TITLE: Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$
QUESTION [12 upvotes]: I am attempting to prove the following problem: 

Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$

I attempted to solve it by induction, but when proving for $n= x+1$ the algebra gets very messy very fast. I was wondering if this is the only way or if there is a quicker way to prove this. I guess I am a little unsure as to how to prove something is an integer. 
I also noticed that letting $f(x) = \frac{x^5}5 + \frac{x^4}2 + \frac{x^3}3 - \frac x{30}$ and deriving $f(x)$ yields a fairly clean result, but I don't know if this helps me at all. Any help would be great.

REPLY [8 votes]: I'm reluctant to add to the noise here, but the following is a systematic (trick-free) way to deal with this type of question: the set of integer valued polynomials  has $\mathbb{Z}$-basis given by the binomial coefficients $$f_k(x)=\left( \begin{matrix} x \\ k \end{matrix} \right)=\frac{x(x-1)(x-2) \cdots (x-k+1)}{k!}.$$ 
Now given an arbitrary polynomial $f(x)$, there is a unique expression 
$$f(x)=\sum a_k f_k(x)$$ as a linear combination of these $f_k(x)$'s, with coefficients $a_k$ that one can get quickly and recursively using the upper-triangularity of the change of basis from $x^k$ to $f_k(x)$. Then $f$ is integer valued iff $a_k \in \mathbb{Z}$ for all $k$. 
Edit: I was going to leave this out but can't resist. Let $\Delta$ be the operator $(\Delta f)(x)=f(x)-f(x-1)$, so that $\Delta(f)$ is a discrete version of the derivative of $f$. Then the coefficients above are given by the following discrete analog of Maclaurin's theorem:
$$a_k=(\Delta^k f)(0).$$ As pointed out by lhf below, to prove that the $a_k$'s are integers it's enough to check that the values $f(m)$ are integers for $\mathrm{deg}(f)+1$ consecutive integers $m$.