Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
Linear systems resulting from pandiagonal magic squares
Received: 26/9/2004
Accepted: 9/4/2006
Saleem Al-Ashhab*
&
Alla Al-Zahawi**
‫ملخص‬
‫ننسس‬
‫ن عيرننم ن ثأرظ ننم ثني ننم يثنمننا م م ن د م نني‬
‫ يسراسب نسس ثن ميين ثت ثنان ي نا ج ي نم‬.‫مدثيسث‬
‫ر‬
‫سننرس ف ننا اننحث ثنما ن‬
‫سسث ب ع‬
‫ثن ع سالت يثن ج ايل يه‬
.‫م‬
‫ال ثنرظ م مش ل‬
Abstract
In this paper we consider linear systems of certain types, having arbitrary
dimensions, and prove a rule about the number of free parameters in their
solution set.
1. Introduction:
A pandiagonal magic square is a n×n matrix
a1
a2
………
an
an+1
an+2
………
a2n
………….
………….
………….
………….
…...
an2
a n 2  n 1 a n 2  n  2
a , a ,..., a
2
n are distinct real numbers (usually integers)
where the entries 1 2
satisfying the following system of equations:
a 1  a 2  ..........  a n
a n 1  a n  2  .....  a 2n
S
S
....
....
a n 2  n 1  a n 2  n  2  ..  a n 2  S
*
**
….………..(1 – a )
Assistant professor, department of mathmatics, Al-albayt University.
Researcher in Mathematics.
Al-Manarah, Vol. 13, No. 6, 2007 .
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Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
a1  a n 1  .........  a n 2  n 1
S
a 2  a n  2  .........  a n 2  n  2  S
.....
.....
a n  a 2n  ..........  a n 2
S
a1  a n  2  ............  a n 2
…...……….(1 – b)
S
a 2  a n  3  ............  a n 2  n 1  S
......
.....
a n  a n 1  ............  a n 2 1
S
……………(1 – c)
a 1  a 2n  ...........  a n 2  n  2  S
a 2  a n 1  ...........  a n 2  n  3  S
....
....
a n  a 2n -1  ...........  a n 2  n 1  S
……………(1 – d)
where S is a real constant (the so-called magic sum). In this system the group
(1 – a) represents the summation of the entries in each row of the matrix. The
group (1 – b) represents the summation of the entries in each column of the
matrix. The group (1 – c) represents the summation of the entries in each right
(extended) diagonal of the matrix. The group (1 – d) represents the summation of
the entries in each left (extended) diagonal of the matrix. We will prove that the
linear system (1) will have a solution, which contains
n*n – 4*n + 3
free parameters, if n is odd, and
n*n – 4*n + 4
free parameters, if n is even.
Al-Manarah, Vol. 13, No. 6, 2007 .
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Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
2. The case of odd pandiagonal magic square:
We consider here the linear system (1), where n is of the form 2k+1 (k = 2, 3, …),
since there is no pandiagonal magic square in the case k = 1. We illustrate first
our theoretical approach by considering the case k = 2, i.e. We consider the square
a1
a2
a3
a4
a5
a6
a7
a8
a9
a10
a11
a12
a13
a14
a15
a16
a17
a18
a19
a20
a21
a22
a23
a24
a25
In this case the system (1) takes the form:
a1  a 2  a 3  a 4  a 5  S
a 6  a 7  a 8  a 9  a 10  S
………..(2 – a )
a 11  a 12  a 13  a 14  a 15  S
a 16  a 17  a 18  a 19  a 20  S
a 21  a 22  a 23  a 24  a 25  S
a 1  a 6  a 11  a 16  a 21  S
a 2  a 7  a 12  a 17  a 22  S
a 3  a 8  a 13  a 18  a 23  S
.……….(2 – b)
a 4  a 9  a 14  a 19  a 24  S
a 5  a 10  a 15  a 20  a 25  S
a 1  a 7  a 13  a 19  a 25  S
a 2  a 8  a 14  a 20  a 21  S
a 3  a 9  a 15  a 16  a 22  S
…..……(2 – c)
a 4  a 10  a 11  a 17  a 23  S
a 5  a 6  a 12  a 18  a 24  S
Al-Manarah, Vol. 13, No. 6, 2007 .
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Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
a 1  a 10  a 14  a 18  a 22  S
a 2  a 6  a 15  a 19  a 23  S
a 3  a 7  a 11  a 20  a 24  S
………(2 – d)
a 4  a 8  a 12  a 16  a 25  S
a 5  a 9  a 13  a 17  a 21  S
In the system (2 – a), …, (2 – d), there are three equations, which are linearly
dependent on the other equations. These equations are: the last equation of the
group (2 – b), the last equation of the group (2 – c) and the last equation of the
group (2 – d). Indeed, the last equation of the group (2 – b) is the result of
subtraction of the sum of all other equations of the group (2 – b) from the sum of
all equations of the group (2 – a). The last equation of the group (2 – c) is the
result of subtraction of the sum of all other equations of the group (2 – c) from the
sum of all equations of the group (2 – a). The last equation of the group (2 – d) is
the result of subtraction of the sum of all other equations of the group (2 – d) from
the sum of all equations of the group (2 – a).
In order to prove that there are no other dependent equations, we write down
the matrix of coefficients of the system after removing the previous mentioned
equations:
1
1  1

