TABLE OF RIGHT DIAGONALS
GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF COMPLEX SQUARES (Part IB)
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
complex squares, are generated using the formula c^{2}
= 2b^{2} + 2 and placed into table T below. The first number in each tuple
all a start with +√2i 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 +√2i.
The initial simple tuple is (√2i,1,2) and thus is the first example.
Table T
√2i  1  2 
√2i  7  10 
√2i  41  58 
√2i  239  338 
√2i  1393  1970 
√2i  8119  11482 
√2i  47321  66922 
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. The initial row, however,
of table I is identical to the first row of table II. On the other hand, there are other examples where this is not true.
 To accomplish this we set a condition. 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. Every first number, however, remains a 1.
 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 (√2i)^{2} + (en + 1)^{2} +
(gn + 2)^{2}
− 3(en +1)^{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 + √2i)^{2} +
(f + en
+ 1)^{2} + f + gn + 2)^{2}
− 3(f + en +1)^{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 II
√2i  1  2 
√2i + f  1+e + f 
2+g + f 


 This explains why when both n and f are both equal to 0
that the first row of both tables are equal.
 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:


Table I
√2i  1  2 
√2i  3  6 
√2i  5  10 
√2i  7  14 


f = S/d
0 
4(√2i) 
12(√2i) 
24(√2i) 


Table II
√2i  1  2 
5√2i  3+4√2i  6+4√2i 
13√2i  5+12√2i  10+12√2i 
25√2i  7+24√2i  14+24√2i 

Table III
√2i  1  2 
50  69−72√2i  4+48√2i 
338  263+120√2i  188+240√2i 
1250  1103+336√2i  956+672√2i 


Δ
3 
27+24√2i 
75+120√2i 
147+336√2i 

To obtain e, g, f
and d the algebraic calculations are performed as follows:
 The condition we set is g = 2e
 Generate the equation: ((√2i)^{2} + (en + 1)^{2}
+ (gn + 2)^{2}
− 3(en +1)^{2} (a)
 Add f to the numbers in the previous equation:
(f ± √2i)^{2} +
(f + en + 1)^{2} +
(f + gn + 2)^{2}
− 3(f + en + 1)^{2}
(b)
 Expand the equation in order to combine and eliminate terms:
(f^{2} ± 2√2if − 2) +
(f^{2} + 2enf +
2f + e^{2}n^{2} +
2en + 1)
+ (f^{2} + 2gnf +
4f + g^{2}n^{2}
+ 4gn + 4) +
(−3f^{2} − 6enf
− 6f − 3e^{2}n^{2}
− 6en − 3) = 0 (c)

(±2√2if
+ (2gnf −
4enf)
+ (g^{2}n^{2}
−2e^{2}n^{2}) + (4gn
− 4en) = 0 (d)
 Move f to the other side of the equation and
since g = 2e then
∓ 2√2f =
(4e^{2}n^{2}
−2e^{2}n^{2}) +
(8en − 4en)
(e)
∓ 2√2if =
2e^{2}n^{2} +
4en (f)
 At this point the divisor d is equal to the coefficient of f,
i.e. d = ∓ 2√2i.
For ∓ 2√2i to divide the right side of
the equation we find the lowest value of e and g
which would satisfy the equation.
To obtain e and g we must first multiply by the factor
(± 2√2i) ⁄ (± 2√2i) which gives (g) followed by
(h).
And where the + of the first factor in the denominator is multiplied
by the − of the second factor in the denominator and vice versa.
 f = (4e^{2}n^{2} +
4en) ⁄ (± 2√2i) × (∓2√2i) ⁄ (∓
2√2i) (g)
f = [(2e^{2}n^{2} +
4en) × (± 2√2i)] ⁄ 8 (h)
 At this point setting e = 2 and therefore, g = 4 affords
f = (n^{2} + n)(± 2√2i)
(i)
 Therefore substituting these values for e, f and g into the requisite two equations affords:
for Table I: ((√2i)^{2} + (2n + 1)^{2}
+ (4n + 2)^{2}
− 1(2n +3)^{2} (j)
for Table II:
((2n^{2} + 2n)(± √2i)
± √2i)^{2} +
((n^{2} + n)(± √2i) + 2n + 1)^{2}
+ ((2n^{2} + 2n)(± √2i) + 4n + 2)^{2}
− 3((2n^{2} + 2n)(± √2i) + 2n + 1)^{2}
(k)
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.
 Three examples are listed on for each of the latter three entries in table III where the radix part can be either + or −. The first has the tuple
(±5√2, 3 ± 4√2, 6 ± 4√2), the second
(±13√2, 5 ± 12√2, 10 ± 12√2) and the third
(±25√2, 7 ± 24√2, 14 ± 24√2)
Magic square A
92 ± 336√2i  19 ∓ 322√2i  4 ± 48√2i 
73 ∓ 264√2i  23 ± 24√2i  119 ± 312√2i 
50  65 ± 360√2i  46 ∓ 288√2i 

 
Magic square A (Unsquared)
(14 ± 12√2i)^{2}  19 ∓ 312√2i  (2 ± 4√2i)^{2} 
73 ∓ 264√2i  (3 ± 4√2i)^{2}  (13 ± 12√2i)^{2} 
( ±5√2i)^{2}  65 ± 360√2i  46 ∓ 288√2i 

Magic square B
28 ∓ 24√2i  949 ± 144√2i  188 ± 240√2i 
1099 ± 384√2i  263 ± 120√2i  47 ∓ 144√2i 
338  103 ± 96√2i  1024 ± 264√2i 

 
Magic square B (Unsquared)
(6 ∓ 2√2i)^{2}  949 ± 144√2i  (10 ± 12√2i)^{2} 
1099 ± 384√2i  (5 ± 12√2i)^{2}  (9 ∓ 8√2i)^{2} 
( ±13√2i)^{2}  103 ± 996√2i  1024 ± 264√2i 

Magic square C
124 ± 360√2i  4141 ∓ 24√2i  956 ± 672√2i 
4435 ± 648√2i  1103 ± 336√2i  23 ± 24√2i 
1250√2i  271 ± 696√2i  4288 ± 312√2i 

 
Magic square C (Unsquared)
(18 ± 10√2i)^{2}  4141 ∓ 24√2i  (14 ± 24√2i)
^{2} 
4435 ± 648√2i  (7 ± 24√2i)^{2}  (3 ± 4√2i)^{2} 
(±25√2i)^{2}  271 ± 696√2i  4228 ± 312√2i 

 The following equations were used for calculating squares and square roots and are described in the page noncomplex
square root equations:
Multiplication = (a + b√2i) × (a + √2i) = a^{2} − 2b^{2} + 2ab√2i
Division = (a + b√2i) / (a + b√2i)
Square root of (a + b√2i) :
r = sqrt(a^{2} − 2b^{2})
r1 = x + y√2i
r2 = −x − y√2i
where,
x = sqrt((r − a) / 2) and
y = b / 2x
This concludes Part IB.
Part IIA continues using the √7 in complex square root numbers.
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Copyright © 2011 by Eddie N Gutierrez. EMail: Fiboguti89@Yahoo.com