Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
Received 11/02/17
Magic Squares with
Perfect Square Number Sums
Inder J. Taneja(1)
Abstract
This short paper shows how to create magic squares in such a way that total
sum of their numbers becomes a perfect square. This has been done in two ways:
Firstly, Take the sum of odd numbers, and secondly, take the numbers in a
sequential way. In the first case, for all orders of magic squares, one can always
have a perfect square sum. In the second case, magic squares with perfect square
magic sums exist, but only for odd order magic squares. For the even order magic
squares, such as 4, 6, 8, etc. it is not possible to write sequential number magic
squares with perfect square sums. For the odd order magic squares, at least three
examples of sequential numbers are given in each case. This is done from the 3rd to
the 25th orders of magic squares. For the prime number orders, examples are
considered up to order 11. Further results for order 13, 17, etc. are presented
along similar lines. Finally, we reach a conclusion that we can always
create a magic square, so that, if the order of magic square is k, then the
number of elements are k2, the magic sum is k3, and the sum of all numbers
on the square is k4, for all k=3,4,5,…
An equation means nothing to me unless it expresses a thought of God.
- S. Ramanujan.
______________________
(1) Formerly, Professor of Mathematics, Universidade Federal de Santa Catarina, 88.040-900 Florianópolis, SC,
Brazil. E-mail: [email protected]; Twitter: @IJTANEJA; Web-site: https://numbersmagic.wordpress.com
RGMIA Res. Rep. Coll. 20 (2017), Art. 11, 34 pp.
1
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
1. Introduction
Recently, William Walkington started an interesting discussion as to how to create magic
squares with cells that had the same areas as their numbers. Below is a graphic design for a 2017
seasonal greetings card, showing a magic square with approximate areas that was constructed
by William Walkington (2016):
Figure 1
Lee Sallows (2017) also constructed another magic square representing the areas as rectangles.
See below:
Figure 2
The sum of all the numbers is given by
2
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
1  2  3  4  5  6  7  8  9  45 .
(1)
The number 45 is not a perfect square. If we make a slight change, then we can transform the
sum into a perfect square:
5  6  7  8  9  10  11  12  13  81  92
(2)
Using this number sequence, Walter Trump (2017) was able to construct the following area
magic square:
Figure 3
Adding 4 to each number in (1), we obtain the numbers in (3). Observing area-wise the figures 1
and 3, there is a considerable difference: For example, from numbers 1 to 2, the cell area is
doubled, while from numbers 5 to 6, there is proportionally less increase between the cell areas.
According to Lee Sallows’s approach, the area-wise magic square for equation (2) is as follows:
Lee Sallows-Style
Figure 4
3
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
In order to construct a magic square with cell areas that are in proportion to their numbers it is
not necessary that the numbers always sum to a perfect square. Below is another example
constructed by William Walkington (2017) with sequential numbers from 3 to 11:
Figure 5
We observe that in Figures 1, 3 and 5, the number cell areas are proportional and aligned in
both directions. In Figures 2 and 4, the proportionality of the areas is only present in one
direction, which is horizontal. In figure 4 we cannot have proportionality in both directions,
because the magic square sum is S33 : 27 , whilst the sum of numbers is 81  27  3 (and not
27  27 ).
Below are two examples of classical order 4 magic squares with cell areas that are proportional
to their numbers:
Figure 7
Figure 6
4
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
More examples of similar kinds of order 4 area magic squares, together with order 6 area magic
squares, can be seen in William Walkington’s pages [8].
From equation (2), the question arises, how to create higher order magic squares such that the
sum of numbers are always a perfect square. Based on this idea, in this paper we establish
some formulas, so that sum of the numbers in magic squares always becomes a perfect square.
2. Series and Magic Square Sums
This section brings some basic idea of series and magic square sums.
2.1
Sum of Natural Numbers Series
It is well-known that the positive natural number series sum is given by
T(n) := 1 + 2 + 3 + ... + n =
n(n + 1)
, n  1.
2
(3)
The sequence T(n) , is also famous Pascal’s triangle values.
2.2
Sum of Odd Numbers Series
It is well-known that sum of odd-number series is given by
Fn := 1+ 3 + 5 + ... +(2n  1) = n2 , n  1.
(4)
Since we are working with magic squares, let us write, n = k 2 , then we have
Fk2 := 1+ 3 + 5 + ... + (2k 2  1) = k 4 , k  1.
(5)
The number k is the order of a magic square. Thus for all k  3 , we can always have a odd
numbers magic square with sum of all numbers a perfect square.
2.3
Sequential Perfect Square Sum
Below is a general formula to write sequential magic squares.
Let us consider
 k -1 
K := T(n) - T(n - k) = k  n , n > k .
2 

5
(6)
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
a) For even order magic squares, k  2 p :
2p  1 

K := k  n 
.
2 

(7)
2 p 1
is never a natural number. Thus, there is no sequential
2
magic square with sum of numbers as a perfect square for even order magic squares.
In this case, the expression n 
b) For odd order magic squares, k  2 p  1 :
2p +1  1 

K := T(n) - T(n - k) := k  n 
 = k n  p  .
2


(8)
In this case we can always find a natural number, such that n  p a perfect square with
n  p  k . The minimum value is n  p  k . This gives K := k 2 and the sum of all numbers is
Fk 2 := k 4 .
2.4
Magic Square Sum and Sum of Numbers
It is well known that magic sum of a magic square of order k has total elements, 1, 2, 3, ..., k 2 is
given by
k  k 2 + 1
.
(9)
Sk×k :=
2
Sum of all members is given by
Fk 2 :=
k 2  k 2 + 1
2
.
(10)
Thus, from (5) and (8), we conclude that at least there are two ways of writing magic squares
with sum of all members a perfect square:
(i) Magic squares formed by odd order numbers;
(ii) For magic square of odd orders, such as, 3, 5, 7,…, one can always find sequential numbers
such that the sum of all numbers is a perfect square.
Based on this idea, sections below give examples magic squares with numbers sum a perfect
square.
6
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
3. Magic Square of Order 3
According to (5) and (8), one can have a perfect square sum magic square of order 3 in two
ways. One using odd order numbers and secondly having sequential values. Below are
examples of both types.
3.1
First Approach – Odd Numbers
Take k  3 in (5), we get
1  3  5  ...  (2  32  1)  34 ,

1 + 3 + 5 + 7 + 9 +11 +13 +15 +17 = 92 = 81.
(11)
According to numbers given in (11), the magic square of order 3 is given by
27
3 13 11 27
17 9 1 27
7 5 15 27
27 27 27 27
Example 1
3.2
Second Approach – Sequential
According to (8),
K :
n(n  1) (n  9)(n  8)

 9(n  4) .
2
2
Take n  13 , we get a perfect square, i.e.,
K : T (13)  T (4) 
13 14 4  5

 9  9  81 .
2
2
Simplifying, one get
5 + 6 + 7 + 8 + 9 +10 +11+12 +13 = 81.
(12)
Again, there is a perfect square for the magic square of order 3 Take the nine numbers in a
sequence from 5 to 13. In this case the magic square of order 3 is given by
27
6 11 10 27
13 9 5 27
8 7 12 27
27 27 27 27
Example 2
In both the examples given above the magic sum is S3×3 := 27 = 33 and sum of all numbers is
F9 := 81 = 34 .
7
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
We observe that the minimum number bigger than 9, and giving n  4 as a perfect square is
n  13 . Thus we can say that the sum of numbers starting from 5 to 13 is a perfect square sum
with least possible value. Still, there are more sequences with the same property. See examples
below:
T(20)  T(11) = 12 + 13 + ... + 20 = 144 = 122 ;
T(29)  T(20) = 21 + 22 + ... + 29 = 225  15 2 ;
T(40)  T(31) = 32 + 33 + ... + 40 = 324  18 2 .
4. Magic Square of Order 4
According to (8), we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 4. Take k  4 in (5), we get
1  3  5  ...  (2  42  1)  44 .

1 + 3 + 5 + 7 + 9 +11 +13 +15 +17 +
+19 + 21 + 23 + 25 + 27 + 29 + 31 = 256 = 162.
(13)
According to values given in (13), the pan diagonal magic square of order 4 is given by
64
64
64
64
64 64 64
13 23 1 27
3 25 15 21
31 5 19 9
17 11 29 7
64 64 64 64
Example 3
64
64
64
64
64
64
In this case, the magic sum is S4×4 := 64 = 43 , and sum of numbers is F16 := 256 = 162 = 44 .
5. Magic Square of Order 5
According to (5) and (8), one can have a perfect square sum magic square of order 5 in two
ways. One using odd order numbers and secondly having sequential values. Below are
examples of both type.
5.1
First Approach – Odd Numbers
Take k  5 in (5), we get
1  3  5  ...  (2  52  1)  54 .

