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I apologize in advance if this question is not of sufficient level. Define a perfect magic hypercube of side length $k$ and dimension $n$ to be one in which the cells are filled with consecutive integers and the sum of numbers over cells in any geometric line is equal to the appropriate constant depending on $n$ and $k$.

From the density Hales-Jewett theorem it follows that for fixed $k$, there cannot exist perfect magic hypercubes of fixed side length $k$ and arbitrarily large $n$. My question is: what are simpler ways to prove the nonexistence of perfect magic hypercubes of fixed $k$ and arbitrarily large $n$? Thanks very much.

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Here is an argument for $k=3$. For simplicity let the average value be $0$. Take any plane. Let the central element be $a$. The sum of the lines through the center minus the sum of the rows equals both $3a$ and $0$, so the central element of the plane must be $0$. Since there is only one $0$, there can't be more than one plane, so $n \lt 3$.

This doesn't generalize directly to larger odd $k$, since at least if you drop the condition that the elements of a magic square are consecutive integers, there are magic squares whose central element is not $0$.

$$\begin{array}{ccccc}2 & -3 & 2 & -3 & 2 \\\ -3 & 2 & 2 & 2 & -3 \\\ 2 &2 & -8 &2 &2 \\\ -3 & 2 & 2 & 2 & -3 \\\ 2 & -3 & 2 & -3 & 2\end{array} $$

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Thank you very much for your nice answer. –  user22202 Aug 8 '12 at 5:18

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