\section*{Exercise 3 - 211023}

Determine all pairs \((x ; y)\) of integers \(x, y\) for which \(x^{2}-y^{2}=1981\) holds!
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A pair \((x ; y)\) of integers has the required property if and only if it satisfies \((x-y)(x+y)=|1981|\).

Due to the prime factorization \(1981=7 \cdot 283\), there are exactly the possibilities listed in the following table for (1) in integers \(x-y\) and \(x+y\). By adding and subtracting and then halving, one obtains that only the values of \(x\) and \(y\) listed subsequently can result in the specified numbers \(x-y\) and \(x+y\); a check shows in each case that they indeed yield these results.

\begin{tabular}{rrrr}
\(x-y\) & \(x+y\) & \(x\) & \(y\) \\
1 & 1981 & 991 & 990 \\
-1 & -1981 & -991 & -990 \\
1981 & 1 & 991 & -990 \\
-1981 & -1 & -991 & 990 \\
7 & 283 & 145 & 138 \\
-7 & -283 & -145 & -138 \\
283 & 7 & 145 & -138 \\
-283 & -7 & -145 & 138
\end{tabular}

Therefore, exactly the pairs (991; 990), (-991; -990), (991; -990), (-991;990), (145; 138), (-145; -138), \((-145 ; 138),(145 ;-138)\) have the required property.

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