A New Procedure for Magic Squares (Part IB)
Consecutive Knight Break MaskGenerated Squares
A Discussion of the New Method
Magic squares such as the Loubère have a center cell which must always contain the middle number of
a series of consecutive numbers, i.e. a number which is equal to one half the sum of the first and last numbers of the series, or
½(n^{2} + 1). The properties of these regular or associated Loubère squares are:
 That the sum of the horizontal rows,
vertical columns and corner diagonals are equal to the magic sum S.
 The sum of any two numbers that are diagonally equidistant from the center (DENS) is equal to
n^{2} + 1, i.e., or twice the number in the center cell and are complementary to each other.
In this method the numbers on the square are placed consecutively starting from the first leftmost column and entered across every other cell until the end
of the row is reached. A (2 down, 2 right) knight break is used to get to the next line. (This knight break is also equivalent to a slant 2 break in the right down direction).
The next line is then added from left to righty.
The square obtained, which is not magic, is modified into a form
which can be converted into a magic one by the use of a mask. This mask generates numbers which are added to certain cells in the square to produce
a final square composed of numbers which may not be in serial order. For example, negative numbers or numbers greater
than n^{2} may be present in the square.
In addition, it will also be shown that the sums of these squares follow the sum equation shown in the
New block Loubère Method.
S = ½(n^{3} ± an)
Construction of a 5x5 Magic Square
Method: Reading regular from left to right and use of a mask
 Construct the 5x5 Square 1 where 5 = 4n + 1 by adding numbers in a consecutive manner starting at row 1 cell 1, on reaching the end of the row
knight break (2 down, 2 right) and continue adding numbers from left to right (squares 1 and 2).
 Since all sums of all the columns or rows are not equal to 65 add or subtract the numbers in the last row from those numbers in the center row. Then add or subtract the
numbers in the last column to those in the center column.
At this point no duplicates has been generated (Square 3).

⇒ 
2
 60  
1  14  2 
15  3  35  30 
21  9  22 
10  23  85  20 
16  4  17 
5  18  60  5 
11  24  12 
25  13  85  20 
6  19  7 
20  8  60  5 
55  70  60 
75  65  60  
10  5  5 
10  0   

⇒ 
3
 70 
1  14  32 
15  3  65 
21  9  2 
10  23  65 
26  1  27 
5  18  65 
11  24  8 
25  13  65 
6  19  12 
20  8  65 
65  65  65 
65  65  70 

 Generate a mask whereby the sums of the columns and rows are constructed as in the box below. This assures that when each
of these values is added to the corresponding cell in square 4 (as in the de la Hire method) that all sums will equal a magic sum.
 We start by subtracting the diagonals(60,60) from 65 to give both 5 and which will be used as what I call the
"de la Hire constants".
However, this constant is too small and will generate too many duplicates.
The constant to use are both 20 and 5 which will produce 90 as the magic presum.
 The following equations are used such that the following conditions are obeyed:
The right diagonal and left diagonals: 90 = 70 + 20
The rows and columns: 90 = 65 + 20 + 5.
 Generate the mask using the 20 and 5 factors or sums thereof (55) adding these factors to the appropriate cells in square 3 to generate square 4.
 Square 4 has a magic sum equal to 90, i.e., S = 90 = ½(n^{3} + 11n).
3
 70 
1  14  32 
15  3  65 
21  9  2 
10  23  65 
26  1  27 
5  18  65 
11  24  8 
25  13  65 
6  19  12 
20  8  65 
65  65  65 
65  65  70 

+ 
Mask A
  
5  20 
 20  
 5 
20  5  
 
5   20 
 
  5 
20  

⇒ 
4
 90 
1  14  32 
20  23  90 
21  29  2 
10  28  90 
46  4  27 
5  18  90 
16  24  12 
25  13  90 
6  19  17 
40  8  90 
90  90  90 
90  90  90 

Construction of a 7x7 Magic Square
Method: Reading consecutive from left to right and use of a mask
 Construct the 7x7 Square 1 where 7 = 4n + 3 by adding numbers in a consecutive manner starting at row 1 cell 1, on reaching the end of the row
knight break (2 down, 2 right) and continue adding numbers from left to right (squares 5 and 6).
 Since all sums of all the columns or rows are not equal to 175 add or subtract the numbers in the last row from those numbers in the center row. Then add or subtract the
numbers in the last column to those in the center column.
At this point one duplicate has been generated (Square 7).
5
1   2 
 3   4 
15   16 
 17   18 
 5  
6   7  
 19  
20   21  
8   9 
 10   11 
22   23 
 24   25 
 12  
13   14  

⇒ 
6
 189  
1  26  2  27  3  28 
4  91  84 
15  40  16 
41  17  42  18  189  14 
29  5  30 
6  31  7  32  140  35 
43  19  44 
20  45  21  46  238  63 
8  33  9 
34  10  35  11  140  35 
22  47  23 
48  24  49  25  238  63 
36  12  37 
13  38  14  39  189  14 
154  182  161 
189  168  196 
175  189  
21  7  14 
14  7  21  0   

⇒ 
7
 112 
1  26  2  111  3  28 
4  175 
15  40  16 
27  17  42  18  175 
29  5  30 
41  31  7  32  175 
64  12  58 
57  52  0 
46  175 
8  33  9 
69  10  35  11  175 
22  47  23 
15  24  49  25  175 
36  12  37 
1  38  14  39  175 
175  175  175 
175  175  175 
175  112 

+ 
 Generate a mask whereby the sums of the columns and rows are constructed as in the box below. This assures that when each
of these values is added to the corresponding cell in square 7 (as in the de la Hire method) that all sums will equal to
a magic sum.
 We start by subtracting the diagonals(112,112) from 175 to give 63, and which will be used as what I call the
"de la Hire constant".
Addition of 63 to 175 gives 238 a magic presum.
 The following equations are used such that
the following conditions are obeyed:
The right and left diagonals: 238 = 112 + 2(63)
The rows and columns: 238 = 175 + 63
 Generate the mask using the 63 factor, adding this factor to the appropriate cells in square 7 generates square 8.
 Square 8 has a magic sum equal to 238, i.e., S = 238 = ½(n^{3} + 19n).
Mask B
  63 
   
  
  63  
63   
   
  
63    
  
 63   
  
   63 
 63  
   

⇒ 
8
 238 
1  26  65  111  3  28 
4  238 
15  40  16 
27  17  105  18  238 
92  5  30 
41  31  7  32  238 
64  12  58 
6  52  0 
46  238 
8  33  9 
69  73  35  11  238 
22  47  23 
15  24  49  88  238 
36  75  37 
1  38  14  39  238 
238  238  238 
238  238  238 
238  238 

This completes this section on a new Consecutive Knight Break MaskGenerated Squares (Part IB). The next section deals with
Consecutive Knight Break MaskGenerated Squares (Part IC). To return to homepage.
Copyright © 2010 by Eddie N Gutierrez. EMail: Fiboguti89@Yahoo.com