TABLE OF RIGHT DIAGONALS

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

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 + 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. The initial simple tuple (1,1,1) is the only tuple stands on its own. Our second example is then (1,29,41).

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 + 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
    12941
    129+e41+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 + c23b2 = 0.
  6. Table I
    12941
    129+e41+g
    Table II
    12941
    1 + f29+e + f 41+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 four 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
12
Table I
12941
13349
13757
14165
14573
14981
15389
15797
161105
165113
169121
173129
177137
f = S/d
0
7
16
27
40
55
72
91
112
135
160
187
216
Table II
12941
84056
175373
286892
4185113
56104136
73125161
92148188
113173217
136200248
161229281
188260316
217293357
Δ
840
1536
2520
3840
5544
7680
10296
13440
17160
21504
26520
32256
38760

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. Again as was shown previously Bremner's square contains 7 squares. The three diagonal squares are divisible by 5 and thus correspond to the tuple (41,85,113) of table II. Two other examples magic square (A) and (B) produced from the tuple (953, 1013, 1217) of order n = 28 and (1, 29, 41) of order n = 0, are also shown.
Bremner's Square
373228925652
3607214252232
20525272222121
  
Magic square A
5412905088512172
1191528510932792
9532761210483085
  
Magic square B
372−527412
1153292232
12472313

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