\section*{Problem 1 - 231031}

Determine all ordered pairs \((g ; r)\) consisting of an integer \(g\) and a real number \(r\) for which

\[
\frac{r}{r^{2}-6 r+10}=g
\]

holds!
}

1. If \((g ; r)\) is a pair with the required properties, then:

If \(g=0\), then \(r=0\). If \(g \neq 0\), then \(r\) satisfies the equation

\[
r^{2}-\left(6+\frac{1}{g}\right) r+10=0
\]

The discriminant of this equation is nonnegative, i.e.,

\[
\left(3+\frac{1}{2 g}\right)^{2}-10 \geq 0 \quad \Rightarrow \quad\left|3+\frac{1}{2 g}\right| \geq \sqrt{10}
\]

Since \(g\) is an integer and not equal to 0, \(|g| \geq 1\), thus \(\frac{1}{|g|} \leq 1\) and therefore \(\frac{1}{g} \geq-1,3+\frac{1}{2 g} \geq 3-\frac{1}{2}>0\). Hence, \(\left|3+\frac{1}{2 g}\right|=3+\frac{1}{2 g}\) and it follows that

\[
3+\frac{1}{2 g} \geq \sqrt{10} \quad \Rightarrow \quad \frac{1}{2 g} \geq \sqrt{10}-3
\]

Since \(\sqrt{10}-3>0\), on the one hand \(\frac{1}{2 g}>0, g>0\), and on the other hand \(2 g \leq \frac{1}{\sqrt{10}-3}=\sqrt{10}+3<8, g<4\).

Therefore, in the case \(g \neq 0\), the only possibilities are listed in the following table

\[
\begin{array}{llll}
g & r^{2}-\left(6+\frac{1}{g} r\right)+10=0 & \text { Solutions } & \\
1 & r^{2}-7 r+10=0 & r_{1,2}=\frac{7}{2} \pm \sqrt{\frac{9}{4}} & r_{1}=5, r_{2}=2 \\
2 & r^{2}-6.5 r+10=0 & r_{1,2}=\frac{13}{4} \pm \sqrt{\frac{9}{16}} & r_{1}=4, r_{2}=\frac{5}{2} \\
3 & r^{2}-\frac{19}{3} r+10=0 & r_{1,2}=\frac{19}{6} \pm \sqrt{\frac{1}{36}} & r_{1}=\frac{10}{3}, r_{2}=3
\end{array}
\]

Thus, the following ordered pairs can at most satisfy the conditions of the problem:

\[
(0 ; 0), \quad(1 ; 5), \quad(1 ; 2), \quad(2 ; 4), \quad\left(2 ; \frac{5}{2}\right), \quad\left(3 ; \frac{10}{3}\right), \quad(3 ; 3)
\]

2. These pairs \((g ; r)\) satisfy the conditions of the problem; because in each of them, \(g\) is an integer and it holds that:

\begin{tabular}{llll}
\(g\) & \(r\) & \(r^{2}-6 r+10\) & \(\frac{r}{r^{2}-6 r+10}\) \\
0 & 0 & 10 & 0 \\
1 & 5 & \(25-30+10=5\) & 1 \\
1 & 2 & \(4-12+10=2\) & 1 \\
2 & 4 & \(16-24+10=2\) & 2 \\
2 & \(\frac{5}{2}\) & \(\frac{25}{4}-\frac{60}{4}+\frac{40}{4}=\frac{5}{4}\) & 2 \\
3 & \(\frac{10}{3}\) & \(\frac{100}{9}-\frac{180}{9}+\frac{90}{9}=\frac{10}{9}\) & 3 \\
3 & 3 & \(9-18+10=1\) & 3
\end{tabular}

Adapted from \([5]\)