3. At a chess tournament with many draws, two students and several college students participated. Everyone played against everyone. The two students scored a total of 8 points, while all the college students scored an equal number of points. How many college students participated in the tournament and how many points did they score. Find all solutions.
Solution. Let $x$ be the number of students who participated in the tournament and $k$ be the number of points each of them earned. Then

$$
\binom{x+2}{2}=k x+8, \text { i.e. } x^{2}-(2 k-3) x-14=0
$$

Since $x, k \in \mathbb{N}$, the discriminant of the above equation must be a perfect square, i.e.

$$
(2 k-3)^{2}+56=n^{2}
$$

or $(2 k-3)^{2}-n^{2}=-56$. This can be written in the following form

$$
(n-2 k+3)(n+2 k-3)=56=1 \cdot 2 \cdot 2 \cdot 2 \cdot 7
$$

We have the following possibilities:

| $1^{\circ} n-2 k+3=1$ | and | $n+2 k-3=56$ |
| :--- | :--- | :--- |
| $2^{\circ} n-2 k+3=2$ | and | $n+2 k-3=28$ |
| $3^{\circ} n-2 k+3=4$ | and | $n+2 k-3=14$ |
| $4^{\circ} n-2 k+3=8$ | and | $n+2 k-3=7$ |
| $5^{\circ} n-2 k+3=56$ | and | $n+2 k-3=1$ |
| $6^{\circ} n-2 k+3=28$ | and | $n+2 k-3=2$ |
| $7^{\circ} n-2 k+3=14$ | and | $n+2 k-3=4$ |
| $8^{\circ} n-2 k+3=7$ | and | $n+2 k-3=8$ |

By direct verification, it is seen that only the systems under $2^{\circ}$ and $3^{\circ}$ give solutions that meet the given conditions.

In $2^{\circ}$, we get $n=15, k=8$ and $x=14$

In $3^{\circ}$, we get $n=9, k=4$ and $x=7$