TABLE OF RIGHT DIAGONALS

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

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 − 7 and placed into table T below. The first number in each tuple all a start with +√7 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+2/bn or cn+2/cn converges on (1 + √2)2 as the b's or c's get larger. This means that moving down every other row on the table each integer value takes on the previous bn or cn multiplied by (1 + √7)2, i.e., 5.8284271247...

Furthermore, only seven of the tuples will be used in Table T in which all a start with +√7 (the rest are listed as previously shown. The initial simple tuple is (√7,2,1). Our second example is then (√7,46,65).

Table T
√721
√745
√7811
√72231
√74665
√7128181
√7268379

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. 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.
  2. 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.
  3. Table I
    √74665
    √746+e65+g
  4. 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 (√7)2 + (en + 46)2 + (gn + 65)2 − 3(en +46)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.
  5. This number f when added to the square of each member in the tuple (1,b,c) generates (f + √7)2 + (f + en + 46)2 + f + gn + 65)2 − 3(f + en +46)2 producing the resulting tuple in table II. This is the desired tuple obeying the rule a2 + b2 + c23b2 = 0.
  6. Table I
    √74665
    √746+e65+g
    Table II
    √74665
    √7 + f46+e + f 65+g + f
  7. This explains why when both n and f are both equal to 0 that the first row of both tables are equal.
  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
Table I
√74665
√784141
√7122217
√7160293
f = S/d
0
2(54±2√7)
6(54±2√7)
12(54±2√7)
Table II
√74665
108±5√7192±4√7249±4√7
324±13√7446±12√7541±12√7
648±25√7808±24√7941±24√7
Table III
721164225
11839±1080√736976±1536√762113±1992√7
106159±8424√7199924±10704√7293689±12984√7
424279±32400√7656896±38784√7889513±45168√7
Δ
2067
25137±456√7
93765±2280√7
222617±6384√7

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.


  1. Three examples are listed on for each of the latter three entries in table III where the radix part can be either + or −. For the unsquared tuples, the first has the tuple (108 ± 5√2, 192 ± 4√2, 249 ± 4√2), the second (323 ± 13√2, 446 ± 12√2, 541 ± 12√2) and the third (648 ± 25√2, 808 ± 24√2, 941 ± 24√2).
  2. Magic square A
    5041 ± 4416√7-16226 ∓ 1800√762113 ± 1992√7
    34048 ∓ 888√736976 ± 1536√739904 ± 3960√7
    11839 ± 1080√7 90178 ± 4872√78911 ∓ 1344√7
      
    Magic square A (Unsquared)
    (23 ± 96√7)2-16226 ∓ 1800√7(249 ± 4√7)2
    721 ± 264√7(192 ± 4√7)2(198 ± 10√7)2
    (108 ± 5√7)790178 ± 4872√78911 ∓ 1344√7
    Magic square B
    95593 ± 2472√7210490 ± 16656√7293689 ± 12984√7
    398020 ± 21216√7199924 ± 10704√71828 ± 192√7
    106159 ± 8424√7 189358 ± 4752√7304255 ± 18936√7
      
    Magic square B (Unsquared)
    (309 ± 4√7)2210490 ± 16656√7(541 ± 12√7)2
    398020 ± 21216√7(446 ± 16√7)2(6 ± 16√7)2
    (324 ± 13√7)2189358 ± 4752√7304255 ± 18936√7
    Magic square C
    237433 ± 7728√7843742 ± 63436√7889513 ±45168√7
    1308976 ± 76224√7656896 ± 38784√74826 ± 1344√7
    424279 ± 32400√7 470050 ± 14112√71076359 ± 69840√7
      
    Magic square C (Unsquared)
    (21 ± 184√7)2843742 ± 63436√7(941 ± 24√7)2
    1308976 ± 76224√7(808 ± 24√7)2(28 ± 24√7)2
    (648 ± 25√7)2470050 ± 14112√71076359 ± 69840√7
  3. The following equations were used for calculating squares and square roots and are described in the page non-complex square root equations:
    Multiplication = (a + b√7) × (a + b√7) = a2 + 7b2 + 2ab√7
    Division = (a + b√7) / (a + b√7)
    Square root of (a + b√7) :
    r = sqrt(a2 − 7b2)
    r1 = x + y√7
    r2 = −x − y√7
    where,
    x = sqrt((a ± r) / 2) and y = b / 2x
    and where r may take either a + or − value.

This concludes Part IIIB.
To go to Part IIB using √7i complex numbers.
Go back to homepage.


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