Auswahlklausur /, 2014

Solution:
Equation (1) is symmetric in $x$ and $y$, so that we may first assume $x \geq y$ and obtain, for every solution $(x, y)$ with $x \neq y$, also $(y, x)$ as a solution. With
$d = x - y \geq 0$ it follows that $\sqrt[3]{7 d^{2} + x y} = d + 1$. Raising to a power gives $x^{2} - d x + \left(-d^{3} + 4 d^{2} - 3 d - 1\right) = 0$ with discriminant $D = d^{2} - 4\left(-d^{3} + 4 d^{2} - 3 d - 1\right) = (d-2)^{2}(4 d + 1) \geq 0$. Thus $x_{1/2} = \frac{d \pm (d-2) \sqrt{4 d + 1}}{2}$. For $x$ to be an integer, $4 d + 1$ must be a perfect square, and in fact the square of an odd number. The ansatz $4 d + 1 = (2 m + 1)^{2}$ with $m \in \{0; 1; 2; \ldots\}$ gives $d = m^{2} + m$, so that we obtain $x_{1/2} = \frac{1}{2}\left[\left(m^{2} + m\right) \pm \left(m^{2} + m - 2\right)(2 m + 1)\right]$. This leads to $\left(x_{1}, y_{1}\right) = \left(m^{3} + 2 m^{2} - m - 1, m^{3} + m^{2} - 2 m - 1\right)$ and, for $m \neq 1$, to $\left(x_{2}, y_{2}\right) = \left(-m^{3} - m^{2} + 2 m + 1, -m^{3} - 2 m^{2} + m + 1\right)$ and the solutions $\left(y_{1}, x_{1}\right)$ resp. $\left(y_{2}, x_{2}\right)$ for $m > 0$. A check confirms that these pairs are indeed solutions.