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

*

*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.
 A: Hint: since $f_n$ is a polynomial in $n$  of degree $\:\!4$ with integer coef's, iteratively taking differences $\,f_{n+1}-f_n\,$ shows  it satisfies a monic recurrence of order $5$ with integer coef's, i.e.
$$f_{n+5} =  a_4 f_{n+4} + \cdots + a_1 f_{n+1} + a_0 f_n,\,\ {\rm for\ some}\ a_i\in \Bbb Z\quad$$
By induction all $\,f_{k}\,$ have form $\,c_4 f_{4} + \cdots + c_1 f_{1} + c_0 f_0\,$ for some $\,c_i\in\Bbb Z,\,$ so by $\rm\color{#c00}{Euclid}$
$$\begin{align} &\gcd(\color{#90f}{f_0},\color{#0a0}{f_1},f_2,f_3,\color{#c00}{f_4},\ \ldots,\ c_4 \color{#c00}{f_4} + \cdots + c_1 \color{#0a0}{f_1} + c_0 \color{#90f}{f_0},\ \ldots)\\[.3em]
=\ & \gcd(f_0,f_1,f_2,f_3,f_4),\,\ {\rm by}\ \ c_4\color{#c00}{f_4}\equiv 0\!\!\!\!\pmod{\!\!\color{#c00}{f_4}},\rm\ etc
\end{align}$$
is the largest integer dividing all $\,f_k$.
Note $ $  See the Remark in this answer for the same method applied to $\,f_n = a^n+b^n+c^n + d^n$ and see also  here for a simpler case of a second order recurrence, for $\,f_n = 5^3\, 25^n + 3^3\, 6^n$.
A: 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:

*

*$24|(n-1)n(n+1)(n+2)$
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.
A: Alternatively.
Let $p$ be an odd prime.  If $p|n$ but $p^2 \not \mid n$ then we have $p\not \mid n-1, n-2, 5n + 2$ so we need never have any odd prime $p^2|n(n^2-1)(5n+2)$.
If $p=3$ we must have one of $n-1, n, n+1$ be divisible by $3$ so we must have $3|n(n^2-1)(5n+2)$ but we need not have $3^2|n(n^2-1)(5n+2)$.
If $p=5$ we will not have $5|5n+2$ and as we have $5$ options for $n \pmod 5$ we can avoid $n \equiv 0, 1, 4$ (by having $n\equiv 2,3\pmod 5$ and thus have $n, n+1, n-1\not \equiv 0\pmod 5$ and so $5\not \mid n(n^2-1)(5n+2)$.
If $p > 5$ we can have more than $4$ options for $n\pmod 5$ and so we can have $n \not \equiv 0, 1,-1, -2\cdot 5^{-1} \pmod p$. ANd that way we can avoid $p|n(n^2-1)(5n+2)$.
So no odd prime other than $3$ need divide $n(n^2-1)(5n+2)$ and $3^2$ need not divide it.
And as what power of $2$ must divide $n(n^2-1)(5n+2)$.
One of $n,n+1$ must be even so $2$ must divide.  If $n$ is odd then $n-1$ and $n+1$ is even and so $4$ must divide.  But one of $n-1$ or $n+1$ must be divisible by $4$ so $8$ must divide.
If $n$ is even then $5n+2$ is even so $4$ must divide.  If $n$ is not divisible by $4$ then $n = 4k + 2$ for some $k$ and $5n+2 = 20k + 12$ is divisible by $4$ and so $8$ must divide.  And if $n$ is divisible by $4$ then as $5n + 2$ is even $8$ must divide.
That exhausts all cases so $8$ must divide $n(n^2-1)(5n+2)$.
Now no higher power of $2$ must divide as we can chose $n$ odd so that $4$ but not $8$ divides one of $n\pm 1$.  For example if $n = 5$ and so $n+1 = 6$ and $n-1 =4$ and only $8$ need divide.  Or we can choose an even $k$ where $n = 4k + 2$ and $5n+2 = 20k + 12= 4(5k + 3)$ is divisible by $4$ but not by $8$.
So we must have $3$ and $8$ divide $n(n^2-1)(5n+2)$ but no other prime or higher power of $3$ or higher power of $2$ need divide.
SO $24$ is the largest number that must divide.