1
 1
1


1  1
1




1  1
1

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
1
1
1
1
 





1 0
1
0
1
0
1
0

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
0 1
0 01
0  0
1
1
 


 1


1


1 0
1
1
0
 0
1 0 0

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
0
0
1
0
1 0
0 1
1
 
1
1



1






1 0
1 0
0
1 0 0 0

1

Al-Manarah, Vol. 13, No. 6, 2007 .
182























 ……(3)








Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
We then prove that this matrix has full rank. To establish this we consider the
transpose of this matrix:
1
1
1
1









1

1
1
1


0
0
0
1

                                          


 1

1
0
0 1




1






1


1
0
 1

 1

0
1
1
0


                                          


1
1
0
1
0 0 1






0



1
1


 1

1

1


0
10
0 1
 1

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 












                                          


1 1
0 1
0







1


1





1
1
0


1
0
1
0
1


and prove that this matrix has full rank. In order to do this we study the
equation of linear dependence between the columns of the last matrix:
b1 * C1  b 2 * C 2  ...  b17 * C17  0
where C1 , C2 ,..., C17 denotes the columns of the matrix and b1 , b 2 ,..., b17 are
real numbers. The componentwise form of this equation is:
b1  b 6  b10  b14
0
b1  b 7  b11  b15
0
b1  b 8  b12  b16
0
b1  b 9  b13  b17
0
b1
0
Al-Manarah, Vol. 13, No. 6, 2007 .
…………..(4 – a)
183
Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
b 2  b 6  b15
0
b 2  b 7  b10  b16  0
…………..(4 – b)
b 2  b 8  b11  b17  0
b 2  b 9  b12
0
b 2  b13  b14
0
b 3  b 6  b13  b16  0
b 3  b 7  b17
0
b 3  b 8  b10
0
b 3  b 9  b11  b14
0
b 3  b12  b15
0
…………..(4 – c)
b 4  b 6  b12  b16  0
b 4  b 7  b13
0
b 4  b 8  b14
0
b 4  b 9  b10  b15
0
b 4  b11  b16
0
b 5  b 6  b11
0
b 5  b 7  b12  b14
0
b 5  b 8  b13  b15
0
b 5  b 9  b16
0
b 5  b10  b17
0
…………..(4 – d)
…………..(4 – e)
We conclude from the last equation of the group (4 – a) that b1  0 . Substituting
this value in the other equations of the group (4 – a), and adding up leads to:
17
b
j 6
j
………….(5)
0
Since the summation of the equations in (4 – b) yields
17
5b 2   b j  0 .
j 6
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Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
Hence, we deduce from (5) that b 2  0 . In the same manner we obtain
b3  b4  b5  0 .
Using our knowledge about b 1 , b 2 ,..., b 5 we are capable of rewriting the
system (4) like this:
b 6  b 10  b 14
0
b 7  b 11  b 15
0
b 8  b 12  b 16
0
b 9  b 13  b 17
0
b 6  b 15
…………..(6 – a)
0
b 7  b 10  b 16  0
b 8  b 11  b 17  0
b 9  b 12
0
b 13  b 14
0
b 6  b 13  b 16
0
b 7  b 17
0
b 8  b 10
0
b 9  b 11  b 14
0
b 12  b 15
0
…………..(6 – b)
…………..(6 – c)
b 6  b 12  b 16  0
b 7  b 13
0
b 8  b 14
0
b 9  b 10  b 15
0
b 11  b 16
0
b 6  b 11
0
b 7  b 12  b 14
0
b 8  b 13  b 15
0
b 9  b 16
0
b 10  b 17
0
Al-Manarah, Vol. 13, No. 6, 2007 .
…………..(6 – d)
…………..(6 – e)
185
Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
When comparing the left side of the first equation in (6 – b) with the left side
of the first equation in (6 – e) we obtain
b11  b15
In the same manner we get the following relations
b 10  b 14
b 12  b 16
b 13  b 17
We substitute now these values in the system (6) obtaining the system
b 6  2b 14
0
b 7  2b 15
0
b 8  2b 16
0
b 9  2b 17
0
……………(7)
If we compare the first equation in (7) with the first equation in (6 – b), we get
b 15  2b 14 . If we compare the second equation in (7) with the second equation
in (6 – c) and so on, we obtain the following relations
b 14  2b 16 , b 16  2b 17 , b 17  2b 15
We can then rewrite them as the linear system
 2b 14  b 15  0
 2b 15  b 17  0
 2b 16  b 14  0
 2b 17  b 16  0
The matrix of coefficients of this system is
  2 1 0 0


