TABLE OF RIGHT DIAGONALS
GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part IIIA)
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} − 2 and placed into table T below. The first number in each tuple
all a start with +√2 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 +√2.
The initial simple tuple is (√2,1,0). Our first example is then (√2,99,140).
Table T
√2  1  0 
√2  3  4 
√2  17  24 
√2  99  140 
√2  577  816 
√2  3363  4756 
√2  19601  27720 
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.
Table I
√2  99  140 
√2  99+e  140+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 (√2)^{2} + (en + 99)^{2} +
(gn + 140)^{2}
− 3(en + 99)^{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 + √2)^{2} +
(f + en
+ 99)^{2} + f + gn + 140)^{2}
− 3(f + en + 99)^{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
√2  99  140 
√2  99+e  140+g 

 ⇒ 
Table II
√2  99  140 
√2 + f  99+e + f 
140+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
√2  99  140 
√2  181  304 
√2  263  468 
√2  345  632 


f = S/d
0 
2(116±2√2) 
6(116±2√2) 
12(116±2√2) 


Table II
√2  99  140 
232±5√2  413±4√2  536±4√2 
696±13√2  959±12√2  1164±12√2 
1392±25√2  1737±24√2  2024±24√2 

Table III (Table II Squared)
2  9801  19600 
53874±2320√2  170601±3304√2  287328±4288√2 
484754±18096√2  919969±23016√2  1355184±27936√2 
1938914±69600√2  3018321±83376√2  4097728±97152√2 


Δ
9799 
116727±984√2 
435215±4920√2 
1079407±13776√2 

To obtain e, g, f
and d the algebraic calculations are performed as follows:
 The condition we set is g = 2e
 Generate the equation: ((√2)^{2} + (en + 99)^{2}
+ (gn + 140)^{2}
− 3(en + 99)^{2} (a)
 Add f to the numbers in the previous equation:
(f ± √2)^{2} +
(f + en + 99)^{2} +
(f + gn + 140)^{2}
− 3(f + en + 99)^{2}
(b)
 Expand the equation in order to combine and eliminate terms:
(f^{2} ± 2√2f + 2) +
(f^{2} + 2enf +
198f + e^{2}n^{2} +
198en + 9801)
+ (f^{2} + 2gnf +
280f + g^{2}n^{2}
+ 280gn + 19600) +
(−3f^{2} − 6enf
− 594f − 3e^{2}n^{2}
− 594en − 29403) = 0 (c)

(±2√2f − 116f)
+ (2gnf −
4enf)
+ (g^{2}n^{2}
−2e^{2}n^{2}) + (280gn
− 396en) = 0 (d)
 Move f to the other side of the equation and
since g = 2e then
116f ± 2√2f =
(4e^{2}n^{2}
−2e^{2}n^{2}) +
(560en − 396en)
(e)
116f ± 2√2f =
2e^{2}n^{2} +
164en (f)
 At this point the divisor d is equal to the coefficient of f,
i.e. d = 116 − 2√2.
For 116 − 2√2 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
(116 ∓ 2√2) ⁄ (116 ∓ 2√2) 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} +
116en) ⁄ (116 ± 2√2) × (116 ∓2√2) ⁄ (116 ∓
2√2) (g)
f = [(2e^{2}n^{2} +
4en) × (116 ± 2√2)] ⁄ 13448 (h)
 At this point setting e = 82 and therefore, g = 164 affords
f = (n^{2} + n)(116 ± 2√2)
(i)
 Therefore substituting these values for e, f and g into the requisite two equations affords:
for Table I: ((√2)^{2} + (82n + 99)^{2}
+ (164n + 140)^{2}
− 3(82n + 99)^{2} (j)
for Table II:
((n^{2} + n)(116 ± 2√2)
± √2)^{2} +
((n^{2} + n)(116 ± 2√2) + 82n + 99)^{2}
+ ((n^{2} + n)(116 ± 2√2) + 164n + 140)^{2}
− 3((n^{2} + n)(116 ± 2√2) + 82n + 99)^{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
(232 ± 5√2, 413 ± 4√2, 536 ± 4√2), the second
(696 ± 13√2, 959 ± 12√2, 1164 ± 12√2) and the third
(1392 ± 25√2, 1737 ± 24√2, 2024 ± 24√2)
Magic square A
119081 ± 1468√2  105394 ± 4156√2  287328 ± 4288√2 
338848 ± 2383√2  170601 ± 3304√2  2354 ± 480√2 
53874 ± 2320√2  235808 ± 2452√2  222121 ± 1399√2 

 
Magic square A (Unsquared)
(3 ± 244√2)^{2}  105394 ± 4156√2  (536 ± 4√2)^{2} 
338848 ± 2383√2  (413 ± 4√2)^{2}  (48 ± 5√2)^{2} 
(232 ± 5√2)^{2}  235808 ± 2452√2  222121 ± 1399√2 

Magic square B
523336 ± 57120√2  881387 ∓ 16008√2  1355184 ± 27936√2 
1751817 ∓ 6168√2  919969 ± 23016√2  88121 ± 52200√2 
484754 ± 18096√2  958551 ± 62040√2  1316602 ∓ 11088√2 

 
Magic square B (Unsquared)
(56 ± 510√2)^{2}  881387 ∓ 16008√2  (1164 ± 12√2)^{2} 
1751817 ∓ 6168√2  (959 ± 12√2)^{2}  (261 ± 100√2)^{2} 
(696 ± 13√2)^{2}  958551 ± 62040√2  1316602 ∓ 11088√2 

Magic square C
1243409 ∓ 61464√2  3713826 ± 214440√2  4097728 ± 97152√2 
5872640 ± 241992√2  3018321 ± 83376√2  164002 ∓ 75240√2 
1938914 ± 69600√2  2322816 ∓ 47688√2  4793233 ± 228216√2 

 
Magic square C (Unsquared)
(39 ∓ 788√2)^{2}  6162 ± 4296√2  (2024 ± 24√2)^{2} 
5872640 ± 241992√2  (1737 ± 24√2)^{2}  (380 ∓ 99√2)^{2} 
(1392 ± 25√2)^{2}  2322816 ∓ 47688√2  4793233 ± 228216√2 

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