Auswahlwettbewerb zur Internationalen Mathematik-Olympiade

Solution:

Transforming (1) gives $2 z = (m+n-1)^2 + m - n - 1 = 2 m - 2 + (m+n-1)(m+n-2)$, from which $z+1 = m + \frac{(m+n-1)(m+n-2)}{2}$ follows. We set $m+n-1 = k$ and obtain $z+1 = m + \frac{k(k-1)}{2}$, where $0 < m \leq k$ and $k \in \mathbb{Z}, k \geq 1$ holds. Furthermore, $\frac{k(k-1)}{2} \in \mathbb{Z}$.

Obviously, for every positive integer $z+1$ there exists exactly one $k$ with
$$\frac{k(k-1)}{2} < z+1 \leq \frac{k(k-1)}{2} + k = \frac{(k+1)k}{2}$$
Then $m = z+1 - \frac{k(k-1)}{2}$ and $n = k+1-m$ are also uniquely determined positive integers, so that the equation yields a unique representation of every positive integer $z+1$, and hence of every nonnegative integer $z$, by means of $m$ and $n$.

Solution:

Transforming (1) gives $2 z = (m+n)(m+n-1) - 2 n$. We note that the right-hand side is always even and set $m+n = k$. For constant $k$ ($k \geq 2$), $n$ can take the values $1, 2, \ldots, k-1$, so that the right-hand side yields respectively distinct even numbers from $k(k-1)-2$ to $k(k-1)-2(k-1) = (k-1)(k-2)$. For $k = 2, 3, \ldots$ this produces a complete, disjoint decomposition of the set of all nonnegative even integers. The number $2 z$ therefore lies in exactly one interval determined by $k$, at a position determined by $n$, whereby $m$ is also uniquely determined for it.