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It's known that the numbers 1 to 9 can only form eight different magic squares, which are all rotations and reflections of each other.

Is there any set of 9 distinct numbers that can form two essentially different magic squares, which means that you can't get from one to the other using rotations and reflections? If not, is there a way to prove it?

(The numbers from 1 to 9 only have one essentially different magic square configuration.)

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Note that any two magic squares with the same numbers must have the same row/column sum. The middle number must be 1/3 of the sum (because the sum of the four lines passing through the center is the sum of all the numbers plus three times the center), so it is the same in both squares. Subtract this number from all the numbers, so that the middle cell and the row/column sum are both 0. Then diametrically opposite cells must be negatives of each other.

If the corner cells of one square contain the same numbers as the corner cells of the other, they must be rotations or reflections of each other. So if there are two essentially different squares, one number must be a corner in one square and an edge in another.

Suppose one of the squares looks like this:$$\begin{array}{ccc}a&b&c\\-d&0&d\\-c&-b&-a\\\end{array}$$

Either $a$ and $c$ both become edge cells in the second square, or one is an edge and the other is a corner. If one remains as a corner, then the only possibility (up to certain symmetries) that does not immediately require two numbers to be the same is this: $$\begin{array}{ccc}a&c&b\\d&0&-d\\-b&-c&-a\\\end{array}$$However, if we try to solve this, we find that $a$ must be 0, which makes $d=b=-c$, and the squares are not essentially different.

Otherwise, the configuration of the second square must be something like this:$$\begin{array}{ccc}-b&a&d\\c&0&-c\\-d&-a&b\\\end{array}$$ If we solve this, we find that all the numbers must be 0.

Therefore, there are no sets of 9 numbers that can produce two essentially different magic squares.

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  • 2
    $\begingroup$ Can you elaborate on that "the middle square must be exactly 1/3 of the row/column sum" part? $\endgroup$ – John Dvorak Nov 2 '15 at 9:02
  • $\begingroup$ if the square is as follow (by rows) a b c d e f g h i, then sum of all rows (a + b + c + d+ e + f+ g+ h+ i) = 3 x sum. We know that the sum of the lines crossing through center = 4x sum = a + e + i + b + e + h + c + e + g + d + e + f = (a + b + c+ d + e+ f + g + h + i)+(3e) . Second equation minus first: sum = 3e , or e = 1/3 of the row sum. $\endgroup$ – Tim Couwelier Nov 2 '15 at 16:58

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