A New Procedure for Magic Squares (Part II)
Loubère Block Modified Squares
Continuation using 7x7 Squares
As shown in the previous page a new Loubère method Modified Loubère Block Modified 5x5 Squares was introduced where
numbers are added consecutively starting with 1 and ending with n^{2}. The final square is transformed into a modified
Loubère, having numbers greater than n^{2} by modifying a block or grid of numbers. The initial number is added either at the normal Loubère
sites, i.e., (the center of the last row or last column) or on the main diagonal.
This is done by taking the nonmagic squares
and converting them into magic ones using a variety of means. These squares are depicted below in methods I and II.
The generation of these magic squares involves an almost normal approach:
Numbers are added consecutively starting out with 1 and ending with numbers greater
than n^{2} after modification. A break involves
translational moves (up, down or sideways).
Moreover, it must be stated here that the magic sum has been modified from the known equation
S = ½(n^{3} + n) to
the general equation:
>
S = ½(n^{3} ± an)
which takes into account these new squares. The variable a, an odd number, is equal to 1,3,5,7 or ... and may take on + or 
values. For example when a = 1 the normal magic sum S is implied.
When a takes on different odd values S gives the magic sum of a modified magic square.
It will be shown that the addition or subtraction of n^{2} to some of the cells in
the square gives rise to a new magic square.
Construction of 7x7 Block Modified Loubère Squares
Method I: Start at lower left hand corner (1, 2, 3 ⇒ break)
 Place a 1 into the lower left hand corner.
 Add consecutive numbers up to the middle cell.
 Move, i.e. break, one cell down.
 Repeat the process until the square is filled as shown below in squares 12.
 As shown below this square is not magic because the three columns and three rows (in grey) sum to 147 instead of 196.
 Where the sums 147 cross corresponds to the small square in blue.
 Since the numbers 1, 2 and 3 sit on a diagonal which sums to 196, these numbers remain unchanged.
 Adding n^{2} = 49 to (34, 10 and 4) or to (42, 43, 12)
(color change to light green)
gives the two magic squares 3 and 4 where S = ½(n^{3} + 7n).

⇒ 
2
 196 
9  11  20  29 
38  40  49  196 
17  19  25  37 
39  48  8  196 
18  27  36  45 
47  7  16  196 
26  35  44  46 
6  15  24  196 
34  43  3  5 
14  23  25  147 
42  2  4  13 
22  31  33  147 
1  10  12  21 
30  32  41  147 
147  147  147 
196  196  196 
196  196 

⇒ 
3
 196 
9  11  20  29 
38  40  49  196 
17  19  25  37 
39  48  8  196 
18  27  36  45 
47  7  16  196 
26  35  44  46 
6  15  24  196 
83  43  3  5 
14  23  25  196 
42  2  53  13 
22  31  33  196 
1  59  12  21 
30  32  41  196 
196  196  196 
196  196  196 
196  196 

+ 
4
 196 
9  11  20  29 
38  40  49  196 
17  19  25  37 
39  48  8  196 
18  27  36  45 
47  7  16  196 
26  35  44  46 
6  15  24  196 
34  92  3  5 
14  23  25  196 
91  2  4  13 
22  31  33  196 
1  10  61  21 
30  32  41  196 
196  196  196 
196  196  196 
196  196 

Method II: Start at lower left hand corner (1, 2, 3, 4 ⇒ break)
 Place a 1 into the lower left hand corner.
 Add 4 consecutive numbers to past the center cell.
 Move, i.e. break, one cell down.
 Repeat the process until the square is filled as shown below in squares 12.
 As shown below this square is not magic because the two columns and two rows (in grey) sum to 154 instead of 203.
 Where the sums 154 cross corresponds to the small square in blue.
 Since the numbers 1, 2, 3 and 4 sit on a diagonal which sums to 154, at least one of these numbers must be changed and since the left diagonal also
sums to 154 the number to change is 4.
 Adding n^{2} = 49 to either (4, 10, 11, 34) or to (4, 12, 42, 43) gives the magic squares 3 and 4
where S = ½(n^{3} + 9n).

