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.

For all larger magic squares with an odd numbered side length, I can find a general counter-example using only ones and zeroes.  If every number must be different, than add the counter-example to a magic square filled with even numbers to get your new counter-example.

To get the general counter example, start with a 5 x 5 square like this:

01000
00001
00100
10000
00010

The magic constant is 1, but the center is not 1/5.

To generate a 7 x 7 counter example, insert new rows between the 2nd and 3rd rows, and the 4th and 5th rows.  Insert two new columns in the same places.  Fill the rows with zeroes, except for the upper right and lower left intersections, which will have ones.

0100000
0000001
0000010
0001000
1000000
0010000
0000100

Noticed that the positions of the new rows and columns are:
1) between the middle row/column and the one before it
2) and between the last row/column and the one before it

Before regenerating a new square, turn this one around 180 degrees, then follow the same steps as before.  Continue to regenerate new squares this way, turning upside-down, filling the new columns and rows the same way as before.  This should give a counter-example for all odd-numbered squares above 5.

I haven't checked, so I could be wrong on this.  There may be a problem later on with having diagonals add up, and it might not work after all.

Added in:
After trying it a few times, I have become convinced that it works, which is wonderful, except I don't have prooof.

Edited on April 16, 2004, 6:01 pm

Posted by Tristan on 2004-04-16 17:47:12