TITLE: Integrality of $\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$
QUESTION [6 upvotes]: I've been asked to provide a proof for

If $n$ is an integer then
$$\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$$
is also an integer.

Any help would be appreciated

REPLY [12 votes]: Hint $\ \ \displaystyle \frac{n^3}{6} + \frac{n^2}{2} + \frac{n}{3}\, =\, \frac{n^3+3n^2+2n}6\,=\, \frac{(n+2)(n+1)n}6\,  =\, {n+2\choose 3}\in\Bbb Z$
Remark $\ \ $ This is a special case of a classical result of Polya and Ostrowski $(1920).$ Namely, $\,f(x) \in \mathbb Q[x]$ is an integer-valued polynomial, i.e.  $f(\mathbb Z)\subset \mathbb Z\:,\:$ iff $f(x)$ is an integral linear combination of binomial coefficients ${x\choose k},\:$ for $\: k\le \deg f$. For a proof see e.g. Polya And Szego, Problems and theorems in analysis, vol II, Problem 85 p. 129 and its solution on p. 320.  This theorem has been extended to more general rings (e.g. Dedekind domains) by Cahen and others.