An order 6n$$\pm$$1
pandiagonal magic square
has an odd magic constant, however, the generation process of a magic
square differs from the case
4n+2. This is
because there is one more uneven number and the rows, columns and
diagonals have uneven lengths as well. This allows the sum of any square
to always be an integer, unlike with an order 4n+2 magic square. To prove
this, each box in a square (order 7) gets a name and is expressed in the
sum of each row, column and diagonal, which all add up to the same value
s1.
| a(1) |
a(2) |
a(3) |
a(4) |
a(5) |
a(6) |
a(7) |
| a(8) |
a(9) |
a(10) |
a(11) |
a(12) |
a(13) |
a(14) |
| a(15) |
a(16) |
a(17) |
a(18) |
a(19) |
a(20) |
a(21) |
| a(22) |
a(23) |
a(24) |
a(25) |
a(26) |
a(27) |
a(28) |
| a(29) |
a(30) |
a(31) |
a(32) |
a(33) |
a(34) |
a(35) |
| a(36) |
a(37) |
a(38) |
a(39) |
a(40) |
a(41) |
a(42) |
| a(43) |
a(44) |
a(45) |
a(46) |
a(47) |
a(48) |
a(49) |
table 11: 7 by 7 square with 49 imaginary numers
All columns, rows and diagonals in a pandiagonal magic square of order
6n$$\pm$$ 1 must equal to the same magic constant. To show this, the
following linear equations were written.
Rows
a( 1) + a( 2) + a( 3) + a( 4) + a( 5) + a( 6) + a( 7) = s1
a( 8) + a( 9) + a(10) + a(11) + a(12) + a(13) + a(14) = s1
a(15) + a(16) + a(17) + a(18) + a(19) + a(20) + a(21) = s1
a(22) + a(23) + a(24) + a(25) + a(26) + a(27) + a(28) = s1
a(29) + a(30) + a(31) + a(32) + a(33) + a(34) + a(35) = s1
a(36) + a(37) + a(38) + a(39) + a(40) + a(41) + a(42) = s1
a(43) + a(44) + a(45) + a(46) + a(47) + a(48) + a(49) = s1
Columns
a( 1) + a( 8) + a(15) + a(22) + a(29) + a(36) + a(43) = s1
a( 2) + a( 9) + a(16) + a(23) + a(30) + a(37) + a(44) = s1
a( 3) + a(10) + a(17) + a(24) + a(31) + a(38) + a(45) = s1
a( 4) + a(11) + a(18) + a(25) + a(32) + a(39) + a(46) = s1
a( 5) + a(12) + a(19) + a(26) + a(33) + a(40) + a(47) = s1
a( 6) + a(13) + a(20) + a(27) + a(34) + a(41) + a(48) = s1
a( 7) + a(14) + a(21) + a(28) + a(35) + a(42) + a(49) = s1
Diagonals top-left to bottom-right
a( 1) + a( 9) + a(17) + a(25) + a(33) + a(41) + a(49) = s1
a( 2) + a(10) + a(18) + a(26) + a(34) + a(42) + a(43) = s1
a( 3) + a(11) + a(19) + a(27) + a(35) + a(36) + a(44) = s1
a( 4) + a(12) + a(20) + a(28) + a(29) + a(37) + a(45) = s1
a( 5) + a(13) + a(21) + a(22) + a(30) + a(38) + a(46) = s1
a( 6) + a(14) + a(15) + a(23) + a(31) + a(39) + a(47) = s1
a( 7) + a( 8) + a(16) + a(24) + a(32) + a(40) + a(48) = s1
Diagonals top-right to bottom-left
a( 7) + a(13) + a(19) + a(25) + a(31) + a(37) + a(43) = s1
a( 6) + a(12) + a(18) + a(24) + a(30) + a(36) + a(49) = s1
a( 5) + a(11) + a(17) + a(23) + a(29) + a(42) + a(48) = s1
a( 4) + a(10) + a(16) + a(22) + a(35) + a(41) + a(47) = s1
a( 3) + a( 9) + a(15) + a(28) + a(34) + a(40) + a(46) = s1
a( 2) + a( 8) + a(21) + a(27) + a(33) + a(39) + a(45) = s1
These solutions for s1 can be manipulated through substitution with the
following solutions as a result.
