\section*{Problem 2 - 231022}

Determine all ordered pairs \([g ; r]\) consisting of an integer \(g\) and a real number \(r\) that satisfy the equation:

\[
\frac{3}{3 r^{2}+1}=g
\]
I. If a pair of numbers \([g ; r]\) has the required properties, then:

Since the square of any real number \(\geq 0\), it follows that

\[
3 r^{2}+1 \geq 1
\]

From this, on the one hand, \(3 r^{2}+1>0\); and since \(3>0\),

\[
\frac{3}{3 r^{2}+1}>0
\]

On the other hand, it follows from (1) that

\[
\frac{3}{3 r^{2}+1} \leq \frac{3}{1}=3
\]

Thus, the integer \(g\) satisfies the inequality \(0<g \leq 3 \quad\) (2).

From \(\frac{3}{3 r^{2}+1}=g\) it follows that

\[
3 g r^{2}+g=3 \quad ; \quad r^{2}=\frac{3-g}{3 g}
\]

This equation, for the integer solutions of (2), is as given in the table:

\[
\begin{array}{ll}
g & r^{2}=\frac{3-g}{3 g} \\
1 & r^{2}=\frac{2}{3} \\
2 & r^{2}=\frac{1}{6} \\
3 & r^{2}=0
\end{array}
\]

Therefore, only the pairs

\[
\left[1 ; \sqrt{\frac{2}{3}}\right],\left[1 ; \sqrt{-\frac{2}{3}}\right],\left[2 ; \sqrt{\frac{1}{6}}\right],\left[2 ; \sqrt{-\frac{1}{6}}\right]
\]

can have the required properties.

These pairs have these properties because 1, 2, and 3 are integers and \(\pm \sqrt{\frac{2}{3}}, \pm \sqrt{\frac{1}{6}}, 0\) are real numbers. A check by substituting the numbers confirms the solution.