TITLE: Integer decomposition of dilated integral polytopes
QUESTION [8 upvotes]: For $n > 0$, let $P$ be an integral polytope, that is, the convex hull in $\mathbb{R}^n$ of points in $\mathbb{Z}^n$. Suppose that $\dim(P) = n$.
Question: Given $d > n + 2$ is it true that
$$ dP \cap \mathbb{Z}^n \stackrel{?}{=} (n+2)P \cap \mathbb{Z}^n + (d-(n+2))P \cap \mathbb{Z}^n \ ? $$
That is for any integer point $u$ in the dilation $dP$ is there an integer point $v$ in the dilation $(n+2)P$ such that $u-v$ is an integer point of the dilation $(d-(n+2))P$? If so, is the construction explicit? If not are there conditions on $P$ where it does hold?

REPLY [3 votes]: Somewhat belatedly (four years later), I noticed that the Shapley-Folkman lemma (https://en.wikipedia.org/wiki/Shapley%E2%80%93Folkman_lemma) yields a strong version of the result. SF says (as a special case) that if $S$ is a subset of $n$-dimensional Euclidean space, and we set $P$ to be its convex hull, then when $d > n$, for all $x$ in $dP$, there exists $y $ in $nP$ such that $x - y$ is a sum of $n-d$ elements of $S$. 
So let $S = P \cap {\bf Z^n}$. For $x$ in $dP \cap {\bf Z^n}$, SF implies there is a decomposition $x = y + w$ where $w$ is a sum of $d-n$ elements of $S$ and $y \in dP$. Then $y = x-w$ is in ${\bf Z^n}$ (since $x$ and $w$ are). Hence $y \in nP \cap {\bf Z^n}$, yielding a stronger result: $dP \cap {\bf Z^n} = nP \cap {\bf Z^n} + ( S + \dots + S)$  with $d-n$  copies of $S$.