<s>[INST] Prove that 360 divides (a-2)(a-1)a.a.(a+1)(a+2) there's a question which asks to prove that
360 | a2(a2-1)(a2-4)
I attempted it in the following manner.
a2(a2-1)(a2-4) = (a-2)(a-1)(a)(a+1)(a+2)(a)
The first 5 terms represent the product of 5 consecutive terms. Hence,
One of them will have a factor of 5
One of them will have a factor of 4
One of them will have a factor of 3
One of them will have a factor of 2
=> 5 x 4 x 3 x 2 = 120 will divide the given expression for sure.
Now, how do i bring a factor of 3. I know that i have missed one power of a. But, i think i am missing something. Help will be appreciated
Thank you
 [/INST] $$
\begin{align}
(a-2)(a-1)a\cdot a(a+1)(a+2)
&=5!\,a\binom{a+2}{5}\\
&=120[(a+3)-3]\binom{a+2}{5}\\
&=120(a+3)\binom{a+2}{5}-360\binom{a+2}{5}\\
&=120\cdot6\binom{a+3}{6}-360\binom{a+2}{5}\\
&=360\left[2\binom{a+3}{6}-\binom{a+2}{5}\right]
\end{align}
$$
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