Factorize $n^{5}-5 n^{3}+4 n$. What can we conclude in terms of divisibility?
The idea is to try to make squares and differences of squares appear in order to factorize the proposed expression.

$$
\begin{aligned}
n^{5}-5 n^{3}+4 n & =n\left(n^{4}-5 n^{2}+4\right) \\
& =n\left(\left(n^{4}-4 n^{2}+4\right)-n^{2}\right) \\
& =n\left(\left(n^{2}-2\right)^{2}-n^{2}\right) \\
& =n\left(n^{2}-n-2\right)\left(n^{2}+n-2\right) \\
& =(n-2)(n-1) n(n+1)(n+2)
\end{aligned}
$$

Thus, for any integer $n, n^{5}-5 n^{3}+4 n$ is the product of 5 consecutive integers. Among these five integers, at least one of them will be divisible by 3, at least one by 5, at least one by 2, and another by 4. In this way, $n^{5}-5 n^{3}+4 n$ will always be divisible by $3 \times 5 \times 2 \times 4=120$.

Remark: More generally, it can be shown (for example using binomial coefficients or Legendre's formula) that the product of $k$ consecutive integers is divisible by $k!$, that is, by $1 \times 2 \times 3 \times \cdots \times k$.