Editing Magic square (section)

===Bordering method for order 5===
Consider the fifth-order square. For this, we have a 3×3 magic core, around which we will wrap a magic border. The bone numbers to be used will be ±5, ±6, ±7, ±8, ±9, ±10, ±11, and ±12. Disregarding the signs, we have 8 bone numbers, 4 of which are even and 4 of which are odd. In general, for a square of any order ''n'', there will be 4(''n'' &minus; 1) border cells, which are to be filled using 2(''n'' &minus; 1) bone numbers. Let the magic border be given as
 
{| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;width:10em;height:10em;table-layout:fixed;"
|-
| ''u'' || ''a'' || ''b'' || ''c'' || ''v''
|-
| ''d''{{sup|&lowast;}} || || || || ''d''
|-
| ''e''{{sup|&lowast;}} || || || || ''e''
|-
| ''f''{{sup|&lowast;}} || || || || ''f''
|-
| ''v''{{sup|&lowast;}} || ''a''{{sup|&lowast;}} || ''b''{{sup|&lowast;}} || ''c''{{sup|&lowast;}} || ''u''{{sup|&lowast;}}
|}

As before, we should

* ''place a bone number and its complement opposite to each other, so that the magic sum will be zero.''

It is sufficient to determine the numbers ''u, v, a, b, c, d, e, f'' to describe the magic border. As before, we have the two constraint equations for the top row and right column:

: ''u'' + ''a'' + ''b'' + ''c'' + ''v'' = 0
: ''v'' + ''d'' + ''e'' + ''f'' + ''u*'' = 0.

Multiple solutions are possible. The standard procedure is to

* ''first try to determine the corner cells, after which we will try to determine the rest of the border.''

There are 28 ways of choosing two numbers from the set of 8 bone numbers for the corner cells ''u'' and ''v''. However, not all pairs are admissible. Among the 28 pairs, 16 pairs are made of an even and an odd number, 6 pairs have both as even numbers, while 6 pairs have them both as odd numbers.

We can prove that the corner cells ''u'' and ''v'' cannot have an even and an odd number. This is because if this were so, then the sums ''u'' + ''v'' and ''v'' + ''u''{{sup|&lowast;}} will be odd, and since 0 is an even number, the sums ''a'' + ''b'' + ''c'' and ''d'' + ''e'' + ''f'' should be odd as well. The only way that the sum of three integers will result in an odd number is when 1) two of them are even and one is odd, or 2) when all three are odd. Since the corner cells are assumed to be odd and even, neither of these two statements are compatible with the fact that we only have 3 even and 3 odd bone numbers at our disposal. This proves that ''u'' and ''v'' cannot have different parity. This eliminates 16 possibilities.

Using similar type reasoning we can also draw some conclusions about the sets {''a'', ''b'', ''c''} and {''d'', ''e'', ''f''}. If ''u'' and ''v'' are both even, then both the sets should have two odd numbers and one even number. If ''u'' and ''v'' are both odd, then one of the sets should have three even numbers while the other set should have one even number and two odd numbers.

As a running example, consider the case when both ''u'' and ''v'' are even. The 6 possible pairs are: (6, 8), (6, 10), (6, 12), (8, 10), (8, 12), and (10, 12). Since the sums ''u'' + ''v'' and ''v'' + ''u''{{sup|&lowast;}} are even, the sums ''a'' + ''b'' + ''c'' and ''d'' + ''e'' + ''f'' should be even as well. The only way that the sum of three integers will result in an even number is when 1) two of them are odd and one is even, or 2) when all three are even. The fact that the two corner cells are even means that we have only 2 even numbers at our disposal. Thus, the second statement is not compatible with this fact. Hence, it must be the case that the first statement is true: two of the three numbers should be odd, while one be even.

Now let ''a, b, d, e'' be odd numbers while ''c'' and ''f'' be even numbers. Given the odd bone numbers at our disposal: ±5, ±7, ±9, and ±11, their differences range from ''D'' = {±2, ±4, ±6} while their sums range from ''S'' = {±12, ±14, ±16, ±18, ±20}. It is also useful to have a table of their sum and differences for later reference. Now, given the corner cells (''u'', ''v''), we can check its admissibility by checking if the sums ''u'' + ''v'' + ''c'' and ''v'' + ''u''{{sup|&lowast;}} + ''f'' fall within the set ''D'' or ''S''. The admissibility of the corner numbers is a necessary but not a sufficient condition for the solution to exist.

For example, if we consider the pair (''u'', ''v'') = (8, 12), then ''u'' + ''v'' = 20 and ''v'' + ''u*'' = 6; and we will have ±6 and ±10 even bone numbers at our disposal. Taking ''c'' = ±6, we have the sum ''u'' + ''v'' + ''c'' to be 26 and 14, depending on the sign of ±6 taken, both of which do not fall within the sets ''D'' or ''S''. Likewise, taking ''c'' = ±10, we have the sum ''u'' + ''v'' + ''c'' to be 30 and 10, both of which again do not fall within the sets ''D'' or ''S''. Thus, the pair (8, 12) is not admissible. By similar process of reasoning, we can also rule out the pair (6, 12).