1 + 3 + 5 + 7 + 9 +11 +13 +15 +17 +19 + 21 + 23 + 25 +
+27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 = 25 2 = 625.
8
(14)
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
According to 25 values given in (14), the pan diagonal magic square of order 5 is given by
125
125
125
125
5.2
125 125 125 125 125
1
17 23 39 45 125
33 49
5
11 27 125
15 21 37 43
9 125
47
3
19 25 31 125
29 35 41
7
13 125
125 125 125 125 125 125
Example 4
Second Approach – Sequential
According to (8),
K :
n(n  1) (n  25)(n  24)

 25(n  12) .
2
2
Take, n  37 , we get a perfect square, i.e.,
37  38 12 13

 37 19  6 13
2
2
 703  78  625  252  54.
K : T (37)  T (12) 
Simplifying, one get
13 + 14 + ... + 37 = 54 = 252 = 625 .
(15)
This gives a perfect square sum for 25 numbers in a sequential way from 13 to 37. In this case
the pan diagonal magic square of order 5 is given by
125
125
125
125
125 125 125 125 125
13 21 24 32 35 125
29 37 15 18 26 125
20 23 31 34 17 125
36 14 22 25 28 125
27 30 33 16 19 125
125 125 125 125 125 125
Example 5
In both the examples given above the magic sum is S5×5 := 125 = 53 and sum of all numbers is
F25 := 625 = 252 = 54 .
Above magic square is for minimal number of sequential values. Still, there are more
possibilities of getting sequential numbers for the magic square of order 5, where magic sum is
a perfect square. See below more examples,
T(48)  T(23) = 24 + 25 + ... + 48 = 900 = 302 ;
T(61)  T(36) = 37 + 38 + ... + 61 = 1225 = 352 ;
T(76)  T(51) = 52 + 53 + ... + 76 = 1600 = 40 2
9
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
6. Magic Square of Order 6
According to (7) we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 6. Take k  6 in (5), we get
1  3  5  ...  (2  62  1)  64

1 + 3 + 5 + 7 + ... + 67 + 69 + 71 = 36 2 = 1296.
(16)
Thus, there is a perfect square sum for the magic square of order 6 of odd numbers. According
to values given in (16), the magic square of order 6 is given by
1
45 55 67 33 15
57 13 69 27 41
9
23 11 25 53 61 43
63 31
7
47 19 49
37 65 21
5
59 29
35 51 39 17
3
71
216 216 216 216 216 216
Example 6
216
216
216
216
216
216
216
216
In this case, the magic sum is S6×6 := 216 = 63 , and sum of number is F36 := 1296 = 362 = 64 .
7. Magic Square of Order 7
According to (5) and (8), one can have a perfect square sum magic square of order 7 in two
ways. One using odd order numbers and secondly having sequential values. Below are
examples of both type.
7.1
First Approach – Odd Numbers
Take k  7 in (5), we get
1  3  5  ...  (2  7 2  1)  7 4 ,

1 + 3 + 5 + 7 + ... + 93 + 95 + 97 = 49 2 = 2401.
According to values given in (17), a pan diagonal magic square of order 7 is given by
343
343
343
343
343
343
1
79
45
25
89
69
35
343
343
17
95
61
41
7
71
51
343
343
33
13
77
43
23
87
67
343
343
49
15
93
59
39
5
83
343
343
65
31
11
75
55
21
85
343
343
81
47
27
91
57
37
3
343
343
97
63
29
9
73
53
19
343
343
343
343
343
343
343
343
343
343
Example 7
In this case, the magic sum is S7×7 := 343 = 73 , and sum of all number is F49 := 2401 = 74 .
10
(17)
Inder J. Taneja
7.2
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
Second Approach – Sequential
According to (8),
K :
n(n  1) (n  49)(n  48)

 49  n  24 
2
2
Take n  73 , we get a perfect square sum, i.e.,
73  74 24  25

 73  37  12  25
2
2
 2701  300  2401  492  7 4.
K : T (73)  T (24) 
Simplifying, one get
25 + 26 + ... + 72 + 73 = 74 = 492 = 2401 .
(18)
This gives a perfect square sum for 49 values put in a sequence starting from 25. According to
values given in (18), a pan diagonal magic square of order 7 is given by
343
343
343
343
343
343
343 343 343 343 343 343
25 33 41 49 57 65 73
64 72 31 32 40 48 56
47 55 63 71 30 38 39
37 45 46 54 62 70 29
69 28 36 44 52 53 61
59 60 68 27 35 43 51
42 50 58 66 67 26 34
343 343 343 343 343 343 343
Example 8
343
343
343
343
343
343
343
343
343
Still, there are more possibilities of getting sequential numbers magic square of order 7 with
perfect square sum. See below examples,
T(88)  T(39) = 40 + 41 + ... + 88 = 3136 = 56 2 ;
T(105)  T(56) = 57 + 88 + ... + 105 = 3969 = 63 2 ;
T(124)  T(79) = 80 + 81 + ... + 124 = 4900 = 70 2 .
In this case, also the magic sum is S 7×7 := 343 , and sum of number is F49 := 2401 = 492 = 74 .
8. Magic Square of Order 8
According to (7) we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 8. Take k  8 in (5), we get
1  3  5  ...  (2  82  1)  84

1 + 3 + 5 + 7 + ... + 123 + 125 + 127 = 642 = 4096.
11
(19)
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
According to values given in (19), the magic square of order 8 is given by
512
512
512
512
512
512
512
512 512 512 512 512 512
31 81 71
9
53 123
51 125 107 37 25 87
1
79 89 23 43 101
45 99 117 59
7
73
75
5
19 93 97 47
103 41 63 113 77
3
85 27 13 67 127 49
121 55 33 111 83 29
512 512 512 512 512 512
Example 9
512
109
65
115
95
57
21
39
11
512
512
35
15
61
17
119
91
105
69
512
512
512
512
512
512
512
512
512
512
512
In this case the magic sum is S 8×8 := 512 , and sum of number is F64 := 4096 = 642  84 . The above
magic square is also a bimagic with bimagic sum Sb 8×8 := 43688 .
9. Magic Square of Order 9
According to (5) and (8), one can have a perfect square sum magic squares of order 9 in two
ways. One, using odd order numbers, and second with sequential values.
9.1
First Approach – Odd Numbers
Take k  9 in (5), we get
1  3  5  ...  (2  92  1)  94 ,

1 + 3 + 5 + 7 + ... + 157 + 159 + 161 = 812 = 6561.
(20)
According to values given in (20), the magic square of order 9 is given by
1
65
123
53
93
151
27
85
131
729
35
75
133
9
67
113
37
101
159
729
45
103
149
19
83
141
17
57
115
729
69
109
11
97
161
39
77
135
31
729
79 95 119
143 153 15
21 49 55
59 87 147
117 127 43
13 29 107
105 61 139
145 125 23
47
3
81
729 729 729
Example 10
129
25
89
121
5
63
155
51
91
729
157
41
99
137
33
73
111
7
71
729
729
729
729
729
729
729
729
729
729
729
729
In this case the magic sum is S9×9 := 729 = 93 , and sum of numbers is F81 := 6561 = 812 = 94 . It is
also a bimagic square with bimagic sum is Sb 9×9 := 78729 .
12
Inder J. Taneja
9.2
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
Second Approach – Sequential
According to (8),
K : T (n)  T (n  81),
i.e.,
K :
n  81 ,
n(n  1) (n  81)(n  80)

 81n  3240  81 n  40  .
2
2
Take n  121 , we get a perfect square, i.e.,
121122 40  41

2
2
 121 61  20  41
K : T (121)  T (40) 
 7381  820  6561  812  94.
Simplifying, one get
41 + 42 + ... + 120 + 121 = 94 = 812 = 6561 .
(21)
Thus, we have a perfect square sum for the 81 numbers starting from 41. According to values
given in (21), the magic square of order 9 is given by
41 58 63 75 80 88 100
73 78 92 95 112 117 48
102 107 115 46 51 65 68
67 45 50 89 70 84 114
87 74 82 121 99 104 62
116 97 111 60 47 55 94
54 59 49 79 93 71 110
83 91 69 108 113 103 52
106 120 98 56 64 42 81
729 729 729 729 729 729 729
Example 11
729
105 119 729
53 61 729
85 90 729
101 109 729
43 57 729
72 77 729
118 96 729
66 44 729
86 76 729
729 729 729
In this case also the magic sum is S9×9 := 729  93 , and sum of number is F81 := 6561 = 812 = 94 .
This magic square is also a bimagic with bimagic sum is Sb 9×9 := 63969 .
Still, there are more possibilities of getting sequential numbers magic squares of order 9, where
numbers sum is a perfect square. See below examples,
T(140)  T(59) = 60 + 61 + ... + 140 = 8100 = 90 2 ;
T(161)  T(81) = 82 + 88 + ... + 161 = 9801 = 99 2 ;
T(184)  T(103) = 104 + 105 + ... + 184 = 11644 = 108 2 .
13
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
10. Magic Square of Order 10
According to (8) we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 10. Take n  100 in (5) or k  10 in (6), we
get
1  3  5  ...  (2 10 2  1)  10 4

1 + 3 + 5 + 7 + ... + 197 + 198 + 199 = 100 2 = 10000.
(22)
According to values given in (22), the magic square of order 10 is given by
1
97
31
183 175
47
149
73
119 125
71
23
135 165
59
197
81
107
9
153
123 115
45
99
191
13
77
161 147
29
39
173 109
67
141
85
195
3
131
57
187
41
17
113
89
143 171 139
25
75
169 127
61
157
33
111
19
185
55
83
95
5
159
51
103
69
133
37
181 167
53
189
87
21
137 179 105 155
63
11
145
79
193 129
7
35
43
91
177 101
117 151 163
15
65
121
27
49
93
199
1000 1000 1000 1000 1000 1000 1000 1000 1000 1000
Example 12
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
In this case the magic sum is S10×10 := 1000 = 103 , and sum of numbers is F100 := 10000 = 1002 = 104 .
11. Magic Square of Order 11
According to (5) and (8), one can have a perfect square sum magic square of order 11 in two
ways. One, using odd order numbers, and second with sequential values.
11.1 First Approach – Odd Numbers
Take k  11 in (5), we get
1  3  5  ...  (2 112  1)  114 ,