⇒ 
2
 154 
9  18  20  29 
38  40  49  203 
17  19  28  37 
46  48  8  203 
25  27  36  45 
47  7  16  203 
26  35  44 
4  6  15  24  154 
34  43  3 
5  14  23  32  154 
42  2  11 
13  22  31  33  154 
1  10  12 
21  30  39  41  154 
154  154  154 
154  203  203 
203  154 

⇒ 
3
 203 
9  18  20  29 
38  40  49  203 
17  19  28  37 
46  48  8  203 
25  27  36  45 
47  7  16  203 
26  35  44 
53  6  15  24  203 
83  43  3 
5  14  23  32  203 
42  2  60 
13  22  31  33  203 
1  59  12 
21  30  39  41  203 
203  203  203 
203  203  203 
203  203 

+ 
4
 203 
9  18  20  29 
38  40  49  203 
17  19  28  37 
46  48  8  203 
25  27  36  45 
47  7  16  203 
26  35  44 
53  6  15  24  203 
34  92  3 
5  14  23  32  203 
91  2  11 
13  22  31  33  203 
1  10  61 
21  30  39  41  203 
203  203  203 
203  203  203 
203  203 

Method III: Start at lower left hand corner (1, 2, 3, 4, 5, 6 ⇒ break)
 Place a 1 into the lower left hand corner.
 Add 5 consecutive numbers to past the center cell.
 Move, i.e. break, one cell down.
 Repeat the process until the square is filled as shown below in squares 12.
 As shown below this square is not magic because the two columns and two rows (in grey) sum to 70 or 168 instead of 217.
 Where the sums 168 cross corresponds to the small square in blue.
 Since the numbers 1, 2, 3, 4, 5 and 6 sit on a diagonal which sums to 70, at least two of these numbers must be changed and since the left diagonal also
sums to 167 the number to change is 4.
 Adding n^{2} = 49 to either (4, 6, 11, 25,30, 43) or to (1, 4, 7, 19, 43, 46) gives
the magic squares 3 and 4 where S = ½(n^{3} + 13n).

⇒ 
2
 70 
9  18  27  29 
38  47  49  217 
17  26  28 
37 
46  6  8  168 
25  34  36 
45 
5  7  16  168 
33  35  44 
4  13  15  24 
168 
41  43  3 
12  14  23  32 
168 
42  2  11 
20  22  31  40 
168 
1  10  19 
21  30  39  48 
168 
168  168  168 
168  168  168 
217  168 

⇒ 
3
 217 
9  18  27  29 
38  47  49  217 
17  26  28 
37 
46  55  8  217 
74  34  36 
45 
5  7  16  217 
33  35  44 
53  13  15  24 
217 
41  92  3 
12  14  23  32 
217 
42  2  60 
20  22  31  40 
217 
1  10  19 
21  79  39  48 
217 
217  217  217 
217  217  217 
217  217 

+ 
4
 217 
9  18  27  29 
38  47  49  217 
17  26  28 
37 
95  6  8  217 
25  34  36 
45 
5  56  16  217 
33  35  44 
53  13  15  24 
217 
41  92  3 
12  14  23  32 
217 
42  2  60 
20  22  31  40 
217 
50  10  19 
21  30  39  48 
217 
217  217  217 
217  217  217 
217  217 

Method IV: Start at the center of the last column (1, 2, 3, 4 ⇒ break))
 Place a 1 into the lower left hand corner.
 Add consecutive numbers 2,3,4 up to the top cell.
 Move, i.e. break, one cell down.
 Repeat the process until the square is filled as shown below in squares 12.
 As shown below this square is not magic because the two columns and two rows (in grey) sum to 154 instead of 203.
 Where the sums 154 cross corresponds to the small square in blue.
 Since the numbers 35,3 and 13 on the left diagonal which sums to 154 at least one of these numbers must be changed.
 Two examples are obtained by adding n^{2} = 49 to either
(1, 5, 11, 35) or to (1, 2, 3, 4) giving the magic squares 3 and 4 where S = ½(n^{3} + 9n).

⇒ 
2
 203 
35  44  4  6 
15  24  26  154 
43  3  5 
14  23  32  34  154 
2  11  13 
22  31  33  42  154 
10  12  21 
30  39  41  1  154 
18  20  29 
38  40  49  9 
203 
19  28  37 
46  48  8  17 
203 
27  36  45 
47  7  16  25 
203 
154  154  154 
203  203  203 
154  154 

⇒ 
3
 203 
84  44  4  6 
15  24  26  203 
43  3  54 
14  23  32  34  203 
2  60  13 
22  31  33  42  203 
10  12  21 
30  39  41  50  203 
18  20  29 
38  40  49  9 
203 
19  28  37 
46  48  8  17 
203 
27  36  45 
47  7  16  25 
203 
203  203  203 
203  203  203 
203  203 

+ 
4
 203 
35  44  53  6 
15  24  26  203 
43  52  5 
14  23  32  34  203 
51  11  13 
22  31  33  42  203 
10  12  21 
30  39  41  50  203 
18  20  29 
38  40  49  9 
203 
19  28  37 
46  48  8  17 
203 
27  36  45 
47  7  16  25 
203 
203  203  203 
203  203  203 
203  203 

This completes this section on the new block Loubère Method (Part I). The next section deals with
A New Procedure for Méziriac type magic squares (Part III). To return to Part I.
To return to homepage.
Copyright © 2009 by Eddie N Gutierrez. EMail: Fiboguti89@Yahoo.com