As another example, if we consider the pair (''u'', ''v'') = (10, 12), then ''u'' + ''v'' = 22 and ''v'' + ''u''{{sup|&lowast;}} = 2; and we will have ± 6 and ± 8 even bone numbers at our disposal. Taking ''c'' = ±6, we have the sum ''u'' + ''v'' + ''c'' to be 28 and 16. While 28 does not fall within the sets ''D'' or ''S'', 16 falls in set ''S''. By inspection, we find that if (''a'', ''b'') = (&minus;7, &minus; 9), then ''a'' + ''b'' = &minus;16; and it will satisfy the first constraint equation. Also, taking ''f'' = ± 8, we have the sum ''v'' + ''u''{{sup|&lowast;}} + ''f'' to be 10 and -6. While 10 does not fall within the sets ''D'' or ''S'', &minus;6 falls in set ''D''. Since &minus;7 and &minus;9 have already been assigned to ''a'' and ''b'', clearly (''d'', ''e'') = (-5, 11) so that ''d'' + ''e'' = 6; and it will satisfy the second constraint equation.

Likewise, taking ''c'' = ±8, we have the sum ''u'' + ''v'' + ''c'' to be 30 and 14. While 30 does not fall within the sets ''D'' or ''S'', 14 falls in set ''S''. By inspection, we find that if (''a'', ''b'') = (&minus;5, &minus;9), then ''a'' + ''b'' = &minus;14. Also, taking ''f'' = ± 6, we have the sum ''v'' + ''u''{{sup|&lowast;}} + ''f'' to be 8 and -4. While 8 does not fall within the sets ''D'' or ''S'', &minus;4 falls in set ''D''. Clearly, (''d'', ''e'') = (&minus;7, 11) so that ''d'' + ''e'' = 4, and the second constraint equation will be satisfied.

Hence the corner pair (''u'', ''v'') = (10, 12) is admissible; and it admits two solutions: {{tmath|1=(a, b, c, d, e, f) = (-7, -9, -6, -5, 11, -8)}} and {{tmath|1=(a, b, c, d, e, f) = ( -5, -9, -8, -7, 11, -6)}}. The finished skeleton squares are given below. The magic square is obtained by adding 13 to each cells.

{{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:10em;height:10em;table-layout:fixed;"
|-
| 10 || -7 || -9 || -6 || 12
|-
| 5 || || || || -5
|-
| -11 || || || || 11
|-
| 8 || || || || -8
|-
| -12 || 7 || 9 || 6 || -10
|}
{{col-break|valign=bottom|gap=1em}}
{| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;width:10em;height:10em;table-layout:fixed;"
|-
| 23 || 6 || 4 || 7 || 25
|-
| 18 || || || || 8
|-
| 2 || || || || 24
|-
| 21 || || || || 5
|-
| 1 || 20 || 22 || 19 || 3
|}
{{col-break|valign=bottom|gap=1em}}
{| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;width:10em;height:10em;table-layout:fixed;"
|-
| 10 || -5 || -9 || -8 || 12
|-
| 7 || || || || -7
|-
| -11 || || || || 11
|-
| 6 || || || || -6
|-
| -12 || 5 || 9 || 8 || -10
|}
{{col-break|valign=bottom|gap=1em}}
{| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;width:10em;height:10em;table-layout:fixed;"
|-
| 23 || 8 || 4 || 5 || 25
|-
| 20 || || || || 6
|-
| 2 || || || || 24
|-
| 19 || || || || 7
|-
| 1 || 18 || 22 || 21 || 3
|}
{{col-end}}

Using similar process of reasoning, we can construct the following table for the values of ''u, v, a, b, c, d, e, f'' expressed as bone numbers as given below. There are only 6 possible choices for the corner cells, which leads to 10 possible border solutions.

{| class="wikitable" style="margin-left:auto;margin-right:auto;text-align:center;width:18em;height:10em;table-layout:fixed;"
|-
! ''u, v'' 
! ''a, b, c'' 
! ''d, e, f'' 
|-
| 12, 10 || -6, -7, -9 || -11, 5, 8
|-
| 12, 10 || -5, -8, -9 || -11, 6, 7
|-
| 11, 5 || 6, -10, -12 || -9, 7, 8
|-
| 10, 6 || 5, -9, -12 || -11, 7, 8
|-
| 10, 6 || 7, -11, -12 || -9, 5, 8
|-
| 9, 7 || 5, -10, -11 || -12, 6, 8
|-
| 9, 7 || 6, -10, -12 || -11, 5, 8
|-
| 8, 6 || 7, -10, -11 || -12, 5, 9
|-
| 8, 6 || 9, -11, -12 || -10, 5, 7
|-
| 7, 5 || 9, -10, -11 || -12, 6, 8
|}

Given this group of 10 borders, we can construct 10×8×(3!)<sup>2</sup> = 2880 essentially different bordered magic squares. Here the bone numbers ±5, ..., ±12 were consecutive. More bordered squares can be constructed if the numbers are not consecutive. If non-consecutive bone numbers were also used, then there are a total of 605 magic borders. Thus, the total number of order 5 essentially different bordered magic squares (with consecutive and non-consecutive numbers) is 174,240.<ref>http://oz.nthu.edu.tw/~u9621110/IT2010/txt/0929/canterburypuzzle00dudeuoft.pdf
The Canterbury Puzzles and Other Curious Problems, Henry Ernest Dudeney, 1907
</ref><ref>http://budshaw.ca/howMany.html, Bordered Square Numbers, S. Harry White, 2009</ref> See history.<ref>http://www.law05.si/iwms/presentations/Styan.pdf Some illustrated comments on 5×5 golden magic matrices and on 5×5 ''Stifelsche Quadrate'', George P. H. Styan, 2014.</ref> The number of fifth-order magic squares constructible via the bordering method is about 26 times larger than via the superposition method.