Largest number that divides $n^2(n^2 - 1)(n^2 - n - 2)$ for all $n$ Obtain the greatest natural that divides $n^2(n^2 - 1)(n^2 - n - 2)$ for all natural numbers $n$.
What should be the approach in these type of questions? 
Should I equate with prime factorization $2^a 3^b 5^c \cdots$ etc?

\begin{align}
f(n) & = n^2(n^2-1)(n^2-n-2)\\
& = n^2(n+1)(n-1)(n+1)(n-2)\\
& = n^2(n+1)^2(n-1)(n-2)
\end{align}
$n=3$ gives us $9 \times 16 \times 2 = 288$. So the natural number dividing $n^2(n+1)^2(n-1)(n-2)$ for all $n$ should be a divisor of $288$.
Also, note that $24$ has to divide $n^2(n+1)^2(n-1)(n-2)$ since $24$ divides $(n-2)(n-1)n(n+1)$ since it is a product of $4$ consecutive integers.
Further, $2$ divides $n(n+1)$. Hence, we know that at-least $48$ divides $f(n)$. Hence, the desired number should be a multiple of $48$ and must divide $288$.
Can you finish it from here by looking at $f(n)$ for couple of other values?
Move the mouse over the gray area for the answer.

 Since the largest number must be a multiple of $48$ and must divide $288$. The possible options are $$\{48, 96, 144, 288\}$$
 $$\text{$f(4) = 2400$. And $144,288$ doesn't divide $2400$. Hence, the largest number cannot be $144$ or $288$.}$$
 $$\text{$f(5) = 10,800$. And $96$ doesn't divide $10,800$. Hence, the largest number cannot be $96$. Hence, the largest number is $48$.}$$