1 + 3 + 5 + 7 + ... + 239 + 241 = 1212 = 14641.
According to values given in (23), the magic square of order 11 is given by
14
(23)
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331 1331 1331 1331 1331 1331 1331 1331 1331 1331
1
41
59
77
95
113 153 171 189 207 225
201 241
17
35
53
71
89
129 147 165 183
159 177 217 235
11
29
47
87
105 123 141
117 135 175 193 211 229
5
23
63
81
99
75
93
111 151 169 187 205 223
21
39
57
33
51
69
109 127 145 163 181 199 239
15
233
9
27
45
85
103 121 139 157 197 215
191 209 227
3
43
61
79
97
115 133 173
149 167 185 203 221
19
37
55
73
91
131
107 125 143 161 179 219 237
13
31
49
67
65
83
101 119 137 155 195 213 231
7
25
1331 1331 1331 1331 1331 1331 1331 1331 1331 1331 1331
Example 13
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
In this case the magic sum is S11×11 := 1331 = 113 , and sum of numbers is F121 := 14641 = 1212 = 114 .
In this case the magic square is pan diagonal.
11.2 Second Approach – Sequential
According to (8), we have
K : T (n)  T (n  121),
i.e.,
K :
n  121 ,
n(n  1) (n  121)(n  120)

 121n  7260  121 n  60  .
2
2
Take n  181 , we get a perfect square sum, i.e.,
181182 60  61

2
2
 181 91  30  61
K : T (181)  T (60) 
 16471  1830  14641  1212  114.
Simplifying,
61 + 62 + ... + 180 + 181 = 114 = 1212 = 14641 .
(24)
Thus, we have a perfect square sum for the 121 numbers starting from 61. According to values
given in (24), the magic square of order 11 is given by
15
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331 1331 1331 1331 1331 1331 1331 1331 1331 1331
61
81
90
99
108 117 137 146 155 164 173
161 181
69
78
87
96
105 125 134 143 152
140 149 169 178
66
75
84
104 113 122 131
119 128 148 157 166 175
63
72
92
101 110
98
107 116 136 145 154 163 172
71
80
89
77
86
95
115 124 133 142 151 160 180
68
177
65
74
83
103 112 121 130 139 159 168
156 165 174
62
82
91
100 109 118 127 147
135 144 153 162 171
70
79
88
97
106 126
114 123 132 141 150 170 179
67
76
85
94
93
102 111 120 129 138 158 167 176
64
73
1331 1331 1331 1331 1331 1331 1331 1331 1331 1331 1331
Example 14
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
1331
In both the examples 13 and 14, the magic sum is S11×11 := 1331 = 113 , and the sum of numbers is
F121 := 14641 = 1212 = 114 . Still, there are more possibilities of getting sequential number magic
squares for the magic square of order 11, where the numbers sum is a perfect square. See below
examples,
T(204)  T(83) = 84 + 85 + ... + 204 = 17424 = 132 2 ;
T(229)  T(108) = 109 + 110 + ... + 229 = 20449 = 143 2 ;
T(256)  T(135) = 136 + 137 + ... + 256 = 23716 = 154 2 .
12. Magic Square of Order 12
According to (8) we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 12. Take k  12 in (5), we get
1  3  5  ...  (2 12 2  1)  12 4

1 + 3 + 5 + ... + 285 + 287 = 124 = 20736 .
(25)
According to values given in (25), the magic square of order 12 is given by
135 165
3
273
89
211
53
223 109 191
25
251
9
267 141 159
55
221
91
209
35
241 119 181
285
15
153 123 235
65
199
77
263
37
179
97
147 129 279
21
197
79
233
67
169 107 253
47
85
215
49
227 111 189
27
249 137 163
5
271
59
217
95
205
33
243 117 183
7
269 139 161
239
61
203
73
261
39
177
99
283
17
151 125
193
83
229
71
171 105 255
45
149 127 281
19
113 187
29
247 133 167
1
275
87
213
51
225
31
245 115 185
11
265 143 157
57
219
93
207
259
41
175 101 287
13
155 121 237
63
201
75
173 103 257
43
145 131 277
23
195
81
231
69
1728 1728 1728 1728 1728 1728 1728 1728 1728 1728 1728 1728
Example 15
16
1728
1728
1728
1728
1728
1728
1728
1728
1728
1728
1728
1728
1728
1728
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
In this case the magic sum is S12×12 := 1728 = 123 , and sum of numbers is F144 := 20736 = 1442 = 124 .
Each block of order 4 is pan diagonal magic square with magic sum S 4×4 := 576 .
13. Magic Square of Order 14
According to (7) we don’t have sequential magic square with numbers sum a perfect square. We
will obtain only odd numbers magic square of order 14. Take k  14 in (5), we get
1  3  5  ...  (2 142  1)  144

1 + 3 + 5 + ... + 389 + 391 = 144 = 38416 .
(26)
According to values given in (26), the magic square of order 14 is given by
2744
1
189 381 223 275 155 115
63
33
317 291 355 249
97 2744
293
31
219 351 375 253 173 233
7
105 139 165
75
325 2744
251 323
61
273 343 125 159
11
93
51
193 197 297 367 2744
379 221 299
91
309 229
9
153
67
339
45
191 279 133 2744
327 167 109
23
241 171
57
201 265 123 345
49
371 295 2744
207
19
177
73
41
87
267 117 225 2744
331 357 307 389 147
37
69
263 113
5
305 391 329 359 183 243 101 143 203 2744
89
129 141 179
71
3
205
47 2744
385 303 361 335 237 259
59
85
119 145 187 363 333 387 301 213
13
235
43
261 2744
135 287 311
53
149
95
209 185 239 271 365
83
21
341 2744
161 373 349 227 137
35
269 103 169
27
211 285 319
79 2744
277 347
15
321 283 215 231
29
157 369
77
121 107 195 2744
353 257 245 383 111
65
39
127 199 281 163 315 181
25 2744
175 247
55
289 217
17
99
255 131
81
313 377 337 151 2744
2744 2744 2744 2744 2744 2744 2744 2744 2744 2744 2744 2744 2744 2744 2744
Example 16
In
this
case
the
magic
sum
is
S14×14 := 2744 = 143 ,
and
sum
of
numbers
is
F196 := 38416 = 1962 = 144 . The middle block is a pan diagonal magic square of order 4 with
magic sum S 4×4 := 1384 .
14. Magic Square of Order 15
According to (5) and (8), one can have a perfect square sum magic square of order 15 in two
ways. One, using odd order numbers, and second with sequential values.
14.1 First Approach – Odd Numbers
Take k  15 in (5), we get
1  3  5  ...  (2 152  1)  154 ,

1 + 3 + 5 + 7 + ... + 239 + 241 = 2252 = 50625.
17
(27)
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
According to values given in (27), the magic square of order 15 is given by
3375
3
339
165
381
207
61
281
103
443
265
35
307
137
409
239
171
387
183
9
345
113
445
241
71
283
139
419
215
37
317
189
15
351
177
363
251
73
293
115
421
217
47
319
149
395
357
153
369
195
21
295
91
431
253
83
329
125
397
227
49
375
201
27
333
159
433
263
85
271
101
407
229
59
305
127
65
277
107
439
269
33
309
135
411
237
1
341
163
383
205
109
449
245
67
287
141
417
213
39
315
173
385
181
11
343
247
77
289
119
425
219
45
321
147
393
191
13
353
175
361
299
95
427
257
79
327
123
399
225
51
355
151
371
193
23
437
259
89
275
97
405
231
57
303
129
373
203
25
331
161
31
311
133
413
235
5
337
167
379
209
63
279
105
441
267
143
415
211
41
313
169
389
185
7
347
111
447
243
69
285
221
43
323
145
391
187
17
349
179
365
249
75
291
117
423
325
121
401
223
53
359
155
367
197
19
297
93
429
255
81
403
233
55
301
131
377
199
29
335
157
435
261
87
273
99
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375
Example 17
In
this
case
the
magic
sum
is
S15×15 := 3375 = 153 ,
and
sum
of
numbers
is
F225 := 50625 = 2252 = 154 . In this case each block of order 5 is a pan diagonal magic square
forming a magic square of order 3:
3375
1095 1157 1123 3375
1153 1125 1097 3375
1127 1093 1155 3375
3375 3375 3375 3375
Example 18
14.2 Second Approach – Sequential
According to formula (8), we have
K : T (n)  T (n  225),
i.e.,
n  225 ,
n(n  1) (n  225)(n  224)

