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.

Let's call the corners A, B, C and D (A is opposite C) and let the magic constant be S. The cell between A and B must be S-A-B; the cell between B and C, S-B-C; the one between C and D, S-C-D, and the one between D and A, S-D-A.

Considering the A-to-C diagonal, the center cell must be S-A-C. Considering the B-to-D diagonal, the center cell must be S-B-D. Since both values must be the same, it follows that A+C=B+D; let's call that sum K, and the center cell is then S-K.

Consider now the horizontal line through the center; the center cell must be A+B+C+D-S=2K-S. Since 2K-S must equal S-K, it follows that K=2S/3, and in that case the center cell turns out to be S/3.

Posted by e.g. on 2004-04-16 12:28:11