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Max Alekseyev
Max Alekseyev
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I assume that each magic square must be composed of numbers $1,2,\dots,9$ (or $1,2,\dots n^2$$1,2,\dots, n^2$ in general), and that under filling a cube with magic squares we understand assigning a number to each $1\times 1\times 1$ cell at the cube surface such that the numbers at each face form a magic square.

CoveringFilling $3\times 3\times 3$ cube with magic squares is not possible. Notice that the center number of a magic square is $5$, and once a number at a corner of a magic square is fixed to be $a\ne 5$, the number at the opposite corner (ie. on the same diagonal as $a$) is $10-a\ne a$. However, in the cube with corners having coordinates $\pm 1$ (say), there is a triangle with vertices $(-1,-1,-1)$, $(-1,1,1)$, $(1,-1,1)$, whose sides are formed by surface (magic) diagonals, which is impossible.

I assume each magic square must be composed of numbers $1,2,\dots,9$ (or $1,2,\dots n^2$ in general).

Covering $3\times 3\times 3$ cube with magic squares is not possible. Notice that the center number of a magic square is $5$, and once a number at a corner of a magic square is fixed to be $a\ne 5$, the number at the opposite corner (ie. on the same diagonal as $a$) is $10-a\ne a$. However, in the cube with corners having coordinates $\pm 1$ (say), there is a triangle with vertices $(-1,-1,-1)$, $(-1,1,1)$, $(1,-1,1)$, whose sides are formed by surface (magic) diagonals, which is impossible.

I assume that each magic square must be composed of numbers $1,2,\dots,9$ (or $1,2,\dots, n^2$ in general), and that under filling a cube with magic squares we understand assigning a number to each $1\times 1\times 1$ cell at the cube surface such that the numbers at each face form a magic square.

Filling $3\times 3\times 3$ cube with magic squares is not possible. Notice that the center number of a magic square is $5$, and once a number at a corner of a magic square is fixed to be $a\ne 5$, the number at the opposite corner (ie. on the same diagonal as $a$) is $10-a\ne a$. However, in the cube with corners having coordinates $\pm 1$ (say), there is a triangle with vertices $(-1,-1,-1)$, $(-1,1,1)$, $(1,-1,1)$, whose sides are formed by surface (magic) diagonals, which is impossible.

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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

I assume each magic square must be composed of numbers $1,2,\dots,9$ (or $1,2,\dots n^2$ in general).

Covering $3\times 3\times 3$ cube with magic squares is not possible. Notice that the center number of a magic square is $5$, and once a number at a corner of a magic square is fixed to be $a\ne 5$, the number at the opposite corner (ie. on the same diagonal as $a$) is $10-a\ne a$. However, in the cube with corners having coordinates $\pm 1$ (say), there is a triangle with vertices $(-1,-1,-1)$, $(-1,1,1)$, $(1,-1,1)$, whose sides are formed by surface (magic) diagonals, which is impossible.