2
2
: 225n  25200  225  n  112  .
K :
Take n  337 , we get a perfect square, i.e.,
K : T (337)  T (112)  50625  2252  154.
Simplifying, one get
113 + 114 + ... + 336 + 337 = 154 = 2252 = 50625 .
(28)
Thus, we have a perfect square sum for the 225 numbers starting from 113. According to values
given in (28), the magic square of order 15 is given by
18
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
114
282
195
303
216
143
253
164
334
245
130
266
181
317
232
198
306
204
117
285
169
335
233
148
254
182
322
220
131
271
207
120
288
201
294
238
149
259
170
323
221
136
272
187
310
291
189
297
210
123
260
158
328
239
154
277
175
311
226
137
300
213
126
279
192
329
244
155
248
163
316
227
142
265
176
145
251
166
332
247
129
267
180
318
231
113
283
194
304
215
167
337
235
146
256
183
321
219
132
270
199
305
203
118
284
236
151
257
172
325
222
135
273
186
309
208
119
289
200
293
262
160
326
241
152
276
174
312
225
138
290
188
298
209
124
331
242
157
250
161
315
228
141
264
177
299
214
125
278
193
128
268
179
319
230
115
281
196
302
217
144
252
165
333
246
184
320
218
133
269
197
307
205
116
286
168
336
234
147
255
223
134
274
185
308
206
121
287
202
295
237
150
258
171
324
275
173
313
224
139
292
190
296
211
122
261
159
327
240
153
314
229
140
263
178
301
212
127
280
191
330
243
156
249
162
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375
3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375 3375
Example 19
In this case also the magic sum is
S15×15 := 3375 = 153 , and sum of number is
F225 := 50625 = 2252 = 154 . In this case each block of order 5 is a pan diagonal magic square
forming a magic square of order 3:
3375
1110 1141 1124 3375
1139 1125 1111 3375
1126 1109 1140 3375
3375 3375 3375 3375
Example 20
Still, there are more possibilities of getting sequential numbers for the magic square of order 15,
where numbers sum is a perfect square. See below examples,
T(368)  T(143) = 144 + 145 + ... + 368 = 57600 = 240 2 ;
T(401)  T(176) = 177 + 178 + ... + 401 = 65025 = 255 2 ;
T(436)  T(211) = 212 + 213 + ... + 436 = 72900 = 2702 .
15. Magic Square of Order 16
According to (8) we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 16. Take n  256 in (5) or k  16 in (6), we
get
1  3  5  ...  (2 162  1)  164

1 + 3 + 5 + ... + 509 + 511 = 164 = 65536 .
According to values given in (29), the magic square of order 16 is given by
19
(29)
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
1
463
243
317
91
405
169
359
109
419
159
337
55
505
197
267
307
253
449
15
361
167
411
85
351
145
429
99
261
203
503
57
477
19
303
225
391
73
373
187
433
127
323
141
491
37
281
215
239
289
29
467
181
379
71
393
131
333
113
447
217
279
43
485
45
483
223
273
119
441
133
331
65
399
179
381
27
469
233
295
287
209
493
35
325
139
439
121
371
189
385
79
297
231
475
21
497
63
259
205
427
101
345
151
413
83
367
161
455
9
309
251
195
269
49
511
153
343
107
421
175
353
93
403
245
315
7
457
87
409
165
363
13
451
255
305
59
501
201
263
97
431
147
349
357
171
407
89
319
241
461
3
265
199
507
53
339
157
417
111
395
69
377
183
465
31
291
237
487
41
277
219
445
115
335
129
185
375
75
389
227
301
17
479
213
283
39
489
143
321
125
435
123
437
137
327
33
495
211
285
23
473
229
299
77
387
191
369
329
135
443
117
275
221
481
47
293
235
471
25
383
177
397
67
423
105
341
155
509
51
271
193
459
5
313
247
401
95
355
173
149
347
103
425
207
257
61
499
249
311
11
453
163
365
81
415
4096
4096
4096
4096
4096
4096
4096
4096
4096
4096
4096
4096
4096
4096
4096
4096
4096
4096 4096 4096 4096 4096 4096 4096 4096 4096 4096 4096 4096 4096 4096 4096 4096 4096
Example 21
In
this
case
the
magic
sum
is
S16×16 := 4096 = 163 ,
and
sum
of
numbers
is
F256 := 65536 = 2562 = 164 . Each block of order 4 is also a pan diagonal magic square with magic
sum S 4×4 := 1024 . Moreover, the above magic square is also a bimagic with bimagic sum
Sb16×16 := 1398096 .
16. Magic Square of Order 18
According to (7) we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 18. Take k  18 in (5), we get
1  3  5  ...  (2 182  1)  184

1 + 3 + 5 + ... + 645 + 647 = 184 = 104976 .
(30)
According to values given in (30), below are two different ways of writing the magic square of
order 18:
20
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
● First way:
5832
217
273
239
279
253
251
145
201
167
207
181
179
505
561
527
567
541
539
261
229
227
247
281
267
189
157
155
175
209
195
549
517
515
535
569
555
271
285
241
223
237
255
199
213
169
151
165
183
559
573
529
511
525
543
283
243
269
263
221
233
211
171
197
191
149
161
571
531
557
551
509
521
249
257
277
235
275
219
177
185
205
163
203
147
537
545
565
523
563
507
5832 5832
5832
5832
5832
231
225
259
265
245
287
159
153
187
193
173
215
519
513
547
553
533
575
577
633
599
639
613
611
289
345
311
351
325
323
1
57
23
63
37
35
621
589
587
607
641
627
333
301
299
319
353
339
45
13
11
31
65
51
631
645
601
583
597
615
343
357
313
295
309
327
55
69
25
7
21
39
643
603
629
623
581
593
355
315
341
335
293
305
67
27
53
47
5
17
609
617
637
595
635
579
321
329
349
307
347
291
33
41
61
19
59
3
591
585
619
625
605
647
303
297
331
337
317
359
15
9
43
49
29
71
73
129
95
135
109
107
433
489
455
495
469
467
361
417
383
423
397
395
117
85
83
103
137
123
477
445
443
463
497
483
405
373
371
391
425
411
127
141
97
79
93
111
487
501
457
439
453
471
415
429
385
367
381
399
139
99
125
119
77
89
499
459
485
479
437
449
427
387
413
407
365
377
105
113
133
91
131
75
465
473
493
451
491
435
393
401
421
379
419
363
87
81
115
121
101
143
447
441
475
481
461
503
375
369
403
409
389
431
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832
Example 22
In this case the magic sum is S18×18 := 5832= 183 , and sum of all 324 numbers is
F324 := 104976 = 3242 = 184 . Each block of order 6 is a magic square constructed according to
section 6 with different sums:
5832
27
1512 3672 648 5832  6  7 17 3 27
1080 1944 2808 5832
5 9 13 27
3240 216 2376 5832
15 1 11 27
3
5832 5832 5832 5832
Example 23
27 27 27 27
Interestingly, the same happens with the sum of all members of each block of order 6 resulting
again in a magic square of order 3:
34992
27
9072 22032 3888 34992  64  7 17 3 27
6480 11664 16848 34992
5 9 13 27
9072 22032 3888 34992
15 1 11 27
34992 34992
34992 34992
Example 24
27 27 27 27
● Second way:
Redistributing the total numbers 32992 in little different way, we get an alternative way of
writing a magic square of order 18 with the same values given in (30). See below
21
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
5832
7
511
205
565
331
313
5
509
203
563
329
311
15
519
213
573
339
321
403
115
97
277
583
457
401
113
95
275
581
455
411
123
105
285
591
465
493
619
223
61
187
349
491
617
221
59
185
347
501
627
231
69
195
357
601
241
475
421
43
151
599
239
473
419
41
149
609
249
483
429
51
159
295
367
547
169
529
25
293
365
545
167
527
23
303
375
555
177
537
33
133
79
385
439
259
637
131
77
383
437
257
635
141
87
393
447
267
645
17
521
215
575
341
323
9
513
207
567
333
315
1
505
199
559
325
307
413
125
107
287
593
467
405
117
99
279
585
459
397
109
91
271
577
451
503
629
233
71
197
359
495
621
225
63
189
351
487
613
217
55
181
343
611
251
485
431
53
161
603
243
477
423
45
153
595
235
469
415
37
145
305
377
557
179
539
35
297
369
549
171
531
27
289
361
541
163
523
19
143
89
395
449
269
647
135
81
387
441
261
639
127
73
379
433
253
631
3
507
201
561
327
309
13
517
211
571
337
319
11
515
209
569
335
317
399
111
93
273
579
453
409
121
103
283
589
463
407
119
101
281
587
461
489
615
219
57
183
345
499
625
229
67
193
355
497
623
227
65
191
353
597
237
471
417
39
147
607
247
481
427
49
157
605
245
479
425
47
155
291
363
543
165
525
21
301
373
553
175
535
31
299
371
551
173
533
29
129
75
381
435
255
633
139
85
391
445
265
643
137
83
389
443
263
641
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832
5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832 5832
Example 25
In this case also the magic sum is S18×18 := 5832= 183 , and sum of all 324 numbers is
F324 := 104976 = 3242 = 184 . Here also each block of order 6 is a magic square with different
magic sums:
1932 1992 1908
1920 1944 1968
1980 1896 1956
5832 5832 5832
5832
5832  12 
5832
5832
5832
Example 26
161 166 159
160 162 164
165 158 163
486 486 486
486
486
486
486
486
Similar to first way, here also, the sum of all members of each block of order 6 results in a
magic square of order 3:
34992
486
11592 11952 11448 34992  72  161 166 159 486
11520 11664 11808 34992
160 162 164 486
11880 11376 11736 34992
165 158 163 486
34992
34992
34992 34992
Example 27
486 486 486 486
17. Magic Square of Order 20
According to (7) we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 18. Take n  400 in (5) or k  20 in (6), we
get
1  3  5  ...  (2  202  1)  204

