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In Intuitive/direct proof that a rectangle partitioned into rectangles each with at least one integer side must itself have an integer side I asked a question about intuitively proving that a rectangle with axis parallel sides partitioned into subrectangles must have an integer side length if all the subrectangles have an integer side length. But now I'm wondering if there is a generalization for hyperboxes in ${\mathbb R}^d$. I know that if a hyperbox $R$ is partitioned into hyperboxes $R_1,\ldots,R_n$ such that each $R_i$ has at least one integer side length, then $R$ has at least one integer side length. There at least two possible proofs, which go the same way as the proofs for rectangles given in the other post. However, it's also true that if each $R_i$ has $d$ integer sides (i.e. all sides are integers), then $R$ must also have $d$ integer sides. This is trivial to prove. So now I'm wondering, what if each $R_i$ has $k$ integer sides, where $1 < k < d$? Can we prove that $R$ has $k$ integer sides? The proof in terms of complex integrals for $k=1$ seems suggestive. The proof notes that $\int_R \prod_j e^{2 \pi i x_j} dx_j = 0$ if and only if $R$ has an integer side. So if all $R_i$ have an integer side, then the integral is $0$ over each $R_i$ hence also $0$ over $R$. Intuitively, for $k > 1$ the integral is not just $0$ but also $0$ "of order $k$", so if all $R_i$ have $k$ integer sides then it seems like $R$ should too, because the sum of "zeros of order $k$" should give zero of order at least $k$, and "zero of order $k$" for the integral should correspond to having $k$ integer sides. Can someone make this rigorous e.g. maybe with infinitessimals, or give another proof, or a counter-example where the conjecture for $d > k > 1$ is false?

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I believe I have proved if that $R$ has at least $k$ integer sides then

$$\lim_{\delta \to 0^+} \frac{\int_{(1+\delta)R} \prod_j e^{2 \pi i x_j} dx_j}{\delta^k} = L$$

where $L$ is some finite constant, and if $R$ has $k-1$ or fewer integer sides then $L = \infty$. Thus, if $R$ is partitioned into hyperboxes $R_1,\ldots,R_n$ with $k$ integer sides each, then the limit is finite over each $R_i$, and thus the limit is also finite over $R$, so $R$ has at least $k$ integer sides.

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An idea for a proof by induction: (I'm not very good at visualizing $(d-1)$-dimensional cross-sections of $d$-dimensional hyperrectangles so there may be simple errors.)

Assume that for all $1 \leq l \leq d-1$, each $d-1$-dimensional hyperbox partitioned into hyperboxes with at least $l$ integer side lengths has at least $l$ integer side lengths. Consider a $d$-dimensional hyperbox $R$ that is partitioned into hyperboxes with at least $k$ integer side lengths. If $R$ has at most $k-1$ integer side lengths, we may take an intersection of $R$ and a $(d-1)$-dimensional hyperplane $T$ such that the resulting $(d-1)$-dimensional hyperbox $R'$ has at most $k-2$ integer side lengths. On the other hand, the partitioning $d$-hyperboxes $R_i$ that intersect $T$ become $d-1$-hyperboxes $R'_i$ that form a partition of $R'$ and have at least $k-1$ integer side lengths, which is a contradiction. Thus $R$ has at least $k$ integer side lengths.

The boundary cases (e.g. $k=1$ for each $d$) need to be solved separately.

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