Problem from Red Book

Question

The following problem is from The Red Book (list of practice problems for undergraduate mathematics competition). Let $k$ denote a positive integer. Determine the number $N(k)$ of triples $(x,y,z)$ of integers satisfying $$\begin{cases} |x|\leq k, & |y|\leq k, & |z|\leq k,\\ |x-y|\leq k, & |y-z|\leq k, & |z-x|\leq k. \end{cases}$$ A solution is as follows: Let $N(k,i)$ be the number of lattice points satisfying the given conditions, and the further condition $z=i$, $-k \leq i \leq k$. Then $N(k,0)=(2k+1)^2 -k(k+1)$ (the $(2k+1)^2$ lattice points on the $x,y$ plane without the points lying above the line $y=k$ or below the line $y=-k$). If $z=1$, then we need to subtract the bottom row of points (there are $k+1$), and the left column (an additional $k$ of them), for a total of $2k+1$ points. If $z=2$, we need to subtract another $2k+1$ points. In general, for $z=i$ we need to subtract $(2k+1)i$ points. In other words, $$N(k,i)=(2k+1)^2-k(k+1)-(2k+1)i, 1\leq i \leq k.$$ Since the configuration will be symmetric for negative values of $z$, $N(k,i)=N(k,-i)$. Hence $$N(k)=N(k,0)+2\sum_{i=1}^kN(k,i) = (2k+1)^2-k(k+1)+2\sum_{i=1}^k((2k+1)^2-k(k+1)-(2k+1)i)=4k^3+6k^2+4k+1.$$ This proof is not too complicated but not very straightforward either. Here is my question. The answer can be re-written as $(k+1)^4-k^4$. This makes me wonder if there is a really simple proof using some geometric interpretation in four dimensions.