Does anyone have an answer to the three-dimensional analogue of the 2009 Putnam Competition A1 problem, viz., if $f\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ satisfies $\sum\_{i=1}^8 f(a\_i) = 0$ whenever $a\_1, \ldots, a\_8$ are the vertices of a cube, must $f$ be identically zero? 


A few thoughts: Since the cube is not self-dual above dimension $2$, the solution (well, at least, my solution) to the $2$-dimensional problem doesn’t generalize.  To try to show that the answer to the $3$-dimensional problem is “no,” one might try letting $\Omega$ be the set of ordered pairs $(X, f\_X)$ where $X \subseteq \mathbb{R}^3$ is a subset and $f\_X\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ is a function that sums to zero over any eight points of $X$ that form the vertices of a cube.  Ordering $\Omega$ in the usual way, we can invoke Zorn’s Lemma to obtain a maximal $(X, f\_X) \in \Omega$.  If $X \neq \mathbb{R}^3$, then every $x \in \mathbb{R}^3 \setminus X$ must, in two “incompatible” ways, be the eighth vertex of a cube whose remaining vertices lie in $X$.  But I don't see how a contradiction arises from this (and of course one could make the same argument in two dimensions, where a contradiction does not arise).



function that sums to zero over cube vertices
First prove that the sum of the values of your function at the vertices of any regular tetrahedron is zero. You can do this by considering a $2\times 2\times 2$ cube made out of $8$ smaller cubes that you can color in a checkerboard fashion. Sum all the vertices of the black cubes and subtract the sum of the vertices of all the white cubes.


Second you can show that the sum of the values at the vertices of a $x\times x\sqrt{2}$ rectangle is always zero. You need to consider two tetrahedra sharing an edge for this.


From here it is very similar to the Putnam problem you mention. (In fact some solutions apply in the same exact way.)


P.S. This shows that it is enough to have the information only for cubes whose sides are parallel to the axis.