1 + 3 + 5 + ... + 397 + 399 = 20 4 = 160000 .
According to values given in (31), the magic square of order 20 for odd numbers is given by
22
(31)
Inder J. Taneja
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
43
573
259
709
395
5
611
301
667
357
121
495
177
791
473
87
529
223
745
439
269
715
363
53
579
307
677
325
11
621
191
793
441
135
497
225
759
407
89
543
373
59
589
275
683
331
21
627
317
645
455
137
511
193
761
409
103
545
239
727
595
243
693
379
69
637
285
651
341
27
513
161
775
457
151
559
207
729
423
105
699
389
75
563
253
661
347
37
605
291
777
471
153
481
175
743
425
119
527
209
81
535
217
751
433
127
489
183
785
479
3
613
299
669
355
45
571
261
707
397
231
753
401
95
537
185
799
447
129
503
309
675
323
13
619
267
717
365
51
581
415
97
551
233
721
449
143
505
199
767
333
19
629
315
643
371
61
587
277
685
553
201
735
417
111
519
167
769
463
145
635
283
653
339
29
597
245
691
381
67
737
431
113
521
215
783
465
159
487
169
659
349
35
603
293
701
387
77
565
251
7
609
303
665
359
41
575
257
711
393
85
531
221
747
437
123
493
179
789
475
305
679
327
9
623
271
713
361
55
577
227
757
405
91
541
189
795
443
133
499
329
23
625
319
647
375
57
591
273
681
411
101
547
237
725
453
139
509
195
763
639
287
649
343
25
593
241
695
377
71
557
205
731
421
107
515
163
773
459
149
663
345
39
607
289
697
391
73
561
255
741
427
117
525
211
779
469
155
483
173
125
491
181
787
477
83
533
219
749
435
47
569
263
705
399
1
615
297
671
353
187
797
445
131
501
229
755
403
93
539
265
719
367
49
583
311
673
321
15
617
451
141
507
197
765
413
99
549
235
723
369
63
585
279
687
335
17
631
313
641
517
165
771
461
147
555
203
733
419
109
599
247
689
383
65
633
281
655
337
31
781
467
157
485
171
739
429
115
523
213
703
385
79
567
249
657
351
33
601
295
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
Example 28
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
8000
In this case the magic square is pan diagonal and magic sum is S20×20 := 8000= 203 . Also, the sum
of all 400 numbers is F400 := 160000 = 4002 = 204 . Moreover each block of order 5 a pan diagonal
magic square with different magic sums:
8000
8000 8000 8000
1979
8000 1941
8000 2057
8000 2023
2017 1943 2061 8000
8000 8000
8000 8000 8000 8000
Example 29
2063 1977 2019 8000
1939 2021 1983 8000
1981 2059 1937 8000
The second block from up is a pan diagonal magic square of order with magic sum 2017, i.e.,
the year we are.
18. Magic Square of Order 21
According to (5) and (8), one can have a perfect square sum magic square of order 21 in two
ways. One using odd order numbers and secondly having sequential values. Below are
examples of both type.
18.1 First Approach – Odd Numbers
Take k  21 in (5), we get
1  3  5  ...  (2  212  1)  214

1 + 3 + 5 + 7 + ... + 879 + 881 = 4412 = 194481.
23
(32)
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
According to values given in (32), we shall contruct in two different ways of magic square of
order 21:
●First way:
9261
295
373
339
319
383
363
329
197
275
241
221
285
265
231
687
765
731
711
775
755
721
311
389
355
335
301
365
345
213
291
257
237
203
267
247
703
781
747
727
693
757
737
327
307
371
337
317
381
361
229
209
273
239
219
283
263
719
699
763
729
709
773
753
343
309
387
353
333
299
377
245
211
289
255
235
201
279
735
701
779
745
725
691
769
359
325
305
369
349
315
379
261
227
207
271
251
217
281
751
717
697
761
741
707
771
375
341
321
385
351
331
297
277
243
223
287
253
233
199
767
733
713
777
743
723
689
391
357
323
303
367
347
313
293
259
225
205
269
249
215
783
749
715
695
759
739
705
785
863
829
809
873
853
819
393
471
437
417
481
461
427
1
79
45
25
89
69
35
801
879
845
825
791
855
835
409
487
453
433
399
463
443
17
95
61
41
7
71
51
817
797
861
827
807
871
851
425
405
469
435
415
479
459
33
13
77
43
23
87
67
833
799
877
843
823
789
867
441
407
485
451
431
397
475
49
15
93
59
39
5
83
849
815
795
859
839
805
869
457
423
403
467
447
413
477
65
31
11
75
55
21
85
865
831
811
875
841
821
787
473
439
419
483
449
429
395
81
47
27
91
57
37
3
881
847
813
793
857
837
803
489
455
421
401
465
445
411
97
63
29
9
73
53
19
99
177
143
123
187
167
133
589
667
633
613
677
657
623
491
569
535
515
579
559
525
115
193
159
139
105
169
149
605
683
649
629
595
659
639
507
585
551
531
497
561
541
131
111
175
141
121
185
165
621
601
665
631
611
675
655
523
503
567
533
513
577
557
147
113
191
157
137
103
181
637
603
681
647
627
593
671
539
505
583
549
529
495
573
163
129
109
173
153
119
183
653
619
599
663
643
609
673
555
521
501
565
545
511
575
179
145
125
189
155
135
101
669
635
615
679
645
625
591
571
537
517
581
547
527
493
195
161
127
107
171
151
117
685
651
617
597
661
641
607
587
553
519
499
563
543
509
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261
Example 30
In this case the magic sum is S21×21 := 9261 = 213 , and sum of numbers is F441 := 194481 = 4412 = 214 .
Each block of order 7 are pan diagonal magic square with different magic sums forming again
a magic square of order 3:
9261
27
2401 5831 1029 9261
1715 3087 4459 9261
5145 343 3773 9261
 7  7 17 3 27
5 9 13 27
15 1 11 27
9261
9261
9261
3
9261
Example 31
27 27 27 27
Interestingly, the same happens with the sum of all members of each block of order 7 resulting
again in a magic square of order 3:
64827
27
16807 40817 7203 64827  7  7 17 3 27
12005 21609 31213 64827
5 9 13 27
36015 2401 26411 64827
15 1 11 27
4
64827
64827
64827 64827
Example 32
27 27 27 27
●Second way:
Proceeding similar lines of order 18, we can also calculate second way of writing magic square
of order 21:
24
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
9261
7
709
403
223
799
619
313
5
707
401
221
797
617
311
15
717
411
231
807
627
321
151
853
547
367
61
637
457
149
851
545
365
59
635
455
159
861
555
375
69
645
465
295
115
691
385
205
781
601
293
113
689
383
203
779
599
303
123
699
393
213
789
609
439
133
835
529
349
43
745
437
131
833
527
347
41
743
447
141
843
537
357
51
753
583
277
97
673
493
187
763
581
275
95
671
491
185
761
591
285
105
681
501
195
771
727
421
241
817
511
331
25
725
419
239
815
509
329
23
735
429
249
825
519
339
33
871
565
259
79
655
475
169
869
563
257
77
653
473
167
879
573
267
87
663
483
177
17
719
413
233
809
629
323
9
711
405
225
801
621
315
1
703
397
217
793
613
307
161
863
557
377
71
647
467
153
855
549
369
63
639
459
145
847
541
361
55
631
451
305
125
701
395
215
791
611
297
117
693
387
207
783
603
289
109
685
379
199
775
595
449
143
845
539
359
53
755
441
135
837
531
351
45
747
433
127
829
523
343
37
739
593
287
107
683
503
197
773
585
279
99
675
495
189
765
577
271
91
667
487
181
757
737
431
251
827
521
341
35
729
423
243
819
513
333
27
721
415
235
811
505
325
19
881
575
269
89
665
485
179
873
567
261
81
657
477
171
865
559
253
73
649
469
163
3
705
399
219
795
615
309
13
715
409
229
805
625
319
11
713
407
227
803
623
317
147
849
543
363
57
633
453
157
859
553
373
67
643
463
155
857
551
371
65
641
461
291
111
687
381
201
777
597
301
121
697
391
211
787
607
299
119
695
389
209
785
605
435
129
831
525
345
39
741
445
139
841
535
355
49
751
443
137
839
533
353
47
749
579
273
93
669
489
183
759
589
283
103
679
499
193
769
587
281
101
677
497
191
767
723
417
237
813
507
327
21
733
427
247
823
517
337
31
731
425
245
821
515
335
29
867
561
255
75
651
471
165
877
571
265
85
661
481
175
875
569
263
83
659
479
173
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261
Example 33
In this case the magic sum is S21×21 := 9261 = 213 , and sum of numbers is F441 := 194481 = 4412 = 214 .
Each block of order 7 is a pan diagonal magic square with different magic sums forming again
a magic square of order 3:
3073 3143
3059 3087
3129 3031
9261 9261
9261
3045 9261  7  439
3115 9261
437
3101 9261
447
9261 9261
1323
Example 34
449
441
433
1323
1323
435 1323
445 1323
443 1323
1323 1323
Interestingly, the same happens with the sum of all members of each block of order 7 resulting
again in a magic square of order 3:
64827
1323
21511 22001 21315 64827  7 
21413 21609 21805 64827
21903 21217 21707 64827
439
437
447
64827
1323 1323 1323 1323
2
64827
64827 64827
Example 35
449 435 1323
441 445 1323
433 443 1323
18.2 Second Approach – Sequential
According to formula (8), we have
K : T (n)  T (n  441),
i.e.,
25
n  441 ,
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
n(n  1) (n  441)(n  440)

