Competition problem — Number theory

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Number theory Difficulty 5.0 AIME, harder Prove it Greece

If the number $K = \frac{9n^2 + 31}{n^2 + 7}$ is integer, find the possible values of $n \in \mathbb{Z}$ .

Solution

We have
$K = \frac{9n^2 + 31}{n^2 + 7} = \frac{9(n^2 + 7) - 32}{n^2 + 7} = 9 - \frac{32}{n^2 + 7}.$
Since $K$ is integer, it follows that $n^2 + 7$ is a divisor of $32$ and taking in mind that $n^2 + 7 \ge 8$ , we conclude:
$n^2 + 7 \in \{8, 16, 32\} \Leftrightarrow n^2 \in \{1, 9, 25\} \Leftrightarrow n \in \{-1, 1, -3, 3, -5, 5\}.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.