Something Magical about Squares

We start with an n by n grid (n is odd). Then fill the entries of the middle row (or the (n+1)/2-th row) of the grid with the following:

(i.e. the (j+1)-th entry= the j-th entry + (n + 1))

Then fill out the entries as follows:
If the entry in the i-th row, j-th column is x, then the entry in the (i+1)-th row, (j+1)-th column is x + 2n. If x + 2n is greater than n², then subtract x + 2n by n². If the i-th row is the final row, consider the first row of the grid as the (i+1)-th row, likewise for columns.

For example, for n=5, we have:

This method generates a magic square of order n for any odd number n. (In fact, we can replace the 2 in the expression x + 2n by any number m, m ≠ 1, that is co-prime with n to obtain a different magic square.)

For n that is even but not divisible by 4, we can use a different method to construct a magic square of order n.

There is no magic square of order 2.

First we express n as 4m+2, where m is a positive integer. (Since n is even and not divisible by 4, n divided by 4 gives a remainder of 2.)

We start with an empty n-by-n grid and we use the symbol S_i,j to denote the entry on the i-th row, j-th column. (S_{1, 1} is the entry at the top-left corner of the grid.)

We first fill the entries the diagonals. We have S_{i, i} = i × n - (i - 1) and
S_{i, (n - i + 1)} = (i - 1) × n + i.

For example, for n = 10, we have the following:

We then fill the middle 2 columns of the grid as follows:

• The entries on the (2m + 1)-th and (2m + 2)-th rows are already filled.
• We have S_{1, 2m + 1} = 2m + 1 and S_{1, 2m + 2} = (n - 1) × n + (2m + 2).
• S_{n, 2m + 1} = (n - 1) × n + (2m + 1) and S_{n, 2m + 2} = 2m + 2.
• For i = 2, 3,...2m, we have S_{i, 2m + 1} = (n - i + 1) × n + (2m + 1) and S_{i, 2m + 2} = S_{i, 2m + 1} + 1.
• For i = 2m + 3, 2m + 4, ... n - 1 (or 4m + 1), S_{i, j} = n² + 1 - S_{(n + 1 - i), j}

So, for n = 10 (= 2 × 4 + 2, so m = 2), we have the following:

Now we fill the rest of entries in the first and final columns as follows:

• S_{2m + 1, 1} = n²/2 + 1
• S_{2m + 1, n} = n²/2 + n
• S_{2m + 2, 1} = n²/2
• S_{2m + 2, n} = n²/2 - n + 1
• For i = 2 , 4, ... 2m, and i = 2m + 3, 2m + 5 , ... n - 1, S_{i, 1} = (i - 1) × n + 1 and S_{i, n} = n² + 1 - S_{i, 1}
• For i = 3 , 5, ... 2m - 1, and i = 2m + 4, 2m + 6, ... n - 2, S_{i, n} = (i - 1) × n + 1 and S_{i, 1} = n² + 1 - S_{i, n}

So, for n = 10, we have the following:

Next we deal with the middle 2 rows. We fill the blank entries as follows:

• For j = 2, 3, ... 2m, S_{i, j} = (i - 1) × n + j, i = 2m + 1, 2m + 2
• For j = 2m + 3, 2m + 4, ... n - 1, S_{i, j} = n² + 1 - S_{i, (n + 1 - i)}, i = 2m + 1, 2m + 2.

So for n = 10, the middle 2 rows look like this:

We fill the rest of the entries using a method called "Large-Small-Large Rule" (I have yet to come up with a better name). The method goes like this:

1. For each row i, i = 1, 2, ... 2m, we find the smallest j where the i,j-th entry of the grid is empty. Then we set
   S_{i, j} = (n - i) × n + j. (Large)
2. Find the next empty entry on row i and set S_{i, j} = (i - 1) × n + j, where j is the column number. (Small)
3. Find the next empty entry on row i and set S_{i, j} = (n - i) × n + j, where j is the column number. (Large)
4. We continue this pattern until the 2m-th column is filled. (The (2m + 1)-th and (2m + 2)-th columns on row i are already filled.)
5. S_{i, 2m + 3} = S_{i, 2m} + 3
6. We continue to alternate between small and large values until we fill out the row.
7. For i = 2m + 3, 2m + 4, ... n, if the (i,j)-th entry is empty, then S_{i, j} = n² + 1 - S_{n + 1 - i, j}

For example, in the second row of the 10-by-10 grid, the first empty entry is on the 3rd column, so S_{2, 3} = (10 - 2) × 10 + 3 = 83.

Then next empty entry is on the 4th column, so S_{2, 4} = (2 - 1) × 10 + 4 = 14. S_{2, 7} = S_{2, 4} + 3 = 14 + 3 = 17 and S_{2, 8} = 88.

We don't need to go any further as the row is now filled. The row would look as follows:

After the rest of the entries are filled, we have the following:

We can construct a different magic square by switching the words "rows" and "columns" in the above description.

If n is divisible by 4, then we can construct a magic square of order n as follows:

1. We express n in the form: n = 2^ab, where a < 1 (since n is a multiple of 4) and b is odd (b may be equal to 1).
2. Construct a magic square of order b (using any method).
   
3. If a is even, then replace each entry of the magic square with the magic square
   

   1+i	15+i	8+i	10+i
   12+i	6+i	13+i	3+i
   14+i	4+i	11+i	5+i
   7+i	9+i	2+i	16+i

   where i = (the current entry of the magic square - 1) × 16
   This yields a magic square of order 4b.
   If a is odd, then we replace each entry of the magic square with the following:
   

   1+i	27+i	13+i	23+i	48+i	54+i	36+i	58+i
   20+i	10+i	32+i	6+i	61+i	39+i	49+i	43+i
   34+i	60+i	46+i	56+i	15+i	21+i	3+i	25+i
   51+i	41+i	63+i	37+i	30+i	8+i	18+i	12+i
   62+i	40+i	50+i	44+i	19+i	9+i	31+i	5+i
   47+i	53+i	35+i	57+i	2+i	28+i	14+i	24+i
   29+i	7+i	17+i	11+i	53+i	42+i	64+i	38+i
   16+i	22+i	4+i	26+i	33+i	59+i	45+i	55+i

   where i = (the current entry of the magic square - 1) × 64
   This yields a magic square of order 8b.
4. We now continue the process (by expanding a magic square of order m into a magic square of order 4m) until we get a magic square of order n.

For example, we can use the following magic square of order 3 to construct a magic square of order 12.

The resulting magic square of order 12 is:

In general, one can use any magic square of two arbitrary orders, say a,b, to construct a magic square of order a×b using this method.

References and Acknowlegements