In an article published in 1705, Philippe de la Hire describes some methods to create magic squares of order n=4k from two auxiliary squares.
In the first auxiliary square he fills the first half of the top row with any number z from the range from 1 to n, the second half with their complementary number n + 1 − z. In the row below, he swaps the two halves. This is how he proceeds with the remaining numbers in the other pairs of rows until the entire auxiliary square is filled.
The second auxiliary square, on the other hand, is constructed using columns with slightly changed numbers. Now the upper half is filled with an arbitrary number z from the range from 0 to n − 1 and the lower half with its complementary number, which is n − 1 − z in the changed number range, since the number range starts here at 0. In the next column, the two halves are swapped. According to this principle, all other column pairs are then filled with the remaining numbers as in the figure on the right.
8 | 8 | 8 | 8 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 8 | 8 | 8 | 8 |
5 | 5 | 5 | 5 | 4 | 4 | 4 | 4 |
4 | 4 | 4 | 4 | 5 | 5 | 5 | 5 |
7 | 7 | 7 | 7 | 2 | 2 | 2 | 2 |
2 | 2 | 2 | 2 | 7 | 7 | 7 | 7 |
6 | 6 | 6 | 6 | 3 | 3 | 3 | 3 |
3 | 3 | 3 | 3 | 6 | 6 | 6 | 6 |
7 | 0 | 6 | 1 | 5 | 2 | 4 | 3 |
7 | 0 | 6 | 1 | 5 | 2 | 4 | 3 |
7 | 0 | 6 | 1 | 5 | 2 | 4 | 3 |
7 | 0 | 6 | 1 | 5 | 2 | 4 | 3 |
0 | 7 | 1 | 6 | 2 | 5 | 3 | 4 |
0 | 7 | 1 | 6 | 2 | 5 | 3 | 4 |
0 | 7 | 1 | 6 | 2 | 5 | 3 | 4 |
0 | 7 | 1 | 6 | 2 | 5 | 3 | 4 |
The numbers of the second auxiliary square are now multiplied by n=8 and added to the numbers of the first auxiliary square. The result is a magic square.
64 | 8 | 56 | 16 | 41 | 17 | 33 | 25 |
57 | 1 | 49 | 9 | 48 | 24 | 40 | 32 |
61 | 5 | 53 | 13 | 44 | 20 | 36 | 28 |
60 | 4 | 52 | 12 | 45 | 21 | 37 | 29 |
7 | 63 | 15 | 55 | 18 | 42 | 26 | 34 |
2 | 58 | 10 | 50 | 23 | 47 | 31 | 39 |
6 | 62 | 14 | 54 | 19 | 43 | 27 | 35 |
3 | 59 | 11 | 51 | 22 | 46 | 30 | 38 |
Of course, the role of the two auxiliary squares can be interchanged. This means that the rows of the first auxiliary square have been filled with the numbers from 0 to n − 1, while the rows of the second auxiliary square have been filled with the numbers from 1 to n.
6 | 6 | 6 | 6 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 6 | 6 | 6 | 6 |
4 | 4 | 4 | 4 | 3 | 3 | 3 | 3 |
3 | 3 | 3 | 3 | 4 | 4 | 4 | 4 |
7 | 7 | 7 | 7 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 7 | 7 | 7 | 7 |
5 | 5 | 5 | 5 | 2 | 2 | 2 | 2 |
2 | 2 | 2 | 2 | 5 | 5 | 5 | 5 |
6 | 3 | 8 | 1 | 5 | 4 | 7 | 2 |
6 | 3 | 8 | 1 | 5 | 4 | 7 | 2 |
6 | 3 | 8 | 1 | 5 | 4 | 7 | 2 |
6 | 3 | 8 | 1 | 5 | 4 | 7 | 2 |
3 | 6 | 1 | 8 | 4 | 5 | 2 | 7 |
3 | 6 | 1 | 8 | 4 | 5 | 2 | 7 |
3 | 6 | 1 | 8 | 4 | 5 | 2 | 7 |
3 | 6 | 1 | 8 | 4 | 5 | 2 | 7 |
Now of course all numbers of the first auxiliary square must be multiplied by n=8, before the numbers of the second auxiliary square are added. The result of this is the following magic square.
54 | 51 | 56 | 49 | 13 | 12 | 15 | 10 |
14 | 11 | 16 | 9 | 53 | 52 | 55 | 50 |
38 | 35 | 40 | 33 | 29 | 28 | 31 | 26 |
30 | 27 | 32 | 25 | 37 | 36 | 39 | 34 |
59 | 62 | 57 | 64 | 4 | 5 | 2 | 7 |
3 | 6 | 1 | 8 | 60 | 61 | 58 | 63 |
43 | 46 | 41 | 48 | 20 | 21 | 18 | 23 |
19 | 22 | 17 | 24 | 44 | 45 | 42 | 47 |
You will find further variants of this method in the PDF document.