2
2
: 441n  97020  441 n  220  .
K :
Take n  661 , we get a perfect square, i.e.,
K : T (661)  T (220)  194481  4412  214.
Simplifying, one get
221 + 222 + ... + 660 + 661 = 214 = 4412 = 194481 .
(33)
Thus, we have a perfect square sum for the 441 numbers starting from 221 to 661. According to
values given in (33), we can easily construct magic square of order 21 of sequential values
giving sum of all numbers as prefect sum.
368
407
390
380
412
402
385
319
358
341
331
363
353
336
564
603
586
576
608
598
581
376
415
398
388
371
403
393
327
366
349
339
322
354
344
572
611
594
584
567
599
589
384
374
406
389
379
411
401
335
325
357
340
330
362
352
580
570
602
585
575
607
597
392
375
414
397
387
370
409
343
326
365
348
338
321
360
588
571
610
593
583
566
605
400
383
373
405
395
378
410
351
334
324
356
346
329
361
596
579
569
601
591
574
606
408
391
381
413
396
386
369
359
342
332
364
347
337
320
604
587
577
609
592
582
565
416
399
382
372
404
394
377
367
350
333
323
355
345
328
612
595
578
568
600
590
573
613
652
635
625
657
647
630
417
456
439
429
461
451
434
221
260
243
233
265
255
238
621
660
643
633
616
648
638
425
464
447
437
420
452
442
229
268
251
241
224
256
246
629
619
651
634
624
656
646
433
423
455
438
428
460
450
237
227
259
242
232
264
254
637
620
659
642
632
615
654
441
424
463
446
436
419
458
245
228
267
250
240
223
262
645
628
618
650
640
623
655
449
432
422
454
444
427
459
253
236
226
258
248
231
263
653
636
626
658
641
631
614
457
440
430
462
445
435
418
261
244
234
266
249
239
222
661
644
627
617
649
639
622
465
448
431
421
453
443
426
269
252
235
225
257
247
230
270
309
292
282
314
304
287
515
554
537
527
559
549
532
466
505
488
478
510
500
483
278
317
300
290
273
305
295
523
562
545
535
518
550
540
474
513
496
486
469
501
491
286
276
308
291
281
313
303
531
521
553
536
526
558
548
482
472
504
487
477
509
499
294
277
316
299
289
272
311
539
522
561
544
534
517
556
490
473
512
495
485
468
507
302
285
275
307
297
280
312
547
530
520
552
542
525
557
498
481
471
503
493
476
508
310
293
283
315
298
288
271
555
538
528
560
543
533
516
506
489
479
511
494
484
467
318
301
284
274
306
296
279
563
546
529
519
551
541
524
514
497
480
470
502
492
475
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261
9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261 9261
Example 36
In this case, the magic sum is S21×21 := 9261 = 213 , and sum of number is F441 := 50625 = 4412 = 214.
F441 := 194481 = 4412 = 214 . In this case each block of order 7 is a pan diagonal magic square with
different magic sums forming again a magic square of order 3:
2744 4459
2401 3087
4116 1715
9261 9261
9261
27
3
2058 9261  7  8 13 6 27
3773 9261
7 9 11 27
3430 9261
12 5 10 27
9261 9261
27 27 27 27
Example 37
Interestingly, the same happens with the sum of all members of each block of order 7 resulting
again in a magic square of order 3:
26
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
19208 31213
16807 21609
28812 12005
64827
64827
64827
14406 64827  74 
26411 64827
24010 64827
64827 64827
Example 38
27
27
8 13 6
7 9 11 27
12 5 10 27
27 27 27 27
Proceeding similar lines of order 18, we can also calculate second way of writing magic square
of order 21:
Still, there are more possibilities of getting sequential numbers for the magic square of order 21,
where numbers sum is a perfect square. See below examples,
T(704)  T(263) = 264 + 265 + ... + 704 = 213444 = 462 2 ;
T(749)  T(308) = 309 + 310 + ... + 749 = 233289 = 483 2 ;
T(796)  T(355) = 356 + 357 + ... + 796 = 254016 = 504 2 .
19. Magic Square of Order 22
According to (8) we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 22. Take n  484 in (5) or k  22 in (6), we
get
1  3  5  ...  (2  222  1)  224

1 + 3 + 5 + ... + 965 + 967 = 224 = 234256 .
(34)
According to values given in (34), the magic square of order 22 for odd numbers is given by
1
229
403
63
143
563
659
745
611
697
187
47
221
405
107
743
609
695
571
657
151
11
93
223
397
703
569
655
741
607
91 137 183
181 7
53
769
859
941
807
893
465
341
515
293
379
939
805
901
771
853
513
291
377
473
339
903
765
851
937
813
385
471
337
511
289
399
195
55
139
225
653
739
615
701
567
227
401
19
99
185
613
699
565
651
747
9
89
45
135
849
945
815
897
763
335
509
297
383
469
809
895
761
857
947
295
381
467
333
517
773
869
955
821
907
231
331
417
503
281
367
453
35
131
217
83
169
529
625
711
577
663
953
819
905
781
867
499
277
363
463
241
327
413
215
81
167
43
129
709
575
661
537
623
913
779
865
951
817
459
237
323
409
495
287
373
175
41
127
213
79
669
535
621
707
573
863
949
825
911
777
419
505
283
369
455
233
319
125
211
87
173
39
619
705
581
667
533
823
909
775
861
957
365
451
243
329
415
501
279
85
171
37
123
219
579
665
531
617
713
49 97 475 523 395
95 133 351 301 481
141 179 525 387 347
177 5 303 483 521
3
51 389 349 299
285
371
457
235
321
407
507
749
845
931
797
883
601
647
693
729
555
649
685
731
557
603
553
253
427
591
671
831
877
923
959
785
879 21 191 155 101 61
915 111 57 17 197 147
961 193 153 103 67 13
787 59 23 189 149 109
833 145 105 65 15 199
27
889
755
841
927
793
305
391
477
343
527
799
885
751
837
933
325
411
497
275
375
461
239
Example 39
929
795
881
757
843
345
519
307
393
479
839
925
801
887
753
549
639
721
587
673
25
121
207
73
159
719
585
681
551
633
205
71
157
33
119
683
545
631
717
593
165
31
117
203
69
629
725
595
677
543
115
201
77
163
29
589
675
541
637
727
75
161
27
113
209
643 689 735 561 597 873 919 965 791 827
733 559 605 641 687 963 789 835 871 917
715
599
245
429
635
679
539
645
247
421
423
723
583
691
249
251
425
547
627
737
441
317
491
269
355
783
263
437
811
891
489
267
353
449
315
935
829
255
439
855
361
447
313
487
265
899
759
875
257
431
311
485
273
359
445
433
943
803
921
259
271
357
443
309
493
261
435
767
847
967
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
In this case the magic sum S22×22 := 10648= 223 . Also, the sum of all 484 numbers is
F484 := 234256 = 4842 = 224 . There are 12 block of order 4 and 1 of order 7 are magic squares.
These are given by bold face letters. The approach applied to construct this magic square is
based on the procedure of H. White [9]. This approach is by used of computer program.
20. Magic Square of Order 24
According to (7) we don’t have sequential magic square with number sum a perfect square. We
will obtain only odd numbers magic square of order 24. Take k  24 in (5), we get
1  3  5  ...  (2  242  1)  244

