Magic Squares of Order 3
At the beginning, we saw the 33
Being an odd ordered magic square, the lo-shu can be constructed using the method described in the material on odd-ordered magic squares and flipping the square through a horizontal axis. Another way to construct the lo-shu magic square is given in the following illustration.
Since the 33 magic square is small in the number of numbers that appear in it, we can easily establish some special properties.
If 1 is in the center square we then need to find four pairs of integers from the set {2,3,4,5,6,7,8,9} such that each pair sums to 14 (If 1 is in the center, there will be one column containing 1, one row containing 1 and both diagonals.). There are only two such pairs (5,9) and (6,8), so 1 cannot be in the center. If 2 is in the center, then we need four pairs that sum to 13; however, there are only three such pairs (4,9), (5,8) and (6,7). Thus, 2 cannot be in the center square. If 3 is in the center square, then we need four pairs that sum to 12. The pairs that sum to 12 are (4,8), and (5,7). Note: We cannot use the pair (3,9) because 3 is already being used and we can only use 6 once so there is no (6,6) pair available.. Hence, 3 cannot be in the center. Finally, if 4 is in the center, then we need four pairs that sum to 11. There are only three pairs that sum to 11. They are (2,9), (3,8) and (5,6). Therefore, 4 cannot be in the center and that leaves the only possible center number to be 5.
Suppose the upper-left corner contains 3. Then its opposite cell must contain 7, since 3 + 5 + 7 = 15. That means the upper row must be the numbers 3, 4 and 8. If 8 is in the upper-right corner, the summ of that column will be 7 + 8 + ? which will be more than 15. If 4 is in the corner, then we would have 4 + 7 + ? = 15 and the only number that will fit is 4 which is already used. This argument also shows that 7 cannot be in a corner. The above two properties determine that all magic squares of order come from the same square by rotations and reflections. Once we establish that 5 must be in the center and no odd integer can be a corner, we have that the even integers must be the corners and they must be paired so that 2 and 8 are diagonally opposite of each other with 5 in the middle and 4 and 6 must be diagonally opposite of each other with 5 in the middle. This immediately fixes 1, 3, 7, and 9. The 1 must be between 8 and 6, 3 must be betwee 8 and 4. Thus, 7 must be between 2 and 6 and 9 must be between 2 and 4.
Below is the
If we then reflect each of these through the center column, we get
Reflections through the center row give no new arrangements, as one can easily verify. Likewise, no reflection through a diagonal will yield a new square.
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