\section*{Problem 2 - 131032}

Determine all pairs \((x ; y)\) of integers \(x, y\) that satisfy the equation \(2 x^{3}+x y-7=0\)!
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The equation is equivalent to \(x \cdot\left(2 x^{2}+y\right)=7\), so that \(x\) is an integer divisor of 7, and thus only the following four cases are possible:

1. Case: \(x=1\). Then it follows that \(y=\frac{7}{1}-2 \cdot 1^{2}=5\).

2. Case: \(x=-1\). Then it follows that \(y=\frac{7}{-1}-2 \cdot(-1)^{2}=-9\).

3. Case: \(x=7\). Then it follows that \(y=\frac{7}{7}-2 \cdot 7^{2}=-97\).

4. Case: \(x=-7\). Then it follows that \(y=\frac{7}{-7}-2 \cdot(-7)^{2}=-99\).

The checks confirm each of the results, so we obtain four solution pairs, namely \((x, y) \in\) \(\{(-7,-99),(-1,-9),(1,5),(7,-97)\}\).