Question in Number Theory about divisibility Show that $120$ divides $n^5 - 5n^3 + 4n$. 
My approach was that I factorized this equation into its primary factors and I got $(n-2)(n-1)(n)(n+1)(n+2)$. When I plugged in the value of $n>2$ the resultant value was always divisible by $12$. However is there a general way of proving the divisibility after factorization or is there another approach to this question? 
 A: Note that the product of five consecutive integers is always divisible by $5\times 4 \times 3 \times 2\times 1$.
This can be proved in many different ways. The fact that the binomial coefficient $\binom n r$ is an integer shows that the product of $r$ successive integers is divisible by $r!$
Alternatively $(n+1)n(n-1) \dots (n-r+2)-n(n-1)(n-2) \dots (n-r+1) = r n(n-1) \dots (n-r+2)$ which is the product of $r$ and $r-1$ consecutive integers and you can use induction to show that the difference is divisible by $r!$
A: Note that
$$
\frac{(n-2)(n-1)(n)(n+1)(n+2)}{120}=\binom{n+2}{5}
$$
and since binomial coefficients are integers, the result follows.
A: A more systematic method of finding a common factor of a polynomial for all integer arguments is to write the polynomial as a combinatorial polynomial. For example,
$$
\begin{align}
n^5-5n^3+4n
&=120\binom{n}{5}+240\binom{n}{4}+120\binom{n}{3}\\
&=120\left(\binom{n}{5}+2\binom{n}{4}+\binom{n}{3}\right)
\end{align}
$$
Since the GCD of the coefficients is $120$, we know that
$$
n\in\mathbb{Z}\implies120\mid n^5-5n^3+4n
$$

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A: if $$n=2m$$ then we get
$$4(m-1)m(2m-1)(2m+1)$$ thus $$8|(n-2)(n-1)n(n+1)(n+2)$$
if $$n=2m+1$$ then $$(n-2)(n-1)n(n+1)=(2m-1)(2m)(2m+1)(2m+2)(2m+3)$$ and $$8|(n-2)(n-1)n(n+1)$$
if $$n=3m$$ then all is clear,
if $$n=3m+1$$ then $$3|n-1$$
if $$n=3m+2$$ then $$3|n+1$$
can you prove the divisibility of $5$?
