\section*{Problem 1 - 221021}

Determine all pairs \((x ; y)\) of integers that satisfy the equation \(2 x^{3}+x y-7=0\).
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I. If a pair \((x ; y)\) of integers satisfies the equation \(2 x^{3}+x y-7=0\), then \(x\left(2 x^{2}+y\right)=7\). Since \(x\) and \(2 x^{2}+y\) are integers and 7 is a prime number, it follows that

(1) either \(x=1\) and \(2 x^{2}+y=7\)

(2) or \(x=7\) and \(2 x^{2}+y=1\)

(3) or \(x=-1\) and \(2 x^{2}+y=-7\)

(4) or \(x=-7\) and \(2 x^{2}+y=-1\)

From (1) it follows that \(y=5\); from (2) it follows that \(y=-97\); from (3) it follows that \(y=-9\); from (4) it follows that \(y=-99\).

Therefore, the pairs \((x ; y)\) of integers that can satisfy the equation are at most \((1 ; 5),(7 ;-97),(-1 ;-9),(-7 ;-99)\).

Verification confirms that exactly these four pairs have the required properties.