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

We have $360 = 8 \times 9 \times 5$, so it will suffice to check that $8,9,5|a^2(a^2-1)(a^2-4)$. 
For $9$, note that $a^2$ is congruent to either $1$ or $0$ modulo $3$, so both $a(a^2-1)$ and $a(a^2-4)$ are divisible by $3$, and consequently $9|a^2(a^2-1)(a^2-4)$.
Likewise, $a^2$ is congruent to either $1$ or $0$ or $4$ modulo $5$, so again the divisibility $5|a^2(a^2-1)(a^2-4)$ holds.
Finally, we look at divisibility by $8$. If $a$ is even, then $4|a^2$ and $4|a^2 -4$, so you have the divisibility $8|a^2(a^2-1)(a^2-4)$ (even with $16$ in place of $8$). If $a$ is odd, then $a-1,a+1$ are even and one of them is divisible by $4$, so $a^2-1$ is divisible by $8$ and the sought divisibility holds.

As a general rule, if you have a problem like (i.e. "Show that $360|f(a)$ for all $a$") this and don't want to think too much, you can first writhe $360$ as a product of powers of primes (nothing special about $360$ here, of course), and then check that if $P$ is a highest power of a prime dividing $360$ (i.e. $P=8,9,5$ here) you always have $P|f(a)$. This second check can be done by brute force: just check that $P|f(0),f(1),\dots,f(P-1)$ - this requires just the amount of computation that can be done by hand.
