Order 4n+2 pandiagonal magic squares do not exist for
consecutive numbers. When looking at an order 6
pandiagonal magic square
the following magic square can be derived.
| 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) |
table 1: 6 by 6 square with 36 imaginary numbers
A pandiagonal magic square should have all it’s columns, rows and (broken)
diagonals equate to the magic constant (s1 in example below). This would
result in the following equations.
Rows
a( 1) + a( 2) + a( 3) + a( 4) + a( 5) + a( 6) = s1
a( 7) + a( 8) + a( 9) + a(10) + a(11) + a(12) = s1
a(13) + a(14) + a(15) + a(16) + a(17) + a(18) = s1
a(19) + a(20) + a(21) + a(22) + a(23) + a(24) = s1
a(25) + a(26) + a(27) + a(28) + a(29) + a(30) = s1
a(31) + a(32) + a(33) + a(34) + a(35) + a(36) = s1
Columns
a( 1) + a( 7) + a(13) + a(19) + a(25) + a(31) = s1
a( 2) + a( 8) + a(14) + a(20) + a(26) + a(32) = s1
a( 3) + a( 9) + a(15) + a(21) + a(27) + a(33) = s1
a( 4) + a(10) + a(16) + a(22) + a(28) + a(34) = s1
a( 5) + a(11) + a(17) + a(23) + a(29) + a(35) = s1
a( 6) + a(12) + a(18) + a(24) + a(30) + a(36) = s1
(Broken) diagonals top-left to bottom-right
a( 1) + a( 8) + a(15) + a(22) + a(29) + a(36) = s1
a( 2) + a( 9) + a(16) + a(23) + a(30) + a(31) = s1
a( 3) + a(10) + a(17) + a(24) + a(25) + a(32) = s1
a( 4) + a(11) + a(18) + a(19) + a(26) + a(33) = s1
a( 5) + a(12) + a(13) + a(20) + a(27) + a(34) = s1
a( 6) + a( 7) + a(14) + a(21) + a(28) + a(35) = s1
(Broken) diagonals top-right to bottom-left
a( 6) + a(11) + a(16) + a(21) + a(26) + a(31) = s1
a( 5) + a(10) + a(15) + a(20) + a(25) + a(36) = s1
a( 4) + a( 9) + a(14) + a(19) + a(30) + a(35) = s1
a( 3) + a( 8) + a(13) + a(24) + a(29) + a(34) = s1
a( 2) + a( 7) + a(18) + a(23) + a(28) + a(33) = s1
a( 1) + a(12) + a(17) + a(22) + a(27) + a(32) = s1
These equations account for all rows, columns and (broken diagonals) of an
order 6 magic square. These equations can be transformed to the following
equations, through use of substitution and manipulation.
a(31) = s1 - a(32) - a(33) - a(34) - a(35) - a(36)
a(25) = s1 - a(26) - a(27) - a(28) - a(29) - a(30)
a(19) = s1 - a(20) - a(21) - a(22) - a(23) - a(24)
a(17) = 2/3 * s1 - a(18) + a(20) + a(21) - a(23) - a(24) + a(26) + a(27) -
a(29) - a(30) - a(35) - a(36)
a(16) = 2 * s1 + a(18) - a(20) - 2*a(21) - 2*a(22) - a(23) - a(26) -
2*a(27) - 2*a(28) - a(29) - a(34) + a(36)
a(15) = 2/3 * s1 - a(18) - a(33) - a(36)
a(14) = a(18) - a(20) - a(21) + a(23) + a(24) - a(26) - a(27) + a(29) +
a(30) - a(32) + a(36)
a(13) = -4/3 * s1 - a(18) + a(20) + 2*a(21) + 2*a(22) + a(23) + a(26) +
2*a(27) + 2*a(28) + a(29) - a(31) - a(36)
a(12) = 11/6 * s1 - a(20) - a(21) - a(22) - a(26) - 2*a(27) - a(28) -
a(32) - a(33) - a(34)
a(11) = -7/6 * s1 + a(22) + a(23) + a(24) - a(26) + a(28) + a(29) + a(30)
+ a(34) + a(35) + a(36)
a(10) = -7/6 * s1 + a(21) + a(22) + a(23) - a(25) + a(27) + a(28) + a(29)
+ a(33) + a(34) + a(35)
a(9) = -7/6 * s1 + a(20) + a(21) + a(22) + a(26) + a(27) + a(28) - a(30) +
