Rubiks Cube 6 Magic Squares

Okay, a question I haven't really thought about too much, but seems interesting.

Assume we have an n-by-n-by-n SOLVED Rubiks Cube. On the small squares we write all the integers 1 to (6 *n^2), one distinct integer per small square, such that each face of the cube is a valid magic square (sums of rows, columns, and diagonals all add up to same value). The magic squares do not necessarily each have the same sum.

Then we scramble up the cube.

At some point in attempting to solve the cube, we come to a point were each face has a valid magic square on it. (The sums of each magic square are not necessarily the same as before.)

Question: Is the cube necessarily solved in regards to its colors?

(I am not even absolutely sure that there are 6 magic squares that can be constructed from the integers of 1 to (6 *n^2), for all n = positive integer.)

Thanks, Leroy Quet

Some comments below the spoiler protection, but not a full solution.

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The smallest magic square is of order 3, so there's no solution for n = 2. For higher n, number the sides from 0 to n-1 and the magic square on side s then uses consecutive numbers from s + 1 to s + n^2. <*>

For a magic square of order 3, it is easily proved[1] that the sum is always 3 times the center number; and the center squares, of course, do not move. When the conditions described in the puzzle are met, side 0, which has 5 in the center, must have a sum of 15 and thus can only contain the numbers 1 to 9, which are the ones that were there originally. But then this side is definitely all one color.

But now consider only the remaining numbers, starting at 10. Side 1 has a 14 in the center, so its sum must be 42; to form a magic square, the only possible numbers are then 9 lowest, 42. And so on for each side in turn; thus the cube is indeed solved.

However, this is not a proof even for the order-3 cube, because I have quietly assumed that the magic squares MUST be laid out as indicated at <*> above. There might be another, less obvious arrangement of numbers that would also produce 6 magic squares.

My conjecture is that there isn't one, and that this and the center- square property both hold for all odd n, in which case the answer is yes for all odd n. But I could not begin to prove it. And for n even, where there are no fixed centers, the problem is that much harder.

[1] With the obvious lettering of the cells, the sums A+B+C, D+E+F, G+H+I, A+D+G, B+D+H, C+F+I, A+E+I, C+E+G all equal S. So 2S = (A+B+C) + (G+H+I) = (A+G) + (B+H) + (C+I) = (S-E) + (S-E) + (S-E) = 3S - 3E; thus S = 3E.

[ Re: magic squares on faces of a 3x3x3 cube, using 0..53 ]

There is at least one other way to do so (and I suspect many more): number the faces 0..5 and on face k use the values congruent to k mod 6. The values on face 0 are just 6 times the usual magic square, and the values on face k are each k more than a value on face 0.

Cheers,