Find all $n \in \mathbb{Z}$ such that $n+2 \mid n^{2}+3$.

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Let $n$ be a solution. We proceed by Euclidean division, trying to reduce the degree in $n$ of the right-hand side term at each step. Thus, $n+2 \mid n(n+2)$, so $n+2 \mid n^{2} + 2 n - (n^{2} + 3)$, i.e., $n+2 \mid 2 n - 3$. Therefore, $n+2 \mid 2 n - 3 - 2(n+2)$, i.e., $n+2 \mid -7$. We can establish the list of divisors of $7: \{-7, -1, 1, 7\}$. Thus, $n \in \{-9, -3, -1, 5\}$, and conversely, we verify that these are indeed solutions.