National Math Olympiad in Slovenia

The number of pairs of students over all disciplines can be counted in two ways. On the one hand the number of pairs is equal to the product of the number of all pairs and the number of times every pair of students competes together, which is $\binom{n}{2} \cdot 3$. On the other hand it must be equal to the product of the number of pairs competing in each discipline and the number of disciplines, which is $\binom{k}{2} \cdot m$. So,
$$3n(n-1) = mk(k-1). \qquad (11)$$
Similarly, we can count the number of triples in two ways. On the one hand it is equal to $\binom{n}{3} \cdot 2$, on the other hand it must be $\binom{k}{3} \cdot m$. So,
$$2n(n-1)(n-2) = mk(k-1)(k-2).$$
Since $k, n \ge 3$, we can divide one equation by the other and get
$$\frac{2(n-2)}{3} = k-2.$$
This implies that $n-2$ is divisible by 3 so we can write $n = 3l + 2$, where $l \in \mathbb{N}$. Inserting this back into the equation we get $k = 2l + 2$. We can then use (11) to get
$$m = \frac{3n(n-1)}{k(k-1)} = \frac{3(3l+2)(3l+1)}{(2l+2)(2l+1)}.$$
However, $D(3l+1, 2l+1) = D(2l+1, l) = 1$ and $D(3l+2, 2l+1) = D(2l+1, l+1) = D(l+1, l) = 1$. So, $2l+1$ and $(3l+2)(3l+1)$ are relatively prime and $2l+1$ divides 3. This is only possible when $l=1$ and consequently $n=5$, $k=4$ and $m=5$.
Finally, we must check that the required arrangement of students is indeed possible for $n=5$ and $k=4$. Denote the students by $A_1, A_2, \dots, A_5$. Here is one possible arrangement that satisfies the conditions given in the problem:
$\{A_1, A_2, A_3, A_4\}, \{A_1, A_2, A_3, A_5\}, \{A_1, A_2, A_4, A_5\}, \{A_1, A_3, A_4, A_5\}, \{A_2, A_3, A_4, A_5\}.$
Thus, the only solution is $n=5, k=4$.