 0  2 0 1
 1 0  2 0


0
0
1

2


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Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
which is strictly diagonally dominant. Thus, it is invertible (cf. [1]) and, hence,
the last linear system has the trivial solution, only. Due to the previous relations
between the variables b1 , b 2 ,..., b17 we conclude that all of them are zero.
Setting all variables a 1 , a 2 ,..., a 25 to
S
represents a solution of the
5
nonhomogenous linear system (2). Since we have 25 variables and 20 equations,
the solution of the nonhomogenous linear system (2) has according to our analysis
25 – 20 + 3 = 8 free parameters (cf. [2]).
In order to prove this result in general, we follow the same steps as in the case
of 5x5 square. We have now to study the equation of linear dependence between
the columns of the transpose matrix of (3) in general:
b1 * C1  b 2 * C 2  ...  b 4n -3 * C 4n -3  0
where
C1 , C 2 ,..., C 4n - 3
denotes
the
columns
of
the
matrix
and
b1 , b 2 ,..., b 4n - 3 are real numbers. The componentwise form of this equation is:
b1  b n 1  b 2n  b 3n -1
b1  b n  2  b 2n 1  b 3n
....
0
0
b1  b 2n 1  b 3n - 2  b 4n - 3  0
b1
0
.………..(8 – 1)
b 2  b n 1
 b 3n
0
b 2  b n  2  b 2n  b 3n 1  0
........
b 2  b 2n -1  b 3n -3
0
b2
 b 3n -2  b 3n -1  0
.………..(8 – 2)
…
…
…
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Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
b n  b n 1  b 2n 1
0
b n  b n  2  b 2n  2  b 3n -1  0
....
b n  b 2n -1
bn
 b 4n - 4
0
 b 2n  b 4n - 3
0
..…………..(8 – n)
We conclude from the last equation of the group (8 – 1) that b1  0 .
Substituting this value in the other equations of the group (8 – 1) and adding up
leads to:
4n - 3
bj  0
j  n 1
Since the summation of the equations in (8 – 2) yields
nb 2 
4n - 3
bj  0
j  n 1
we deduce that b 2  0 . In the same manner we obtain:
b 3  b 4  ....  b n  0
Using our knowledge about b1 , b 2 ,..., b n we are capable of rewriting the
system (8) like this:
b n 1  b 2n  b 3n-1
0
b n  2  b 2n1  b 3n
0
..…………..(9 – 1)
....
b 2n1  b 3n-2  b 4n-3  0
b n 1
 b 3n
0
b n  2  b 2n  b 3n1  0
........
b 2n-1  b 3n- 3
..…………..(9 – 2)
0
b 3n- 2  b 3n-1  0
…
…
…
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Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
b n 1  b 2n1
0
b n  2  b 2n 2  b 3n-1
0
..………….. (9 – n)
....
b 2n-1
 b 4n-4
0
 b 4n-3
0
b 2n
When comparing the left side of the first equation in (9 – 2) with the left side
of the first equation in (9 – n) we obtain
b 2n 1  b 3n
In the same manner we get the following system
b 2n  b 3n 1
b 2n 1  b 3n
......
......
b 3n  2  b 4n 3
We substitute now these values in the system (9) obtaining the system
b n 1  2b 3n -1  0
b n  2  2b 3n
0
....
b 2n -1  2b 4n - 3  0 ..………….. (10)
If we compare the first equation in (10) with the first equation in (9 – 2), we
get b 3n  2b 3n -1 . If we compare the second equation in (10) with the second
equation in (9 – 3) and so on, we obtain the following system
Al-Manarah, Vol. 13, No. 6, 2007 .
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Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
 2b 3n 1  b 3n
 2b 3n
0
 b 3n  2  0
......
 2b 7n  5  b 4n  3  0
2
 2b 7n  3  b 3n 1  0
2
 2b 7n 1  b 3n 1  0
2
......
 2b 4n  3  b 4n  4  0
The matrix of coefficients of this system is
0 0 0 
 2 1 0 0 0


