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?
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 $r!$
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 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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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 $5$?
