k+1$ are consecutive numbers and it follows that $k(k+1) \vdots 2,(\forall) k \in \mathbb{N}$......

Thus $k(k+1) \vdots 3 \Rightarrow k=3 t$ or $k=3 t+2,(\forall) t \in \mathbb{N}$ 1p

If: $k=3 t \Rightarrow n=\frac{t(3 t+1)}{2}$

$$
k=3 t+2 \Rightarrow n=\frac{(3 t+2)(t+1)}{2}
$$

Therefore, $n \in\left\{\frac{t(3 t+1)}{2}\right\} \cup\left\{\frac{(3 t+2)(t+1)}{2}\right\}$, where $t \in \mathbb{N}$.

Finalization 1p

## Note. For finding values of $\mathbf{n}$, 1 point is awarded.