a(32) + a(33) + a(34)
a(8) = 11/6 * s1 - a(22) - a(23) - a(24) - a(28) - 2*a(29) - a(30) - a(34)
- a(35) - a(36)
a(7) = 11/6 * s1 - a(21) - a(22) - a(23) - a(27) - 2*a(28) - a(29) - a(33)
- a(34) - a(35)
a(6) = -5/6 * s1 - a(18) + a(20) + a(21) + a(22) - a(24) + a(26) + 2*a(27)
+ a(28) - a(30)+a(32)+a(33)+a(34)-a(36)
a(5) = 1/2 * s1 + a(18) + a(19) + a(24) - a(27) - a(28) - a(29) -
a(34) - a(35)
a(4) = 1/6 * s1 - a(18) + a(20) + a(21) - a(28) - a(29) - a(30) + a(31) +
a(32)
a(3) = 1/2 * s1 + a(18) - a(20) - 2*a(21) - a(22) - a(26) - 2*a(27) -
a(28) + a(30) + a(31) + a(35) + 2*a(36)
a(2) = 1/6 * s1 - a(18) + a(21) + a(22) - a(25) - a(26) - a(30) + a(34) +
a(35)
a(1) = 1/2 * s1 + a(18) + a(23) + a(24) - a(25) - a(26) - a(27) - a(31) -
a(32)
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From these newly derived equations it is clearly deductible that no
solutions can be found for consecutive integers 1,2,...36. This is because
the magic constant 111 is an odd number. This would result in numbers that
are not integers, for example, ½*111(s1) = 55.5 (a(5) in the equations
above (bold)). Henceforth, there cannot be a pandiagonal magic square with
consecutive integers. All magic constants for 4n+2 pandiagonal magic
squares are odd magic constants.
To create a 4n+2 pandiagonal magic square with non consecutive numbers
there is a clear pattern to follow. To demonstrate this an order 6 magic
square will be used. If the consecutive integers 1+2+3+4+5+6=21 are used,
an odd sum will appear (21). As proven above a row, column or diagonal
cannot sum up to an odd value. Therefore, for a 6 by 6 magic square the
numbers 1+2+3+5+6+7=24 will be used, as this is the closest to
a consecutive series that is possible for a pandiagonal magic square of order 4n+2. These 6 numbers will be specifically used to show
what the algorithm for constructing an 4n+2 magic square would look like.
Later on, these numbers will change into the a(i) numbers as demonstrated
above. The sum 24 can be divided in half, when splitting the six numbers
into appropriate groups (1,5,6 and 2,3,7). If these two sets of three
numbers are squared, they will both equate to the number 62, adding a
semi-
bimagic
property to the magic square. Because the series with a sum of 21 lacks
this ability to half itself, they can not be arranged equally in the six
by six grid. In return, it would be impossible to create a magic square
from it, as the rows, columns and diagonals, at the starting point of the
algorithm (explained below) would already not equate to the same amount on
the rows, columns and diagonals. To create an order 6 (or 4n+2) non
consecutive pandiagonal magic square, two order 6 squares must be filled
in with numbers {1,2,3,5,6,7}. These squares will work as a matrix. Later
on a multiplication step is needed to add both matrices, in combination
with a matrix completely filled in with 1’s, together to form the final
pandiagonal magic square of order 4n+2 (order 6).
