A bordered magic square is a magic square, which remains magic when its borders are removed. Let us take the depicted magic square of order 8. When we remove its border, we will get a square of order 6.
64 | 4 | 9 | 54 | 63 | 3 | 10 | 53 |
60 | 15 | 16 | 47 | 48 | 49 | 20 | 5 |
7 | 44 | 22 | 42 | 41 | 25 | 21 | 58 |
51 | 33 | 37 | 29 | 30 | 28 | 38 | 14 |
6 | 32 | 34 | 35 | 36 | 31 | 27 | 59 |
8 | 26 | 40 | 24 | 23 | 43 | 39 | 57 |
52 | 45 | 46 | 18 | 17 | 19 | 50 | 13 |
12 | 61 | 56 | 11 | 2 | 62 | 55 | 1 |
15 | 16 | 47 | 48 | 49 | 20 |
44 | 22 | 42 | 41 | 25 | 21 |
33 | 37 | 29 | 30 | 28 | 38 |
32 | 34 | 35 | 36 | 31 | 27 |
26 | 40 | 24 | 23 | 43 | 39 |
45 | 46 | 18 | 17 | 19 | 50 |
This square isn't normalized anymore, because it doesn't contain the numbers 1, 2, … , n^{2} anymore. But still each row, each column and both diagonals sum S=195.
A second condition for bordered magic squares demands that the numbers of the border enclose the numbers of the inner square. Satisfiying this condition means that the numbers 1, … ,2(n−1) and n^{2}−2(n−1) + 1 , … , n^{2} must form the border. All other numbers must be elements of the inner square.
Let's take a bordered magic square of order n=6 as an example. The inner square is of order n=4, and must be formed of 4^{2}=16 numbers. On the other side, there are
6^{2} − 4^{2} = 36 − 16 = 20
numbers for the border, which have to enclose the inner elements. So, ten numbers are less and ten numbers are greater than the inner numbers.
lower numbers of the border: | 1 … 10 |
inner numbers: | 11 … 26 |
upper numbers of the border: | 27 … 36 |
One example of such an arrangement is shown in the following magic square of order 6:
36 | 2 | 3 | 7 | 32 | 31 |
4 | 26 | 13 | 12 | 23 | 33 |
9 | 15 | 20 | 21 | 18 | 28 |
27 | 19 | 16 | 17 | 22 | 10 |
29 | 14 | 25 | 24 | 11 | 8 |
6 | 35 | 34 | 30 | 5 | 1 |