a(43) = s1 - a(44) - a(45) - a(46) - a(47) - a(48) - a(49)
a(36) = s1 - a(37) - a(38) - a(39) - a(40) - a(41) - a(42)
a(29) = s1 - a(30) - a(31) - a(32) - a(33) - a(34) - a(35)
a(22) = s1 - a(23) - a(24) - a(25) - a(26) - a(27) - a(28)
a(21) = s1 - a(22) - a(27) - a(28) + a(31) + a(32) - a(35) - a(36) - a(41)
- a(42) - a(49)
a(20) = s1 - a(26) - a(27) - a(28) + a(30) + a(31) - a(34) - a(40) - a(41)
- a(42) - a(48)
a(19) = s1 - a(25) - a(26) - a(27) + a(29) + a(30) - a(33) - a(39) - a(40)
- a(41) - a(47)
a(18) = s1 - a(24) - a(25) - a(26) + a(29) - a(32) + a(35) - a(38) - a(39)
- a(40) - a(46)
a(17) = s1 - a(23) - a(24) - a(25) - a(31) + a(34) + a(35) - a(37) - a(38)
- a(39) - a(45)
a(16) = s1 - a(22) - a(23) - a(24) - a(30) + a(33) + a(34) - a(36) - a(37)
- a(38) - a(44)
a(15) = s1 - a(16) - a(17) - a(18) - a(19) - a(20) - a(21)
a(14) = - a(21) + a(29) + a(34) - a(38) - a(39) + a(43) + a(48)
a(13) = a(14) + a(15) - a(20) + a(23) - a(27) + a(31) - a(34) + a(39) -
a(41) + a(47) - a(48)
a(12) = a(13) - a(19) + a(21) + a(22) - a(26) + a(30) - a(33) + a(38) -
a(40) + a(46) - a(47)
a(11) = a(12) - a(18) + a(20) - a(25) + a(28) + a(29) - a(32) + a(37) -
a(39) + a(45) - a(46)
a(10) = a(11) - a(17) + a(19) - a(24) + a(27) - a(31) + a(35) + a(36) -
a(38) + a(44) - a(45)
a( 9) = a(10) - a(16) + a(18) - a(23) + a(26) - a(30) + a(34) - a(37) +
a(42) + a(43) - a(44)
a( 8) = s1 - a( 9) - a(10) - a(11) - a(12) - a(13) - a(14)
a( 7) = s1 - a(14) - a(21) - a(28) - a(35) - a(42) - a(49)
a( 6) = s1 - a(13) - a(20) - a(27) - a(34) - a(41) - a(48)
a( 5) = s1 - a(12) - a(19) - a(26) - a(33) - a(40) - a(47)
a( 4) = s1 - a(11) - a(18) - a(25) - a(32) - a(39) - a(46)
a( 3) = s1 - a(10) - a(17) - a(24) - a(31) - a(38) - a(45)
a( 2) = s1 - a( 9) - a(16) - a(23) - a(30) - a(37) - a(44)
a( 1) = s1 - a( 2) - a( 3) - a( 4) - a( 5) - a( 6) - a( 7)
For more information on this proof visit
entertainmentmathematics.nl
From these manipulations it becomes clear that s1 always stays an integer
as it never gets multiplied by a fraction, unlike the previous orders 4n+2
and 4n, unlike with the orders 4n+2. This allows it to use
consecutive
integers. This type of order has three steps to construct the pandiagonal
magic square. An order 7 pandiagonal magic square will be used to
demonstrate this. An order 7 pandiagonal magic square has a
magic constant of
175. To construct a pandiagonal magic square of order 6n$$\pm$$ 1 the
following formla A+(6n$$\pm$$1)$$A^T$$-(6n$$\pm$$1)C must be used.The 6n$$\pm$$ 1 portion of
the equation functions as a way to regulate the numbers that will be
filled in, it will never allow a number above $$n^2$$, as this is the last
number in the consecutive series, and thus, the largest number in the
magic square. If n=1 and $$\pm$$ -> + the resulting magic square would be
of order 7. In the squares A and B, no number larger than 7 appears,
therefore, the largest number that could result from the multiplication of
(6n$$\pm$$1)*$$A^T$$ is 49, which is also the largest number in the consecutive
series of an order 7 magic square.