1 + 3 + 5 + ... + 1149 + 1151 = 24 4 = 331776 .
(35)
There are many ways of constructing magic square of order 24 with odd order numbers, for
example, 8  3, 3  8, 4  6 or 6  4 . Instead of using all the four ways, we shall use only two. In
the case of 8  3 , we can work with bimagic square of order 8 as given in Section 8. See below:
● First way 8×3 :
277 727 637
79
475 1105 979 313 287 737 647
89
485 1115 989 323 273 723 633
75
471 1101 975 309
457 1123 961 331 223 781 583 133 467 1133 971 341 233 791 593 143 453 1119 957 327 219 777 579 129
7
709 799 205 385 907 1033 547
17
719 809 215 395 917 1043 557
3
705 795 201 381 903 1029 543
403 889 1051 529
61
655 853 151 413 899 1061 539
71
665 863 161 399 885 1047 525
57
651 849 147
673
43
169 835 871 421 511 1069 683
53
179 845 881 431 521 1079 669
39
165 831 867 417 507 1065
925 367 565 1015 691
25
187 817 935 377 575 1025 701
35
197 827 921 363 561 1011 687
21
183 813
763 241 115 601 1141 439 349 943 773 251 125 611 1151 449 359 953 759 237 111 597 1137 435 345 939
1087 493 295 997 745 259
97
619 1097 503 305 1007 755 269 107 629 1083 489 291 993 741 255
93
615
275 725 635
77
473 1103 977 311 279 729 639
81
477 1107 981 315 283 733 643
85
481 1111 985 319
455 1121 959 329 221 779 581 131 459 1125 963 333 225 783 585 135 463 1129 967 337 229 787 589 139
5
707 797 203 383 905 1031 545
9
711 801 207 387 909 1035 549
13
715 805 211 391 913 1039 553
401 887 1049 527
59
653 851 149 405 891 1053 531
63
657 855 153 409 895 1057 535
67
661 859 157
671
41
167 833 869 419 509 1067 675
45
171 837 873 423 513 1071 679
49
175 841 877 427 517 1075
923 365 563 1013 689
23
185 815 927 369 567 1017 693
27
189 819 931 373 571 1021 697
31
193 823
761 239 113 599 1139 437 347 941 765 243 117 603 1143 441 351 945 769 247 121 607 1147 445 355 949
1085 491 293 995 743 257
95
617 1089 495 297 999 747 261
99
621 1093 499 301 1003 751 265 103 625
285 735 645
87
483 1113 987 321 271 721 631
73
469 1099 973 307 281 731 641
83
479 1109 983 317
465 1131 969 339 231 789 591 141 451 1117 955 325 217 775 577 127 461 1127 965 335 227 785 587 137
15
717 807 213 393 915 1041 555
1
703 793 199 379 901 1027 541
11
713 803 209 389 911 1037 551
411 897 1059 537
69
663 861 159 397 883 1045 523
55
649 847 145 407 893 1055 533
65
659 857 155
681
51
177 843 879 429 519 1077 667
37
163 829 865 415 505 1063 677
47
173 839 875 425 515 1073
933 375 573 1023 699
33
195 825 919 361 559 1009 685
19
181 811 929 371 569 1019 695
29
191 821
771 249 123 609 1149 447 357 951 757 235 109 595 1135 433 343 937 767 245 119 605 1145 443 353 947
1095 501 303 1005 753 267 105 627 1081 487 289 991 739 253
91
613 1091 497 299 1001 749 263 101 623
Example 40
In this case the magic sum S24×24 := 13824 = 243 . Also, the sum of all 576 numbers is
F576 := 331776 = 5762 = 244 .
The construction of above magic square is based on the procedure of making combinations of
magic square of order 8 with order 3. The procedure used is the one given in section 8 of
bimagic square of order 8 combining with order 3 given in section 3. Below are the details of
each block of order 8 forming a magic square of order 3:
28
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
4592 4672 4560
4576 4608 4640
4656 4544 4624
13824 13824 13824
13824
864
4
13824  2  287 292 285 864
13824
286 288 290 864
13824
291 284 289 864
13824
864 864 864 864
Example 41
The magic squares formed by sum of all members each block of order 8 forms again a magic
square of order 3:
110592
864
36736 37376 36480 110592  27  287 292 285 864
36608 36864 37120 110592
286 288 290 864
37248 36352 36992 110592
291 284 289 864
110592 110592 110592 110592
Example 42
864 864 864 864
● Second way 6×4 :
433
73
1081
577
489
129
1137
633
455
95
1103
599
495
135
1143
639
469
109
1117
613
467
107
1115
611
793
865
145
361
849
921
201
417
815
887
167
383
855
927
207
423
829
901
181
397
827
899
179
395
1
505
649
1009
57
561
705
1065
23
527
671
1031
63
567
711
1071
37
541
685
1045
35
539
683
1043
937
721
289
217
993
777
345
273
959
743
311
239
999
783
351
279
973
757
325
253
971
755
323
251
477
117
1125
621
445
85
1093
589
443
83
1091
587
463
103
1111
607
497
137
1145
641
483
123
1131
627
837
909
189
405
805
877
157
373
803
875
155
371
823
895
175
391
857
929
209
425
843
915
195
411
45
549
693
1053
13
517
661
1021
11
515
659
1019
31
535
679
1039
65
569
713
1073
51
555
699
1059
981
765
333
261
949
733
301
229
947
731
299
227
967
751
319
247
1001
785
353
281
987
771
339
267
487
127
1135
631
501
141
1149
645
457
97
1105
601
439
79
1087
583
453
93
1101
597
471
111
1119
615
847
919
199
415
861
933
213
429
817
889
169
385
799
871
151
367
813
885
165
381
831
903
183
399
55
559
703
1063
69
573
717
1077
25
529
673
1033
7
511
655
1015
21
525
669
1029
39
543
687
1047
991
775
343
271
1005
789
357
285
961
745
313
241
943
727
295
223
957
741
309
237
975
759
327
255
499
139
1147
643
459
99
1107
603
485
125
1133
629
479
119
1127
623
437
77
1085
581
449
89
1097
593
859
931
211
427
819
891
171
387
845
917
197
413
839
911
191
407
797
869
149
365
809
881
161
377
67
571
715
1075
27
531
675
1035
53
557
701
1061
47
551
695
1055
5
509
653
1013
17
521
665
1025
1003
787
355
283
963
747
315
243
989
773
341
269
983
767
335
263
941
725
293
221
953
737
305
233
465
105
1113
609
473
113
1121
617
493
133
1141
637
451
91
1099
595
491
131
1139
635
435
75
1083
579
825
897
177
393
833
905
185
401
853
925
205
421
811
883
163
379
851
923
203
419
795
867
147
363
33
537
681
1041
41
545
689
1049
61
565
709
1069
19
523
667
1027
59
563
707
1067
3
507
651
1011
969
753
321
249
977
761
329
257
997
781
349
277
955
739
307
235
995
779
347
275
939
723
291
219
447
87
1095
591
441
81
1089
585
475
115
1123
619
481
121
1129
625
461
101
1109
605
503
143
1151
647
807
879
159
375
801
873
153
369
835
907
187
403
841
913
193
409
821
893
173
389
863
935
215
431
15
951
519 735
663 303
1023 231
9
945
513 729
657 297
1017 225
43
979
547 763
691 331
1051 259
49
985
553 769
697 337
1057 265
29
965
533 749
677 317
1037 245
71 1007
575 791
719 359
1079 287
Example 43
In this case also the magic sum S24×24 := 13824 = 243 . Also, the sum of all 576 numbers is
F576 := 331776 = 5762 = 244 .
The construction of above magic square is based on the procedure of making combinations of
magic square of order 4 with order 6. The procedure used is the one given in section 4. Each
block of order 4 is perfect pan diagonal magic square of order 4 with different magic sums.
These magic sums again make a magic square of order 6:
29
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
13824
13824
13824
13824
13824
13824
13824
13824 13824 13824 13824 13824 13824 13824
2164
2388
2252
2412
2308
2300
2340
2212
2204
2284
2420
2364
2380
2436
2260
2188
2244
2316
2428
2268
2372
2348
2180
2228
2292
2324
2404
2236
2396
2172
2220
2196
2332
2356
2276
2444
 4
3456
3456
3456
3456
3456
3456
3456
3456 3456 3456 3456 3456 3456 3456
541
597
563
603
577
575
585
553
551
571
605
591
595
609
565
547
561
579
607
567
593
587
545
557
573
581
601
559
599
543
555
549
583
589
569
611
Example 44
The magic squares formed by sum of all members each block of order 4 forms again a magic
square of order 6:
55296
8656
9552
9008
9648
9232
9200
9360
8848
8816
9136
9680
9456
9520
9744
9040
8752
8976
9264
9712
9072
9488
9392
8720
8912
9168
9296
9616
8944
9584
8688
8880
8784
9328
9424
9104
9776
55296
3456
4 
2
55296
55296
55296
55296
55296
55296 55296 55296 55296 55296 55296 55296
541
597
563
603
577
575
585
553
551
571
605
591
595
609
565
547
561
579
607
567
593
587
545
557
573
581
601
559
599
543
555
549
583
589
569
611
3456
3456
3456
3456
3456
3456
3456 3456 3456 3456 3456 3456 3456
Example 45
Interestingly, the magic sum of second magic square in both the cases is sequential digits 3456.
21. Magic Square of Order 25
According to (5) and (8), one can have a perfect square sum magic square of order 25 in two
ways. One using odd order numbers and secondly having sequential values. Below are
examples of both type.
21.1 First Approach – Odd Numbers
Take n  252  625 in (5) or k  25 in (6), we get
1  3  5  ...  (2  252  1)  214

1 + 3 + 5 + 7 + ... + 1247 + 1249 = 6252 = 390625 .
According to values given in (36), the magic square of order 25 is given by
30
(36)
Inder J. Taneja
1
913
325
1237
649
1069
721
133
795
407
877
289
1191
553
215
685
97
959
361
1023
493
1105
517
179
831
337
1249
601
13
925
145
757
419
1071
733
1153
565
227
889
291
961
373
1035
697
59
529
181
843
455
1117
613
25
937
349
1201
421
1083
745
107
769
239
891
253
1165
577
1047
659
61
973
385
805
467
1129
531
193
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
949
301
1213
625
37
707
119
771
433
1095
265
1177
589
241
853
73
985
397
1009
661
1131
543
155
817
479
1225
637
49
901
313
783
445
1057
719
121
591
203
865
277
1189
359
1011
673
85
997
167
829
481
1143
505
885
297
1159
561
223
693
55
967
379
1031
451
1113
525
187
849
19
921
333
1245
607
1077
739
141
753
415
1161
573
235
897
259
979
381
1043
655
67
537
199
801
463
1125
345
1207
619
21
933
103
765
427
1089
741
247
859
261
1173
585
1005
667
79
981
393
813
475
1137
549
151
621
33
945
307
1219
439
1091
703
115
777
273
1185
597
209
861
81
993
355
1017
679
1149
501
163
825
487
907
319
1221
633
45
715
127
789
441
1053
559
211
873
285
1197
367
1029
681
93
955
175
837
499
1101
513
1233
645
7
919
321
791
403
1065
727
139
469
1121
533
195
807
27
939
341
1203
615
1085
747
109
761
423
893
255
1167
579
231
651
63
975
387
1049
545
157
819
471
1133
303
1215
627
39
941
111
773
435
1097
709
1179
581
243
855
267
987
399
1001
663
75
821
483
1145
507
169
639
41
903
315
1227
447
1059
711
123
785
205
867
279
1181
593
1013
675
87
999
351
1107
519
171
833
495
915
327
1239
641
3
723
135
797
409
1061
281
1193
555
217
879
99
951
363
1025
687
183
845
457
1119
521
1241
603
15
927
339
759
411
1073
735
147
567
229
881
293
1155
375
1037
699
51
963
1093
705
117
779
431
851
263
1175
587
249
669
71
983
395
1007
477
1139
541
153
815
35
947
309
1211
623
129
781
443
1055
717
1187
599
201
863
275
995
357
1019
671
83
503
165
827
489
1141
311
1223
635
47
909
405
1067
729
131
793
213
875
287
1199
551
1021
683
95
957
369
839
491
1103
515
177
647
9
911
323
1235
731
143
755
417
1079
299
1151
563
225
887
57
969
371
1033
695
1115
527
189
841
453
923
335
1247
609
11
767
429
1081
743
105
575
237
899
251
1163
383
1045
657
69
971
191
803
465
1127
539
1209
611
23
935
347
677
89
991
353
1015
485
1147
509
161
823
43
905
317
1229
631
1051
713
125
787
449
869
271
1183
595
207
953
365
1027
689
91
511
173
835
497
1109
329
1231
643
5
917
137
799
401
1063
725
1195
557
219
871
283
1039
691
53
965
377
847
459
1111
523
185
605
17
929
331
1243
413
1075
737
149
751
221
883
295
1157
569
65
977
389
1041
653
1123
535
197
809
461
931
343
1205
617
29
749
101
763
425
1087
257
1169
571
233
895
391
1003
665
77
989
159
811
473
1135
547
1217
629
31
943
305
775
437
1099
701
113
583
245
857
269
1171
Example 43
In
this
case
the
magic
sum
is
S25×25 := 15625 = 253 ,
and
sum
of
numbers
is
F625 := 390625 = 6252 = 254 . In this the magic square is pan diagonal and bimagic with bimagic
sum Sb 25×25 := 13020825 . Moreover each block of order 5 is a pan diagonal magic square
constant magic sum S5×5 := 3125 = 55 .
21.2 Second Approach – Sequential
According to formula (7), we have
K : T (n)  T (n  625),
i.e.,
n  625 ,
n(n  1) (n  625)(n  624)

