TABLE OF RIGHT DIAGONALS GENERAL METHOD
GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part IVC)
Square of Squares Tables
Andrew Bremner's article on squares of squares included the 3x3 square:
Bremmer's square
58^{2}  46^{2}  127^{2} 
94^{2}  113^{2}  2^{2} 
97^{2}  82^{2}  74^{2} 
The numbers in the right diagonal as the tuple (97^{2},113^{2},127^{2}) appears to have come out of the blue. But I will show that
this sequence is part of a larger set of tuples having the same property, i.e. the first number in the tuple when added to a
difference (Δ) gives the second square in the tuple and when this same
(Δ) is added to the second square produces a third square.
All these tuple sequences can be used as entries into the right diagonal of a magic square.
We will show a general method for generating the squares for a right diagonal of a magic square. Beginning with the the tuple
(1,b1,
c1) we can generate the tuple (a, b,
c) which when squared gives the diagonal numbers. Initially either b1 or c1 will be equal to ± k
where k is any natural number 1,2,3,4.... Again the end result is that a_{1}^{2} +
b_{1}^{2} +
c_{1}^{2} − 3b_{1}^{2}
≠ 0 = S is converted to a^{2} +
b^{2} +
c^{2} − 3b^{2} = 0 which is
a necessary condition for the square to be magic.
To summarize the tuples of Table II below will be used as entries into a right diagonal of a magic square. Knowing the difference
(b^{2} − a^{2}) or
(c^{2} − b^{2}) will give us a value
Δ which can be used to produce other entries into the magic square.
To date only one magic square containing 7 entries has been found. Most other squares will contain 6 entries.
As to the reason for the picture of a square, the entries to the square occur as three tuples,viz, (a,b,c),
(l,m,n) and (x,y,z) showing their connectivity. In addition,
six or more of these entries are present as their squares.
Generation of Tables where b_{1} = 33
 The object of this exercise is to generate a Table I with a set of tuples that obey the rule:
a_{1}^{2} +
b_{1}^{2} +
c_{1}^{2} − 3b_{1}^{2}
≠ 0
and convert these tuples into a second set of tuples (Table II) that obey the rule:
a^{2} + b^{2} +
c^{2} − 3b^{2} = 0.
 In addition, we need to know two numbers e and
g where
g = 2e which when added to the
b_{1} and c_{1} numbers of Table I,
produce the next line of numbers (n + 1) in the next row of Table I. The number a_{1} will always be 1.
 Two other numbers f and d
are calculated using the equation
f = [2e^{2}n^{2} +
(4c_{1} − 4b_{1}) en +(1 − 2b_{1}^{2} + c_{1}^{2})]
/ {2(2b_{1} − c_{1} − 1)}
where n is the line number of the tables. f can also be generated directly from Table II from
S/d. However, the value of d
is equal to the denominator of the general equation above.
 Finally Δs are calculated by taking the difference in Table II between
(b^{2} − a^{2}) or
(c^{2} − b^{2}),
and the results placed under the Δ column. Both differences must be the same.
 As an example we begin with the tuple (1,33,1), where a_{1} = 1,
b_{1} = 33
and c_{1} = 1 and use the equation to generate f.
f = [2
e^{2}n^{2} +
(4 − 132)
en − 2176]
/ 2× (64) =
[2
e^{2}n^{2} −
128
en − 2176]
/ (128)
Setting
e = 8 and
g = 16 affords
f =
n^{2} − 8
n − 17
Substituting for f in
(b) gives
a =
(n^{2} −8n − 16 )
b =
(n^{2} + 16)
c =
(n^{2} + 8n − 16)
 Substituting the appropriate n into the equations for a, b,
and c produces Table II below. Using a computer program and the
requisite calculations produced the tables below.
As can be seen taking the value of f from the middle table and adding to a_{1},
b_{1}, c_{1},
produced a, b, c, respectively of Table II.
n 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 


Table I
a_{1} 
b_{1}  c_{1} 
1  33  1 
1  41  17 
1  49  33 
1  57  49 
1  65  65 
1  73  81 
1  81  97 
1  89  113 
1  97  129 
1  105  145 
1  113  161 
1  121  177 
1  129  193 


f = S/d 
17 
24 
29 
32 
33 
32 
29 
24 
17 
8 
3 
16 
31 


Table II
a 
b  c 
16  16  16 
23  17  7 
28  20  4 
31  25  17 
32  32  32 
31  41  49 
28  52  68 
23  65  89 
16  80  112 
7  97  137 
4  116  164 
17  137  193 
32  160  224 


Δ 
0 
240 
384 
336 
0 
720 
1920 
3696 
6144 
9360 
13440 
18480 
24576 

The magic square A was found by Bremner and has 7 square terms with the magic sum (S_{m}) 541875.
Two other examples are B and C having
the right diagonal tuple (368, 592, 752) and (2047, 2113, 2177)as their squares. The tuple
(368, 592, 752) is not on the diagonal in Magic Square C since the tuple (2047, 2113, 2177) gives a larger sum.
The magic sum, S_{m}, for these cases are 1051392 and 13394307, respectively and the n is 24.
Magic square A
373^{2}  289^{2}  565^{2} 
360721  425^{2}  23^{2} 
205^{2}  527^{2}  222121 

 
Magic square B
464^{2}  270592  752^{2} 
700672  592^{2}  16^{2} 
368^{2}  656^{2}  485632 

 
Magic square C
592^{2}  8364034  2177^{2} 
8794114  2113^{2}  368^{2} 
2047^{2}  752^{2}  8579074 

This concludes Part IVC. To go back to Part IVB. To continue to Part IVD
which treats tuples of the type (1,65,1).
Go back to homepage.
Copyright © 2012 by Eddie N Gutierrez. EMail: Fiboguti89@Yahoo.com