0 0 0 
 0 2 0 1 0
 .....



0 0
0 0  0 - 2 0 0 0  0 0 1
 0
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 


 1 0 0
0 0  0 0  2 0 0  0 0 0


 0 0 1 0 0  0 0 0  2 0  0 0 0
 .......



0
0 0
0 1  2 
 0 0
……(11)
which is strictly diagonally dominant. Thus, it is invertible (cf. [1]) and, hence,
the last linear system has the trivial solution, only. Due to the previous relations
between the variables b1 , b 2 ,..., b 4n - 3 we conclude that all of them are zero.
Setting all variables a1 , a 2 ,..., a n 2 to
S
represents a solution of the
n
nonhomogenous linear system (1). Since we have n*n variables and 4*n
equations, the solution of the nonhomogenous linear system (1) has according to
our analysis n*n – 4*n + 3 free parameters (cf. [2]).
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3. The case of double-even pandiagonal magic square:
We consider the linear system (1 – a), (1 – b), (1 – c) and (1 – d), where n is
of the form 2k (k = 2, 4, …). In this system there are four equations, which are
linearly dependent on the other equations. These equations are the same three
equations as in the odd squares beside the (n-1)th equation of the group (1 – d),
which is the result of subtraction of the sum of all other odd-ranked equations of
the group (1 – d) from the sum of all odd-ranked equations of the group (1 – a).
Since k = 2 yields a very special case, we start illustrating it: As in the case of
odd squares, we consider the matrix of coefficients of the system after removing
the previous mentioned equations. This matrix will be the same as the matrix (3)
after deleting the last row. We then consider the transpose of this matrix, which
has now one column less. We prove that this matrix has full rank by studying the
equation of linear dependence between the columns of the matrix:
b1 * C1  b 2 * C2  ...  b12 * C12  0
where C1 , C2 ,..., C12 denotes the columns of the matrix and b1 , b 2 ,..., b12
are real numbers. The component wise form of the last equation is:
b 1  b 5  b 8  b 11  0
b 1  b 6  b 9  b 12  0
b 1  b 7  b 10
0
b1
0
……………..(12 – 1)
b 2  b 5  b 12  0
b2  b6  b8
0
b2  b7  b9
0
.……………..(12 – 2)
b 2  b 10  b 11  0
b 3  b 5  b 10
0
b3  b6
0
b 3  b 7  b 8  b 11  0
b 3  b 9  b 12
.……………..(12 – 3)
0
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b4  b5  b9
0
b 4  b 6  b 10  b 11  0
b 4  b 7  b 12
0
b4  b8
0
…….……....(12 – 4)
We conclude from the last equation of the group (12 – 1) that b1  0 .
Substituting this value in the other equations of the group (12 – 1) and adding up
leads to:
12
b
j 5
j
0
….........……….(13)
Since the summation of the equations in (12 – 2) yields
12
4b 2   b j  0
j5
we deduce from (13) that b 2  0 . In the same manner we obtain:
b3  b4  0 .
Using our knowledge about b 1 , b 2 , b 3 and b 4 we are capable of rewriting
the system (12) like this:
b 5  b 8  b 11
0
b 6  b 9  b 12
0
b 7  b 10
0
b 5  b 12
0
b6  b8
0
b7  b9
0
….……………..(14 – 1)
….……………..(14 – 2)
b 10  b 11  0
b 5  b 10
0
b6
0
b 7  b 8  b 11  0
b 9  b 12
….……………..(14 – 3)
0
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b5  b9
0
b 6  b 10  b 11  0
b 7  b 12
0
b8
0
….……………..(14 – 4)
From the group (14 – 3) and the group (14 – 4 ) we conclude that b 6  0 and
b 8  0 . When we substitute these values in the system (14) and do some
comparisons, e. g. when equating the left side of the first equation in each of the
groups (14 – 1), …, (14 – 4) we obtain:
b 9  b 10  b 11  b 12
………………....(15)
Doing the same with the second equation we get:
b 9  b 12  b 10  b 11  0
………………..(16)
We rewrite the equations in (16) in the following manner:
b 12  b 9 , b 10  b 11
Using these relations, we obtain from (15) the equation  2b 9  0 . Thus,
b 9  0 . Due to the previous relations between the variables b1 , b 2 ,..., b12 we
