TABLE OF RIGHT DIAGONALS
GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part VIB)
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.
It was shown previously that these numbers are a part of a sequence of squares and this page is a continuation of that effort.
I will show from scratch, (i.e. from first principles) that these tuples
(a^{2},b^{2},c^{2})
whose sum a^{2} + b^{2} +
c^{2} − 3b^{2} = 0
are generated from another set of tuples that (except for the initial set of tuples) obeys the equation
a^{2} + b^{2} +
c^{2} − 3b^{2} ≠ 0.
Find the Initial Tuples
As was shown in the web page Generation of Right Diagonals, the first seven tuples of
real squares, are generated using the formula c^{2}
= 2b^{2} − 1 and placed into table T below. The first number in each tuple
all a start with +1 which employ integer numbers as the initial entry in the diagonal.
The desired c^{2} is calculated by searching all b numbers
between 1 and 100,000. However, it was found that the ratio of b_{n+1}/b_{n} or c_{n+1}/c_{n} converges
on (1 + √2)^{2} as the b's or c's get larger. This means that moving down each row on the table
each integer value takes on the previous b_{n} or c_{n} multiplied by (1 + √2)^{2}, i.e.,
5.8284271247...
Furthermore, this table contains seven initial tuples in which all a start with −1 instead as of +1 as was shown in
Part VIA. The initial simple tuple (−1,1,1) is the only tuple stands on its own. Our fifth example is then (−1, 33461, 47321).
Table T
a_{n}  b_{n}  c_{n} 
−1  1  1 
−1  5  7 
−1  29  41 
−1  169  239 
−1  985  1393 
−1  5741  8119 
−1  33461  47321 
Construction of two Tables of Right Diagonal Tuples
 The object of this exercise is to generate a table with a set of tuples that obey the rule:
a^{2} + b^{2} +
c^{2} − 3b^{2} ≠ 0
and convert these tuples into a second set of tuples that obey the rule:
a^{2} + b^{2} +
c^{2} − 3b^{2} = 0.
 To generate table I we take the tuple (−1, 33461, 47321) and add 2 to each entry in the tuple to produce
Table I with +1 entries in the first column.
 We also set a condition for table I. We need to know two numbers e and
g where
g = 2e and which when added to the second and third numbers,
respectively, in the tuple of table I produce the two numbers in the next row of table I.
Table I
1  33463  47323 
1  33463+e  47323+g 
 These numbers, e and g are not initially known but a mathematical method will be shown below
on how to obtain them. Having these numbers on hand we can then substitute them into the tuple
equation 1^{2} + (en + 33463)^{2} +
(gn + 47323)^{2}
− 3(en +33463)^{2} along with n (the order), the terms squared
and summed to obtain a value
S which when divided by a divisor
d produces a number f.
 This number f when added to the square of
each member in the tuple (1,b,c) generates
(f + 1)^{2} +
(f + en
+ 33463)^{2} + f + gn + 47323)^{2}
− 3(f + en +33463)^{2}
producing the resulting tuple in table II. This is the desired tuple obeying the rule
a^{2} + b^{2} +
c^{2} − 3b^{2} = 0.
Table I
1  33463  47323 
1  33463+e  47323+g 

 ⇒ 
Table II
1  33463  47323 
1 + f  33463+e + f 
47323+g + f 


 The Δs are calculated, the difference in Table 2 between columns 2 or 3, and the results placed in the last column.
 The final tables produced after the algebra is performed are shown below:
n
0 
1 
2 
3 
4 
5 
6 
… 
18 
… 
25 


Table I
1  33463  47323 
1  33661  47719 
1  33859  48115 
1  34057  48511 
1  34255  48907 
1  34453  49303 
1  34651  49699 
…  …  … 
1  37027  54451 
…  …  … 
1  38413  57223 


f = S/d
−2 
280 
566 
856 
1150 
1448 
1750 
… 
5686 
… 
8248 


Table II
−1  33461  47321 
281  33941  47999 
567  34425  48681 
857  34913  49367 
1151  35405  50057 
1449  35901  50751 
1751  36401  51449 
…  …  … 
5687  42713  60137 
…  …  … 
8249  46661  65471 


Δ
1119638520 
1151912520 
1184759136 
1218183120 
1252189224 
1286782200 
1321966800 
… 
1792058400 
… 
2109202920 

To obtain e, g, f
and d the algebraic calculations are performed as follows:
 The condition we set is g = 2e
 Generate the equation: (1^{2} + (en + 33463)^{2}
+ (gn + 47323)^{2}
− 3(en + 33463)^{2} (a)
 Add f to the numbers in the previous equation:
(f + 1)^{2} +
(f + en + 33463)^{2} +
(f + gn + 47323)^{2}
− 3(f + en + 33463)^{2}
(b)
 Expand the equation in order to combine and eliminate terms:
(f^{2} + 2f + 1) +
(f^{2} + 2enf +
66926f + e^{2}n^{2} +
66926en + 1119772369)
+ (f^{2} + 2gnf +
94646f + g^{2}n^{2}
+ 94646gn + 2239466329) +
(−3f^{2} − 6en
f − 200778f − 3e^{2}n^{2}
− 200778en − 3359317107) = 0 (c)

−39204f + (2gn
f −4en f)
+ (g^{2}n^{2}
− 2e^{2}n^{2}) + (94646gn
− 133852en) − 78408 = 0 (d)
 Move f to the other side of the equation and
since g = 2e then
39204f = (4e^{2}n^{2}
−2e^{2}n^{2}) +
(189292en − 133852en) − 78408 (e)
39204f = 2e^{2}n^{2} +
55540en − 78408 (f)
 At this point the divisor d is equal to the coefficent of f,
i.e. d = 39204.
For 4 to divide the right side of
the equation we find the lowest value of e and g
which would satisfy the equation.
e = 198 and g = 396 are those numbers.
 Thus 39204f = 78408n^{2} + 10977120n − 78408
and (g)
f = 2n^{2} + 280n − 2 (h)
 Therefore substituting these values for e, f and
g into the requisite two equations affords:
for Table I: (1^{2} + (198n + 33463)^{2}
+ (396n + 47323)^{2}
− 3(198n + 33463)^{2} (i)
for Table II: (2n^{2} + 280n − 1)^{2} +
(2n^{2} + 478n + 33461)^{2} +
(2n^{2} + 676n + 47321)^{2}
− 3(2n^{2} + 478n + 33461)^{2}
(j)
Thus the values of the rows in both tables can be obtain by using a little arithmetic as was shown above or we can employ the two mathematical equations to generate
each row. The advantage of using this latter method is that any n can be used. With the former method one calculation
after another must be performed until the requisite n is desired.
Square A and B are examples of magic square of order number n = 18 and n = 25, respectively. They are generated from the produced from the tuples
(5687, 42713, 60137) and (8249, 46661, 65471), respectively, having the magic sums
5473201107 and 6531746763, respectively.
Magic square A
87725^{2}  5838933287  60137^{2} 
2254816487  42713^{2}  76835^{2} 
5687^{2}  97405^{2}  4046874887 

 
Magic square B
16004^{2}  23361425519  65471^{2} 
19143019679  46661^{2}  153289^{2} 
8249^{2}  166481^{2}  2121222599 


This concludes Part VIB.
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Copyright © 2011 by Eddie N Gutierrez. EMail: Fiboguti89@Yahoo.com