\section*{Problem 3 - 171223}

Determine all integers \(x\) for which

\[
f(x)=\frac{3 x^{2}+x-2}{x^{2}-2}
\]

is an integer.
}

By polynomial division, we obtain

\[
f(x)=\frac{3 x^{2}+x-2}{x^{2}-2}=3+\frac{x+4}{x^{2}-2}
\]

Thus, \(f(x)\) can only be an integer if the fraction \(\frac{x+4}{x^{2}-2}\) is an integer, i.e., \(|x+4| \geq\left|x^{2}-2\right|\) or \(x+4=0\) holds. The equation \(|x+4|=\left|x^{2}-2\right|\) has its solutions at \(x_{1}=-2\) and \(x_{2}=3\).

One can reason that the relationship \(|x+4| \geq\left|x^{2}-2\right|\) only holds in the interval \([-2 ; 3]\), as also illustrated by the following graph:

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2032.jpg?height=460&width=651&top_left_y=1455&top_left_x=700)

An additional possibility for the fraction \(\frac{x+4}{x^{2}-2}\) to be an integer arises for \(x=-4\), since a numerator of 0 makes the entire fraction 0 (the denominator does not become 0).

Since \(x\) should be an integer, the only possibilities for \(x\) are \(x \in\{-4,-2,-1,0,1,2,3\}\). Substituting \(x\) into the original function yields

\[
\begin{array}{cc|cc|cc|cc}
x & f(x) & x & f(x) & x & f(x) & x & f(x) \\
-4 & 3 & -2 & 4 & -1 & 0 & 0 & 1 \\
1 & -2 & 2 & 6 & 3 & 4 & &
\end{array}
\]

Since in each case \(f(x)\) is an integer, the solution set is \(x \in\{-4,-2,-1,0,1,2,3\}\).