conclude that all of them are zero.
Setting all variables a 1 , a 2 ,..., a 16 to
S
represents a solution of the
4
nonhomogenous linear system (1). Since we have 16 variables and 16 equations,
the solution of the nonhomogenous linear system (1) has according to our analysis
16 – 16 + 4 = 4 free parameters (cf. [2]).
Now, we treat the general case. As in the case of odd squares we consider the
matrix of coefficients of the system after removing the previous mentioned
dependent equations. This matrix will be the same as the matrix (3) after deleting
the last row. We then consider the transpose of this matrix, which has now one
column less. We prove that this matrix has full rank by studying the equation of
linear dependence between the columns of the matrix:
b1 * C1  b 2 * C2  ...  b 4n - 4 * C4n - 4  0 .……………(17)
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where
C1 , C2 ,..., C4n - 4 denotes the columns of the matrix and
b1, b 2 ,..., b 4n - 4 are real numbers. The componentwise form of equation (17) is:
b1  b n 1  b 2n  b 3n -1
b1  b n  2  b 2n 1  b 3n
0
0
....
b1  b 2n - 2  b 3n - 3  b 4n - 4  0
b1  b 2n -1  b 3n - 2
0
0
b1
b 2  b n 1
 b 3n
b 2  b n  2  b 2n
....
 b 3n 1  0
0
 b 3n - 2  b 3n -1  0
b2
……………..(18 – 1)
.…………..…..(18 – 2)
…
…
…
b n  b n 1  b 2n 1
0
b n  b n  2  b 2n  2  b 3n -1  0
....
 b 2n
bn
0
…….…………..(18 – n)
We conclude from the last equation of the group (18 – 1) that b1  0 .
Substituting this value in the other equations of the group (18 – 1) and adding up
leads to:
4n - 4
bj  0
j  n 1
……...………….(19)
Since the summation of the equations in (18 – 2) yields
nb 2 
4n - 4
bj  0
j  n 1
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Hence, we deduce from (19) that b 2  0 . In the same manner we obtain:
b 3  b 4  ....  b n  0 .
Using our knowledge about b1 , b 2 ,..., b n we are capable of rewriting the
system (18) like this:
b n 1  b 2n  b 3n -1
b n  2  b 2n 1  b 3n
0
0
....
b 2n - 2  b 3n - 3  b 4n - 4  0
b 2n -1  b 3n - 2
0
b n 1
 b 3n
b n  2  b 2n
....
 b 3n 1  0
….……………..(20 – 1)
0
b 3n - 2  b 3n -1  0
...….…………..(20 – 2)
…
…
…
b n 1  b 2n 1
0
b n  2  b 2n  2  b 3n -1  0
....
b 2n
0
From the group (20 
…….…………..(20 – n)
n
) and the group (20 – n) we conclude that b 3  0
2
n
2
and b 2n  0 . When we substitute these values in the system (20) and do some
comparisons, e. g. when equating the left side of the first equation in each of the
groups (20 – 1), …, (20 – n) we obtain:
b 2n 1  b 2n  2  b 2n  3  b 4n - 4  b 2n  4  b 4n - 5
 ....  b 3n - 2  b 3n 1  b 3n  b 3n -1
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………………..(21)
Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
Doing the same with the second equation, we get:
b 2n 1  b 3n  b 2n  2  b 3n -1  b 2n  3  b 2n  4  b 2n  5  b 4n - 4
 b 2n  6  b 4n - 5  ....  b 3n - 2  b 3n  3  b 3n  2  b 3n 1
Continuing these comparisons till, we reach the
….(22)
n
th equation in each of the
2
groups (20 – 1), …, (20 – n) we get:
b 2n 1  b 4n - 4  b 2n  2  b 4n - 5  ...  b 3n - 3  b 3n -1  b 2n  0
…...(23)
Remember that in the case k = 2 this equation will be also the second
equation, and that we have to deal with two equations, only. We rewrite the
equations in (23) in the following manner:
b 3n -1   b 3n - 2
  b 3n - 3
b 3n
....
....
b 4n - 4   b 2n 1
………...(24)
Using these relations, we obtain from the equations in (21), (22) and similar
equations the following linear system:
 2b 2n 1  b 2n 3  0
 2b 2n 3  b 2n  7  0
......
 2b 5
2
 2b 5
2
 2b 5
2
 2b 5
2
n 3
n 1
n 1
n 3
 b 3n-5
0
0
 b 2n 3  0
 b 2n  7  0
......
 2b 3n-3  b 3n-5  0 ………………...(25)
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The matrix of coefficients of this system is
1
0
0
0
0
0 0 
 2


