Coxeter’s question on most-perfect magic squares and magic-faced hypercubes

For each integer n≥0n\ge 0, determine whether there exists a bijection A:{0,1}n→{1,…,2n}A:\{0,1\}^n\to\{1,\ldots,2^n\} such that, for every pair of distinct coordinates i,j∈{1,…,n}i,j\in\{1,\ldots,n\} and every choice of the remaining coordinates, the face sum ∑(εi,εj)∈{0,1}2A(x1,…,xi−1,εi,xi+1,…,xj−1,εj,xj+1,…,xn)\sum_{(\varepsilon_i,\varepsilon_j)\in\{0,1\}^2}A(x_1,\ldots,x_{i-1},\varepsilon_i,x_{i+1},\ldots,x_{j-1},\varepsilon_j,x_{j+1},\ldots,x_n) is independent of the chosen 22-dimensional face. Classify all such arrangements up to the natural transformations described in the source. The cited paper claims existence for every nn and uniqueness up to transformations of size 2n(n+1)!2^n(n+1)! when nn is even and 2nn n!2^n n\,n! when nn is odd.

References

Progress summary

Refreshed
Claimed solved

A new paper claims to classify these magic structures in every dimension, including the previously known four-by-four case, but the claim has not been independently verified.

Coxeter’s question asks for the existence and classification of most-perfect magic squares and their higher-dimensional magic-faced analogues. Manjul Bhargava’s paper claims a complete classification, with uniqueness understood modulo the natural transformation groups it specifies.

Known results

  • Stewart, 1999: most-perfect squares are pandiagonal, every 2×22\times2 subsquare has the magic sum, and entries at the relevant diagonal separation sum to half the magic constant; the reported enumeration includes 4848 inequivalent 4×44\times4 squares and 368640368640 inequivalent 8×88\times8 squares.

September 22, 2026 classification

On September 22, 2026, Manjul Bhargava’s arXiv paper reported a classification of order-two magic-faced hypercubes and their higher-dimensional Weyl-group orbit structure, recovering the 4×44\times4 case. This is a claimed resolution, but independent verification is not recorded.

Current status (as of September 2026): Bhargava’s paper claims to settle the 4×44\times4 question and classify all higher-dimensional cases, but the resolution remains unverified.

Sources

Solutions 0

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