\section*{Problem 1 - 321031}

Determine the number of all pairs \((x ; y)\) of integers \(x, y\) for which the following holds:

\[
19<x^{2}+y^{2}<93
\]
}

Due to \(93>x^{2}+y^{2} \geq x^{2}\), it follows that \(9 \geq x^{2}\). We now give for each \(0 \leq x \leq 9\) the values for \(y\) that satisfy this chain of inequalities:
- \(x=0\), thus \(19<y^{2}<93\), i.e., \(5 \leq|y| \leq 9\) (10 solutions)
- \(x=1\), thus \(18<y^{2}<92\), i.e., \(5 \leq|y| \leq 9\) (10 solutions)
- \(x=2\), thus \(15<y^{2}<89\), i.e., \(4 \leq|y| \leq 9\) (12 solutions)
- \(x=3\), thus \(10<y^{2}<84\), i.e., \(4 \leq|y| \leq 9\) (12 solutions)
- \(x=4\), thus \(3<y^{2}<77\), i.e., \(3 \leq|y| \leq 8\) (12 solutions)
- \(x=5\), thus \(y^{2}<68\), i.e., \(|y| \leq 8\) (17 solutions)
- \(x=6\), thus \(y^{2}<57\), i.e., \(|y| \leq 7\) (15 solutions)
- \(x=7\), thus \(y^{2}<44\), i.e., \(|y| \leq 6\) (13 solutions)
- \(x=8\), thus \(y^{2}<29\), i.e., \(|y| \leq 5\) (11 solutions) and
- \(x=9\), thus \(y^{2}<12\), i.e., \(|y| \leq 3\) (7 solutions).

Thus, there are a total of \(10+10+12+12+12+17+15+13+11+7=119\) different solution pairs that satisfy this chain of inequalities.