Auswahlklausur

Solution:

For an indirect proof we assume that there exists a $k \in \mathbb{N}$ with $m^{2} + \left\lceil \frac{4 m^{2}}{n} \right\rceil = (m + k)^{2}$, i.e. $\left\lceil \frac{(2m)^{2}}{n} \right\rceil = (2m + k)k$. Obviously $k \geq 1$. Thus the equation $\left\lceil \frac{c^{2}}{n} \right\rceil = (c + k)k$ (1) has a positive integer solution $(c, k)$ with even $c$. Without paying attention to the parity of $c$, we consider such a solution of (1) with minimal $k$.

From $\frac{c^{2}}{n} > \left\lceil \frac{c^{2}}{n} \right\rceil - 1 = c k + k^{2} - 1 \geq c k$ and $\frac{(c - k)(c + k)}{n} < \frac{c^{2}}{n} \leq \left\lceil \frac{c^{2}}{n} \right\rceil = (c + k)k$ we obtain $c > n k > n - k$, so that $c = k n + r$ holds with a suitable $0 < r < k$.

Substituting this into (1) gives $\left\lceil \frac{c^{2}}{n} \right\rceil = \left\lceil \frac{(n k + r)^{2}}{n} \right\rceil = k^{2} n + 2 k r + \left\lceil \frac{r^{2}}{n} \right\rceil$ and $(c + k)k = (k n + r + k)k = k^{2} n + 2 k r + k(k - r)$, so that $\left\lceil \frac{r^{2}}{n} \right\rceil = k(k - r)$ (2) follows.

This yields another positive integer solution of (1) with $c' = r$ and $k' = k - r < k$, which contradicts the minimality of $k$.

Variant:

Let $m^{2} + \left\lceil \frac{4 m^{2}}{n} \right\rceil = c^{2}$ for a positive integer $c > m$, from which follows $c^{2} - 1 < m^{2} + \frac{4 m^{2}}{n} \leq c^{2}$ and from this $0 \leq c^{2} n - m^{2}(n + 4) < n$ (3).

We substitute $d = c^{2} n - m^{2}(n + 4)$, $x = c + m$ and $y = c - m$, obtain $c = \frac{x + y}{2}$ as well as $m = \frac{x - y}{2}$ and thereby rewrite (3):

$\left( \frac{x + y}{2} \right)^{2} n - \left( \frac{x - y}{2} \right)^{2}(n + 4) = d$ or $x^{2} - (n + 2) x y + y^{2} + d = 0$ with $0 \leq d < n$.

For fixed $n$ and $d$ we choose an integer solution pair $(x, y)$ for which $x + y$ is minimal. Because of the symmetry of (4) we may assume $x \geq y \geq 1$. As above, another solution $(z, y)$ can be found here as well, for which $z < x$ — contradiction!