Why do magic squares of order 3 can only be filled up in one unique way?
Hi, Please look at the website that is partially re-produced below: In general, when dealing with NxN squares, where N is odd, it’s helpful to subtract (N^2 + 1)/2 from each of the numbers 1 through N^2, since this makes all the common sums zero. For example, with a typical 3×3 square we subtract 5 from each number to give a b c 1 -4 3 d e f = 2 0 -2 g h i -3 4 -1 which makes it immediately obvious that there is only one possible magic square of order 3, up to rotations and reflections. This can also be seen by combining the algebraic conditions on the variables to eliminate all but two of the variables and arrive at the conic 2a^2 + 2ab + b^2 = 10 (1) (See Note 1 for derivation.) This is an ellipse with the eight lattice points (-3, 2) (-3, 4) (-1, 4) ( 1, 2) ( 3,-2) ( 3,-4) ( 1,-4) (-1,-2) each of which represents one of the eight rotations or reflections of a 3×3 magic square. Therefore, equation (1) and the eight lattice points can be regarded as the “solution” of 3×3 magic squares