The formula which will be used is β*A+B-β*C (β is the last
integer in each array/series of numbers). Matrix A and B, will be filled
in with the numbers {1,2,3,5,6,7} and matrix C is the matrix containing
the ones.The β 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. β always needs to be the largest number that
can be found in squares A and B (β=7). In the squares A and B, no number
larger than 7 appears, therefore, the largest number that could result
from the multiplication of β*A is 49, which is also the largest number in
the consecutive series of an order 7 magic square. And because only six
numbers are used (exculding 4) in squares A and B for a 4n+2 pandiagonal
magic square, it will be the largest number in there as well.
Square A
For the first square A the square will be vertically split in two
(indicated with a bold line in table 2 below). The left half will be
starting at the top with {1,5,6}, in the next row the numbers will all
move one space to the left resulting in {5,6,1}. Fill in each next row by
moving the numbers one space to the left after each row till all 6 rows of
length 3 are filled in. The same rule applies to the right half however
the top row starts with {7,3,2} which is followed by the next row with
{3,2,7}. When all 6 rows are filled in on the right side, square A has
been constructed.
| 1 |
5 |
6 |
7 |
3 |
2 |
| 5 |
6 |
1 |
3 |
2 |
7 |
| 6 |
1 |
5 |
2 |
7 |
3 |
| 1 |
5 |
6 |
7 |
3 |
2 |
| 5 |
6 |
1 |
3 |
2 |
7 |
| 6 |
1 |
5 |
2 |
7 |
3 |
table 2: square A for order 6 pandiagonal magic square
Square B
To create square B, it will have to be split horizontally into equal
slices (marked in bold in table 3 below). The top half will start on the
right with {6,1,5} and move down (to the right in array) one for each next
column of length 3 (e.g. {5,6,1}) till all 6 half columns have been filled
in. To complete the bottom half of square B, the sequence {2,7,3} will
start on the left and will also move down (to the right in the array) one
position for each next column. Once this has been completed square B is
finished.
| 6 |
5 |
1 |
6 |
5 |
1 |
| 1 |
6 |
5 |
1 |
6 |
5 |
| 5 |
1 |
6 |
5 |
1 |
6 |
| 2 |
3 |
7 |
2 |
3 |
7 |
| 7 |
2 |
3 |
7 |
2 |
3 |
| 3 |
7 |
2 |
3 |
7 |
2 |
table 3: square B for order 6 pandiagonal magic square
Square C
In addition to squares A and B there is also a square C, which is entirely
filled with the number 1.
| 1 |
1 |
1 |
1 |
1 |
1 |
| 1 |
1 |
1 |
1 |
1 |
1 |
| 1 |
1 |
1 |
1 |
1 |
1 |
| 1 |
1 |
1 |
1 |
1 |
1 |
| 1 |
1 |
1 |
1 |
1 |
1 |
| 1 |
1 |
1 |
1 |
1 |
1 |
table 4: square C for order 6 pandiagonal magic square
Final square
To finally get the pandiagonal magic square for an order 6 pandiagonal magic square
with non consecutive numbers the aformentioned formula β*A+B-β*C will be implemented.
(β is the last integer in each array/series of numbers). For an order 6
pandiagonal magic square the formula would look like 7A+B-7C. For the top
left square this would be 7*1+6-7*1=6. Finally, a pandiagonal magic square
with non consecutive numbers will appear with the highest integer being 49
and the magic constant being 150. The semi-bimagic feature (only rows and
columns) will have a magic constant of 5150.
| 6 |
33 |
36 |
48 |
19 |
8 |
| 29 |
41 |
5 |
15 |
13 |
47 |
| 40 |
1 |
34 |
12 |
43 |
20 |
| 2 |
31 |
42 |
44 |
17 |
14 |
| 35 |
37 |
3 |
21 |
9 |
45 |
| 38 |
7 |
30 |
10 |
49 |
16 |
table 5: order 6 pandiagonal square with non consecutive numbers