TABLE OF RIGHT DIAGONALS

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

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 + c2 - 3b2 = 0 are generated from another set of tuples that (except for the initial set of tuples) obeys the equation a2 + b2 + c2 - 3b20.

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. 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
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 + c2 - 3b20
    and convert these tuples into a second set of tuples that obey the rule: a2 + b2 + c2 - 3b2 = 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
    19851393
    1985+e1393+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 12 + (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.
  5. 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 a2 + b2 + c2 - 3b2 = 0.
  6. Table I
    19851393
    1985+e1393+g
    Table II
    19851393
    1 + f985+e + f 1393+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
4
5
6
7
8
9
10
11
12
Table I
19851393
110091441
110331489
110541537
110811585
111051633
111291681
111531729
111771777
112011825
112251873
112491921
112731969
f = S/d
0
35
72
111
152
195
240
287
336
387
440
495
552
Table II
19851393
3610441476
7311051561
11211681648
15312331737
19613001828
24113691921
28814402016
33715132113
38815882212
44116652313
49617442416
55318252521
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:

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
13322-68299218762
1494208104428282
36216922405648
  
Magic square B
20342-305638914762
-8791091044214762
36222862-1967749
  
Magic square C
20342-167734816922
4999321332217462
828222862-588708

Three squares that can be formed from order numbers n = 3, 4 or 11, respectively. (There may be more).

Magic square D
19522-243353616482
2698241168215682
116222722-1081856
  
Magic square E
45392-1905882317372
-160650631233243712
153247012-17561943
  
Magic square F
18642-18694424162
5404096174428242
4962250422608576

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. E-Mail: Fiboguti89@Yahoo.com