Step 1
The first step in creating an order 6n ± 1 pandiagonal magic square is to
place the number {1,2,3,4,5,6,7} in the first column from top to bottom.
The second column will be filled in with {6,7,1,2,3,4,5} from top to
bottom. Each number moves two spaces further down the array. This process
repeats itself for every next column till all columns are filled in. The
third column would therefore be {4,5,6,7,1,2,3}. The completed square that
exists out of this will be called square A.
| 1 |
6 |
4 |
2 |
7 |
5 |
3 |
| 2 |
7 |
5 |
3 |
1 |
6 |
4 |
| 3 |
1 |
6 |
4 |
2 |
7 |
5 |
| 4 |
2 |
7 |
5 |
3 |
1 |
6 |
| 5 |
3 |
1 |
6 |
4 |
2 |
7 |
| 6 |
4 |
2 |
7 |
5 |
3 |
1 |
| 7 |
5 |
3 |
1 |
6 |
4 |
2 |
table 11: Step 1 square A for pandiagonal magic square of order 6n ± 1
(n=1; ±→+)
Step 2
To make a second square for an order 6n ± 1 pandiagonal magic square, all
that has to be done is transpose square A. This is done by switching rows
with columns. What once was the column {1,2,3,4,5,6,7} and the row
{1,6,4,2,7,5,3}, now becomes the column {1,6,4,2,7,5,3} and row
{1,2,3,4,5,6,7}. The following square B will exist out of this.
| 1 |
2 |
3 |
4 |
5 |
6 |
7 |
| 6 |
7 |
1 |
2 |
3 |
4 |
5 |
| 4 |
5 |
6 |
7 |
1 |
2 |
3 |
| 2 |
3 |
4 |
5 |
6 |
7 |
1 |
| 7 |
1 |
2 |
3 |
4 |
5 |
6 |
| 5 |
6 |
7 |
1 |
2 |
3 |
4 |
| 3 |
4 |
5 |
6 |
7 |
1 |
2 |
table 12: Step 1 square B=$$A^T$$ for pandiagonal magic square of order 6n ±
1 (n=1; ±→+)
Step 3
To finally create the magic square for order 6n ± 1, the formula
A+(6n$$\pm$$1)$$A^T$$-(6n$$\pm$$1)Cmust be used. Square C is again a square of order 6n±1
filled in with the number 1, just like in
table 4. The top left
corner for an order 7 magic square will therefore be 1 + 7*1 - 7*1 = 1.
The following pandiagonal magic square of order 6n±1 with n=1 and ±→+ will
be formed.
| 1 |
13 |
18 |
23 |
35 |
40 |
45 |
| 37 |
49 |
5 |
10 |
15 |
27 |
32 |
| 24 |
29 |
41 |
46 |
2 |
14 |
19 |
| 11 |
16 |
28 |
33 |
38 |
43 |
6 |
| 47 |
3 |
8 |
20 |
25 |
30 |
42 |
| 34 |
39 |
44 |
7 |
12 |
17 |
22 |
| 21 |
26 |
31 |
36 |
48 |
4 |
9 |
table 13: Step 3 completed pandiagonal magic square of order 6n ± 1
(n=1; ±→+)