# 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?

• what you have done is more than you think it was. to prove something is divisible by $j*k*l$ it is good to prove that is divisible by $j$ and by $k$ and by $l$ and that those are cprime. So to prove something is divisible by $120$ show it is divisible by $3$ and by $5$ and by $8$. In this case you have proven it is divisble by 5 consecutive integers always. One of those is divisible by 5 and at least one by 3 and at least 2 by 2. If one is divisible by $2^1$ then the other must be divisible by $2^2$. Ergo, you are down. Commented Dec 21, 2017 at 17:27

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.

• This doesn't really answer the question IMHO, since you either need to prove the binomial coefficient formula, or if you take it as true by definition, you need to prove that binomial coefficients are integers. Commented Dec 21, 2017 at 22:28

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 $5$?