TABLE OF RIGHT DIAGONALS

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

Picture of a square

Square of Squares Tables

Andrew Bremner's article on squares of squares included the first 3x3 square having 7 squares:

Bremmer's square
373228925652
3607214252232
20525272222121

The numbers in the right diagonal as the tuple (2052,4252,5652) 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 third example is then (1,169,239).

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
    1169239
    1169+e239+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
    1169239
    1169+e239+g
    Table II
    1169239
    1 + f169+e + f 239+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. Note that the third column in Table II is identical to column 2 but shifted up seven rows.
  10. 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
19
Table I
1169239
1183267
1197295
1211323
1225351
1239379
1253407
1267435
1281463
1295491
1309519
1323547
1435771
f = S/d
0
22
48
78
112
150
192
238
288
342
400
462
1102
Table II
1169239
23205289
49245343
79289401
113337463
151389529
193445599
239505673
289569751
343637833
401709919
4637851009
110315371873
Δ
28560
41496
57624
77280
100800
128520
160776
197904
240240
288120
341880
401856
1145760

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 squares that can be formed from order number n = 19. (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
10822240825418732
4230382153721582
11032152223319318
  
Magic square B
21862-119961818732
10919021537219062
1103224342-53858
  
Magic square C
21862-358290815222
-12913881082219062
158224342-2437148

This concludes Part IIIA. To continue to Part IIIB which treats tuples of the type (−1,b,c). To continue to Part IVA.
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Copyright © 2011 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com