Find the largest number that $ n(n^2-1)(5n+2) $ is always divisible by? My Solution:
$$ n(n^2-1)(5n+2) = (n-1)n(n+1)(5n+2) $$

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*This number is divisible by 6 (as at least one of 2 consecutive integers is divisible by 2 and one of 3 consecutive integers is divisible by 3.


*$ 5n+2 \equiv 5n \equiv n \mod 2 $ then $n$ and $5n+2$ have the same pairness and at least one of $n+1$ and $5n+2$ is divisible by 2.


*$ n \equiv 5n \equiv 5n+4 \mod 4 \to $
if $ 2\ | \ n+1 \to n - 1 $ or $ n + 1 $ is divisible by 4
if $ 2\ | \ 5n+2 \to n $ or $ 5n + 2 $ is divisible by 4
The expression is divisible by 6 and has 2 even integers and one of them is divisible by 4 $\to$ is divible by 24.

Your proof seems correct to all of us, as it appears in the comments.

I would consider applying a method like this:
$$\begin{align}f(n)&=n(n^2-1)(5n+2)
\\&=n(n^2-1)(4n+n+2)\\
&=\underbrace{4n^2(n-1)(n+1)}_{\equiv ~0~(\text{mod}~~ 48)}
\\
&+\underbrace{(n-1)n(n+1)(n+2)}_{\equiv ~0~(\text{mod}~ 24)}\end{align}$$
If $n=3$, then $5n+2$ is prime and if the largest number to which the function is always divided was greater than $24$, the next factor must be $17$. But, $f(2)$ is not divisible by $17.$ Therefore, the largest number should only be $24$.

Explanations:

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*$24|(n-1)n(n+1)(n+2)$
Because, the product of $4$ consecutive positive integers are always divisible by $24$.
Applying $$n=8k±m, ~0≤m≤4, m\in\mathbb Z$$ shows that, $8|(n-1)n(n+1)(n+2)$ and we already know that, $6|(n-1)n(n+1)(n+2)$. This means $24|(n-1)n(n+1)(n+2)$.

*

*$48|4n^2(n-1)(n+1)$
Because, $48|4n^2(n-1)(n+1)=12|(n-1)n^2(n+1)$
Observing at the cases where $n$ is odd or even completes the proof.
