\section*{Problem 1 - 271011}

How many ordered pairs of integers \((x, y)\) are there in total for which \(x \cdot y=1987\) holds?
}

The number 1987 is not divisible by any of the numbers

\[
2,3,5,6,7,11,17,19,23,29,31,37,41,43
\]

(This can be verified, for example, by calculating the 14 divisions \(1987: 2, 1987: 3, \ldots\) using a calculator). Since \(\sqrt{1987} < 44\), this means that 1987 is not divisible by any prime number \(p\) for which \(p \leq \sqrt{1987}\).

Thus, it is proven: 1987 is a prime number. It follows that there are exactly the following four ordered pairs of integers \((x ; y)\) such that \(x \cdot y = 1987\):

\[
(1,1987) \quad, \quad(-1,-1987) \quad, \quad(1987,1) \quad, \quad(-1987,-1)
\]