Editing Magic square (section)

===Bordering method for order 3===
In this method, the objective is to wrap a border around a smaller magic square which serves as a core. Consider the 3×3 square for example. Subtracting the middle number 5 from each number 1, 2, ..., 9, we obtain 0, ±1, ±2, ±3, and ±4, which we will, for lack of better words, following S. Harry White, refer to as bone numbers. The magic constant of a magic square, which we will refer to as the skeleton square, made by these bone numbers will be zero since adding all the rows of a magic square will give ''nM'' = Σ ''k'' = 0; thus ''M'' = 0.

It is not difficult to argue that the middle number should be placed at the center cell: let ''x'' be the number placed in the middle cell, then the sum of the middle column, middle row, and the two diagonals give Σ ''k'' + 3 ''x'' = 4 ''M''. Since Σ ''k'' = 3 ''M'', we have ''x'' = ''M'' / 3. Here ''M'' = 0, so ''x'' = 0.

Putting the middle number 0 in the center cell, we want to construct a border such that the resulting square is magic. Let the border be given by:

{| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;width:6em;height:6em;table-layout:fixed;"
|-
| ''u'' || ''a'' || ''v''
|-
| ''b''{{sup|&lowast;}} || 0 || ''b''
|-
| ''v''{{sup|&lowast;}} || ''a''{{sup|&lowast;}} || ''u''{{sup|&lowast;}}
|}

Since the sum of each row, column, and diagonals must be a constant (which is zero), we have

: ''a'' + ''a''{{sup|&lowast;}} = 0, 
: ''b'' + ''b''{{sup|&lowast;}} = 0, 
: ''u'' + ''u''{{sup|&lowast;}} = 0,
: ''v'' + ''v''{{sup|&lowast;}} = 0.

Now, if we have chosen ''a'', ''b'', ''u'', and ''v'', then we have ''a''{{sup|&lowast;}} = &minus;''a'', ''b''{{sup|&lowast;}} = &minus;''b'', ''u''{{sup|&lowast;}} = &minus;''u'', and ''v''{{sup|&lowast;}} = &minus;''v''. This means that if we assign a given number to a variable, say ''a'' = 1, then its complement will be assigned to ''a''{{sup|&lowast;}}, i.e. ''a''{{sup|&lowast;}} = &minus;1. Thus out of eight unknown variables, it is sufficient to specify the value of only four variables. We will consider ''a'', ''b'', ''u'', and ''v'' as independent variables, while ''a''{{sup|&lowast;}}, ''b''{{sup|&lowast;}}, ''u''{{sup|&lowast;}}, and ''v''{{sup|&lowast;}} as dependent variables. This allows us to consider a bone number ±''x'' as a single number regardless of sign because (1) its assignment to a given variable, say ''a'', will automatically imply that the same number of opposite sign will be shared with its complement ''a''{{sup|&lowast;}}, and (2) two independent variables, say ''a'' and ''b'', cannot be assigned the same bone number. But how should we choose ''a'', ''b'', ''u'', and ''v''? We have the sum of the top row and the sum of the right column as

: ''u'' + ''a'' + ''v'' = 0,
: ''v'' + ''b'' + ''u''{{sup|&lowast;}} = 0.

Since 0 is an even number, there are only two ways that the sum of three integers will yield an even number: 1) if all three were even, or 2) if two were odd and one was even. Since in our choice of numbers we only have two even non-zero number (±2 and ±4), the first statement is false. Hence, it must be the case that the second statement is true: that two of the numbers are odd and one even.

The only way that both the above two equations can satisfy this parity condition simultaneously, and still be consistent with the set of numbers we have, is when ''u'' and ''v'' are odd. For on the contrary, if we had assumed ''u'' and ''a'' to be odd and ''v'' to be even in the first equation, then ''u''{{sup|&lowast;}} = &minus;''u'' will be odd in the second equation, making ''b'' odd as well, in order to satisfy the parity condition. But this requires three odd numbers (''u'', ''a'', and ''b''), contradicting the fact that we only have two odd numbers (±1 and ±3) which we can use. This proves that the odd bone numbers occupy the corners cells. When converted to normal numbers by adding 5, this implies that the corners of a 3×3 magic square are all occupied by even numbers.

Thus, taking ''u'' = 1 and ''v'' = 3, we have ''a'' = &minus;4 and ''b'' = &minus;2. Hence, the finished skeleton square will be as in the left. Adding 5 to each number, we get the finished magic square.

{{col-begin|width=auto;margin:0.5em auto}}
{{col-break|valign=bottom}}
{| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;width:6em;height:6em;table-layout:fixed;"
|-
| 1 || &minus;4 || 3
|-
| 2 || 0 || &minus;2
|-
| &minus;3 || 4 || &minus;1
|}
{{col-break|valign=bottom|gap=1em}}
{| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;width:6em;height:6em;table-layout:fixed;"
|-
| 6 || 1 || 8
|-
| 7 || 5 || 3
|-
| 2 || 9 || 4
|}
{{col-end}}

Similar argument can be used to construct larger squares. Since there does not exist a 2×2 magic square around which we can wrap a border to construct a 4×4 magic square, the next smallest order for which we can construct bordered square is the order 5.