2
2
: 625n  195000  625  n  312  .
K :
Take n  937 , we get a perfect square, i.e.,
K (937) : T (937)  T (312)  390625  6252  254.
Simplifying, one get
313 + 314 + ... + 936 + 937 = 254 = 6252 = 390625 .
(37)
Thus, we have a perfect square sum for the 625 numbers starting from 313 to 937. According to
values given in (37), we can easily construct magic square of order 25 of sequential values
giving sum of all numbers as prefect sum.
31
Inder J. Taneja
313
769
475
931
637
847
673
379
710
516
751
457
908
589
420
655
361
792
493
824
559
865
571
402
728
481
937
613
319
775
385
691
522
848
679
889
595
426
757
458
793
499
830
661
342
577
403
734
540
871
619
325
781
487
913
523
854
685
366
697
432
758
439
895
601
836
642
343
799
505
715
546
877
578
409
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
787
463
919
625
331
666
372
698
529
860
445
901
607
433
739
349
805
511
817
643
878
584
390
721
552
925
631
337
763
469
704
535
841
672
373
608
414
745
451
907
492
818
649
355
811
396
727
553
884
565
755
461
892
593
424
659
340
796
502
828
538
869
575
406
737
322
773
479
935
616
851
682
383
689
520
893
599
430
761
442
802
503
834
640
346
581
412
713
544
875
485
916
622
323
779
364
695
526
857
683
436
742
443
899
605
815
646
352
803
509
719
550
881
587
388
623
329
785
466
922
532
858
664
370
701
449
905
611
417
743
353
809
490
821
652
887
563
394
725
556
766
472
923
629
335
670
376
707
533
839
592
418
749
455
911
496
827
653
359
790
400
731
562
863
569
929
635
316
772
473
708
514
845
676
382
547
873
579
410
716
326
782
483
914
620
855
686
367
693
524
759
440
896
602
428
638
344
800
506
837
585
391
722
548
879
464
920
626
332
783
368
699
530
861
667
902
603
434
740
446
806
512
813
644
350
723
554
885
566
397
632
333
764
470
926
536
842
668
374
705
415
746
452
903
609
819
650
356
812
488
866
572
398
729
560
770
476
932
633
314
674
380
711
517
843
453
909
590
421
752
362
788
494
825
656
404
735
541
872
573
933
614
320
776
482
692
518
849
680
386
596
427
753
459
890
500
831
662
338
794
859
665
371
702
528
738
444
900
606
437
647
348
804
510
816
551
882
583
389
720
330
786
467
918
624
377
703
534
840
671
906
612
413
744
450
810
491
822
648
354
564
395
726
557
883
468
924
630
336
767
515
846
677
378
709
419
750
456
912
588
823
654
360
791
497
732
558
864
570
401
636
317
768
474
930
678
384
690
521
852
462
888
594
425
756
341
797
498
829
660
870
576
407
733
539
774
480
936
617
318
696
527
853
684
365
600
431
762
438
894
504
835
641
347
798
408
714
545
876
582
917
618
324
780
486
651
357
808
489
820
555
886
567
393
724
334
765
471
927
628
838
669
375
706
537
747
448
904
610
416
789
495
826
657
358
568
399
730
561
867
477
928
634
315
771
381
712
513
844
675
910
591
422
748
454
832
658
339
795
501
736
542
868
574
405
615
321
777
478
934
519
850
681
387
688
423
754
460
891
597
345
801
507
833
639
874
580
411
717
543
778
484
915
621
327
687
363
694
525
856
441
897
598
429
760
508
814
645
351
807
392
718
549
880
586
921
627
328
784
465
700
531
862
663
369
604
435
741
447
898
Example 44
In this case, also the magic sum is
F252
S25×25 := 15625 = 253 , and sum of numbers is
:= 390625 = 6252 = 254 . The magic square is pan diagonal and bimagic with bimagic sum
Sb 25×25 := 10579425 . Moreover each block of order 5 is a pan diagonal magic square constant
magic sum S5×5 := 3125 = 55 .
Still, there are more possibilities of getting sequential numbers for the magic square of order 25,
where the numbers sum is a perfect square. See below examples,
T(988)  T(363) = 364 + 365 + ... + 988 = 422500 = 650 2 ;
T(1041)  T(416) = 417 + 418 + ... + 1041 = 455625 = 6752 ;
T(1096)  T(471) = 472 + 473 + ... + 1096 = 490000 = 7002 .
22. Summary
Summarizing the results obtained above, we lead to following interesting table:
32
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
Order of
Magic Square
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
…
Number of
Elements
Magic
Sum, S k  k
2
3
Sum of all
Numbers, F 2
k
4
3
42
52
62
72
82
92
102
3
43
53
63
73
83
93
103
3
44
54
64
74
84
94
104
112
113
114
122
132
123
133
124
134
142
152
162
172
182
192
20 2
212
143
153
163
173
183
193
203
213
144
154
164
174
184
194
20 4
214
22 2
232
223
233
22 4
234
24 2
252
…
243
24 4
254
...
253
…
Table 1
The above Table 1 is up to magic square of order 25. According to formulas given in Section 2,
we can extend it for higher orders. From the above Table 1, we derive the following general
values:
There exists magic squares following properties:
Order of Magic Square: k
k2
Number of Elements:
k3
Magic Sum:
k4
Sum of all numbers:
This is true for all order magic squares, i.e.,
k = 3, 4, 5, ...
Table 2
33
Inder J. Taneja
RGMIA Research Report Collection, 20 (2017), http://rgmia.org/v20.php
The examples of magic squares written above are the adaptations of author’s work on
intervally distributed magic squares studied in Taneja (2015a, 2015b, 2015c). For other kind of
magic squares refer Taneja (2015d, 2015e).
The examples given for the magic square of orders 8, 9, 16 and 25 are bimagic squares. The
order 14 magic square is based on the approach due to Zhu [10], and the order 22 is based on
the approach due to White [9].
Acknowledgement
The author wishes to thank William Walkington (France) for his comments and suggestions.
References
1. L. SALLOWS (2017), Online discussion, January.
2.
I.J. TANEJA (2015a), Intervally Distributed, Palindromic, Selfie Magic Squares, and Double Colored
Patterns, RGMIA Research Report Collection, Art. 127, Vol. 18, pp. 1-45,
http://rgmia.org/papers/v18/v18a127.pdf.
3. I.J. TANEJA (2015b), Intervally Distributed, Palindromic and Selfie Magic Squares: Genetic Table and
Colored Pattern - Orders 11 to 20, RGMIA Research Report Collection, Art. 140, Vol. 18, pp. 1-43,
http://rgmia.org/papers/v18/v18a140.pdf.
4.
I.J. TANEJA (2015c), Intervally Distributed, Palindromic and Selfie Magic Squares – Orders 21 to 25,
RGMIA Research Report Collection, Art. 151, Vol. 18, pp. 1-33, http://rgmia.org/papers/v18/v18a151.pdf.
5. I.J. TANEJA (2015e), Selfie Palindromic Magic Squares, RGMIA Research Report Collection, Art. 98,
Vol. 18, pp. 1-15, http://rgmia.org/papers/v18/v18a98.pdf.
6. I.J. TANEJA (2015d), Multi-Digits Magic Squares, RGMIA Research Report Collection, Art. 159, Vol.
18, pp. 1-22, http://rgmia.org/papers/v18/v18a159.pdf.
7.
W. TRUMP (2017), Online discussion, January, web-site: http://www.trump.de/magic-squares.
8. W. WALKINGTON (2017), Online discussion, January. For more details:
a. https://carresmagiques.blogspot.com.br/2017/01/area-magic-squares-and-tori-of-order-3.html
b. https://carresmagiques.blogspot.com.br/2017/01/area-magic-squares-of-order-6.html
c. http://www.primepuzzles.net/puzzles/puzz_865.htm
9. H. WHITE (2016), SODLS Construction Methods, http://budshaw.ca/addenda/SODLSmethods.html,
10. L. ZHU (1984), Orthogonal Diagonal Latin Squares of Order Fourteen, J. Austral. Math. Soc. (Series
A), 36, 1-3.
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34