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Find all natural numbers $n$ such that $n^{2}+11$ divides $n^{3}+13$.
Suppose $n \in \mathbb{N}$ is such that $n^{2}+11$ divides $n^{3}+13$. Then $n^{2}+11$ divides $n\left(n^{2}+11\right)-\left(n^{3}+13\right)=$ $11 n-13$, and conversely if $n^{2}+11$ divides $11 n-13$ then $n^{2}+11$ divides $n\left(n^{2}+11\right)-(11 n-13)=$ $n^{3}+13$: we have equivalence. Therefore, it is a matter of finding the $n$ such that $n^{2}+11$ divides $11 n-13$.

Now $n^{2}+11$ is of degree 2 in $n$ and $11 n-13$ is of degree 1: for large $n$, $011 n-13>0$ and we need to test the remaining values.

$11 n-13>0$ is satisfied for $n \geq 2$. Then, $n^{2}+11>11 n-13$ is equivalent to $n^{2}-11 n+24>0$, which is equivalent to $(n-3)(n-8)>0$, that is, $n>8$ or $n<3$.

It only remains to test the values of $n$ from 0 to 8.

| $n$ | $n^{2}+11$ | $11 n-13$ | $n^{2}+11$ divides $11 n-13 ?$ |
| :---: | :---: | :---: | :---: |
| 0 | 11 | -13 | no |
| 1 | 12 | -2 | no |
| 2 | 15 | 9 | no |
| 3 | 20 | 20 | YES |
| 4 | 27 | 31 | no |
| 5 | 36 | 42 | no |
| 6 | 47 | 53 | no |
| 7 | 60 | 64 | no |
| 8 | 75 | 75 | YES |

Finally, there are two solutions: $n=3$ and $n=8$.