-2
0
1
0
0
0
0 
 0
 ......



0
0
0
0  0 2 0
0 0 0  0
1
0
 0
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 


 0
0
0
0
0  0 0 -2
0 0 0  0
0
0


- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
 0
1
0
0
0  0 0 0 -2 0 0  0
0
0


0
0
1
0  0 0
0
0 -2 0  0
0
0
 0
 ......



 0
0
0
0
0
0
1  2 

which is strictly diagonally dominant. Thus, it is invertible and, hence, the last
linear system has the trivial solution, only. Due to the previous relations between
the variables b1, b 2 ,..., b 4n - 4 we conclude that all of them are zero.
Setting all variables a1 , a 2 ,..., a n 2 to
S
represents a solution of the
n
nonhomogenous linear system (1). Since we have n*n variables and 4*n
equations, the solution of the nonhomogenous linear system (1) has according to
our analysis n*n – 4*n + 4 free parameters (cf. [2]).
4. The case of simple even pandiagonal magic square:
We consider the linear system (1), where n is of the form 2k (k = 3, 5, …):
Following the same steps as in the case of double-even squares we reach the same
conclusion about the values of b1 , b 2 ,..., b n . Further, we obtain the system (20)
again for their values, from which we deduce the fact that b 3
2
n
 0 and
b 2n  0 . Now, by comparing the left side of the equations in the groups (20 – 1),
(20 – 2), …, (20 – n) we get the same relations (21), (22) and other similar
relations. But, the comparisons of the
n
th equation in each group yields now:
2
b 2n1  b 4n-4  b 2n 2  b 4n-5  ...  b 3n-2  b 3n-1  0
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……..……....(26)
Linear systems resulting from……......….………..Saleem Al-Ashhab & Alla Al-Zahawi
We rearrange the equations in (26) as follows:
b 3n-1   b 3n-2
  b 3n-3
b 3n
....
....
b 4n-4   b 2n 1
We substitute these relations in the sets of equations (21), (22) and the similar
sets of equations. This will generate a different linear system than in the case of
double-even squares. In fact, we obtain now the following linear system:
 2b 2n 1  b 2n 3  0
 2b 2n 3  b 2n  7  0
......
 2b 5
2
 2b 5
2
 2b 5
2
n 2
n
n2
 b 3n-3
0
 b 2n 1  0
 b 2n 5  0
......
 2b 3n-3  b 3n-5  0
When we write the coefficients of this system in the form of a matrix, we
obtain the matrix (11). This is a strictly diagonally dominant matrix. Like in the
first case we are so done with the proof.
Notice: This paper is extracted from the Master thesis “A Study on Pandiagonal
Magic Squares” by the student Alla Al - zahawi and supervisor Saleem Alashhab, Al-albayt University, May 2004.
Al-Manarah, Vol. 13, No. 6, 2007 .
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References:
[1] W.S. Andrews, Magic Squares and Cubes, 1908.
[2] W.W. Rouse Ball, Mathematical Recreations and Essays, 1911.
[3] Anton H., Elementary Linear Algebra, 7th edition, Jon Wiley, New York,
1994.
[4] Horen, R. and Johnson, C., Matrix Analysis, 9th edition, Cambridge University
Press, New York, 1996.
[5] S. Al-ashhab, The Magic and Semi Magic 4x4 Squares Problem, Dirasat;
October 1998,Volume 25, No. 3, pp 445-450.
[6] S. Al-ashhab, The book “Theory of Magic Squares / Mathematically and
Programming” in Arabic, 2000.
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