Prove that the central cell (the number in the middle cell) of any 3x3 magic square is always one-third the magic constant (the sum of any side, either 2 major diagonals, or either center row in the magic square).
Show that in any larger square (n x n), the central cell does not need to be 1/n the magic constant.
(The layout of variables was similar to another comment on the site -- I don't remember where)
Call the top-left square x+a and the bottom right square x-a, (thus, x is their arithmetic mean), label the central cell z, and the magic sum M, which must equal 2x+z. Call the other cells (clockwise) x+b, x+c, and x+d and then x-b, x-c, and x-d so the magic sum is reached in all cases:
x+a, x+b, x+c
x-d, z, x+d
x-c, x-b, x-a
Since the sum of the top and bottom row equals 6x=2M, x=M/3, and since 2x+z=M, z=x=M/3.
Posted by Gamer
on 2006-12-21 15:42:54