TABLE OF RIGHT DIAGONALS

GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part VIB)

Picture of a square

Square of Squares Tables

Andrew Bremner's article on squares of squares included the 3x3 square:

Bremmer's square
5824621272
942113222
972822742

The numbers in the right diagonal as the tuple (972,1132,1272) 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 (a2,b2,c2) whose sum a2 + b2 + c23b2 = 0 are generated from another set of tuples that (except for the initial set of tuples) obeys the equation a2 + b2 + c23b20.

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 c2 = 2b2 − 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 c2 is calculated by searching all b numbers between 1 and 100,000. However, it was found that the ratio of bn+1/bn or cn+1/cn 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 bn or cn 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
anbncn
−111
−157
−12941
−1169239
−19851393
−157418119
−13346147321

Construction of two Tables of Right Diagonal Tuples

  1. The object of this exercise is to generate a table with a set of tuples that obey the rule: a2 + b2 + c23b20
    and convert these tuples into a second set of tuples that obey the rule: a2 + b2 + c23b2 = 0.
  2. 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.
  3. 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.
  4. Table I
    13346347323
    133463+e47323+g
  5. 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 12 + (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.
  6. 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 a2 + b2 + c23b2 = 0.
  7. Table I
    13346347323
    133463+e47323+g
    Table II
    13346347323
    1 + f33463+e + f 47323+g + f
  8. The Δs are calculated, the difference in Table 2 between columns 2 or 3, and the results placed in the last column.
  9. The final tables produced after the algebra is performed are shown below:
n
0
1
2
3
4
5
6
18
25
Table I
13346347323
13366147719
13385948115
13405748511
13425548907
13445349303
13465149699
13702754451
13841357223
f = S/d
−2
280
566
856
1150
1448
1750
5686
8248
Table II
−13346147321
2813394147999
5673442548681
8573491349367
11513540550057
14493590150751
17513640151449
56874271360137
82494666165471
Δ
1119638520
1151912520
1184759136
1218183120
1252189224
1286782200
1321966800
1792058400
2109202920

To obtain e, g,  f  and d the algebraic calculations are performed as follows:

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
877252-5838933287601372
-2254816487427132768352
56872974052-4046874887
Magic square B
160042-23361425519654712
-191430196794666121532892
824921664812-2121222599

This concludes Part VIB.
Go back to homepage.


Copyright © 2011 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com