Find all integers $n$ such that $\frac{70 n+200}{n^{2}+1}$ is an integer.

Initial text: 150
Solution: Since $\left(n^{2}+1\right) \mid 10(7 n+20)$, then,
$$
\begin{array}{l}
\left(n^{2}+1\right)\left\{\left[10 \cdot 7^{2}\left(n^{2}+1\right)-10(7 n+20)(7 n-20)\right] .\right. \\
\text { Also, } 10 \cdot 7^{2}\left(n^{2}+1\right)-10(7 n+20)(7 n-20) \\
=4490=2 \times 5 \times 449,
\end{array}
$$

Since 449 is a prime number, then
$$
\left(n^{2}+1\right) \in\{1,2,5,449,2 \times 5,2 \times 449,5 \times 449,4490\} \text {, }
$$

which means $n \in\{0, \pm 1, \pm 2, \pm 3, \pm 67\}$
(This is only a necessary condition).
After checking each one, we find that only $n=0, \pm 1, \pm 2, \pm 3$, -67 satisfy the divisibility requirement, so the solutions to this problem are
$$
n=0, \pm 1, \pm 2, \pm 3,-67 \text {. }
$$