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).