TABLE OF RIGHT DIAGONALS
GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part IVA)
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.
The initial simple tuple (1,1,1) is the only tuple stands on its own. Our fourth example is then (1,985,1393).
Table T
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. 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
1  985  1393 
1  985+e  1393+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 + 29)^{2} +
(gn + 41)^{2}
 3(en +29)^{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
+ 29)^{2} + f + gn + 41)^{2}
 3(f + en +5)^{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  985  1393 
1  985+e  1393+g 

 ⇒ 
Table II
1  985  1393 
1 + f  985+e + f 
1393+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:
n
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 


Table I
1  985  1393 
1  1009  1441 
1  1033  1489 
1  1054  1537 
1  1081  1585 
1  1105  1633 
1  1129  1681 
1  1153  1729 
1  1177  1777 
1  1201  1825 
1  1225  1873 
1  1249  1921 
1  1273  1969 


f = S/d
0 
35 
72 
111 
152 
195 
240 
287 
336 
387 
440 
495 
552 


Table II
1  985  1393 
36  1044  1476 
73  1105  1561 
112  1168  1648 
153  1233  1737 
196  1300  1828 
241  1369  1921 
288  1440  2016 
337  1513  2113 
388  1588  2212 
441  1665  2313 
496  1744  2416 
553  1825  2521 


Δ
970224 
1088640 
1215696 
1351680 
1496880 
1651584 
1816080 
1990656 
2175600 
2371200 
2577744 
2795520 
3024816 

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 + 985)^{2}
+ (gn + 1393)^{2}
 3(en +985)^{2} (a)
 Add f to the numbers in the previous equation:
(f + 1)^{2} +
(f + en + 985)^{2} +
(f + gn + 1393)^{2}
 3(f + en +985)^{2}
(b)
 Expand the equation in order to combine and eliminate terms:
(f^{2} + 2f + 1) +
(f^{2} + 2enf +
1970f + e^{2}n^{2} +
1970en + 970225)
+ (f^{2} + 2gnf +
2786f + g^{2}n^{2}
+ 2786gn + 1940449) +
(−3f^{2} − 6en
f −
5910f − 3e^{2}n^{2}
− 5910en  85683) = 0 (c)

1152f + (2gnf  4en
f)
+ (g^{2}n^{2}
2e^{2}n^{2}) + (2786gn
 3940en) = 0 (d)
 Move f to the other side of the equation and
since g = 2e then
1152f = (4e^{2}n^{2}
2e^{2}n^{2}) +
(5572en  3940en)
(e)
1152f = 2e^{2}n^{2} +
1632en (f)
 At this point the divisor d is equal to the coefficent of f, i.e. d = 1152.
For 1152 to divide the right side of
the equation we find the lowest value of e and g
which would satisfy the equation.
e = 24 and g = 48 are those numbers.
 Thus 1152f = 1152n^{2} + 39168n
and (g)
f = n^{2} + 34n (h)
 Therefore substituting these values for e, f and g into the requisite two equations affords:
for Table I: (1^{2} + (24n + 985)^{2}
+ (48n + 1393)^{2}
 3(24n + 985)^{2} (i)
for Table II: (n^{2} + 34n + 1)^{2} +
(n^{2} + 58n> + 985)^{2} +
(n^{2} + 82n + 1393)^{2}
 3(n^{2} + 58n + 985)^{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.
Three squares that can be formed from order number n = 1. (There may be more.) These magic squares (A) and (B) are
produced from the tuple (1103, 1537, 1873). A third square formed from the two other tuples from magic squares (A) and (B) as shown
in magic square (C).
Magic square A
1332^{2}  682992  1876^{2} 
1494208  1044^{2}  828^{2} 
36^{2}  1692^{2}  405648 

 
Magic square B
2034^{2}  3056389  1476^{2} 
879109  1044^{2}  1476^{2} 
36^{2}  2286^{2}  1967749 

 
Magic square C
2034^{2}  1677348  1692^{2} 
499932  1332^{2}  1746^{2} 
828^{2}  2286^{2}  588708 

Three squares that can be formed from order numbers n = 3, 4 or 11, respectively. (There may be more).
Magic square D
1952^{2}  2433536  1648^{2} 
269824  1168^{2}  1568^{2} 
116^{2}  2272^{2}  1081856 

 
Magic square E
4539^{2}  19058823  1737^{2} 
16065063  1233^{2}  4371^{2} 
153^{2}  4701^{2}  17561943 

 
Magic square F
1864^{2}  186944  2416^{2} 
5404096  1744^{2}  824^{2} 
496^{2}  2504^{2}  2608576 

This concludes Part IVA. To continue to Part IVB which treats tuples of the type (−1,b,c).
To continue to Part V.
Go back to homepage.
Copyright © 2011 by Eddie N Gutierrez. EMail: Fiboguti89@Yahoo.com