TABLE OF RIGHT DIAGONALS

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

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 fifth example is then (1, 5741, 8119).

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
    157418119
    15741+e8119+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
    157418119
    15741+e8119+g
    Table II
    157418119
    1 + f5741+e + f 8119+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
24
Table I
157418119
158238283
159058447
159878611
160698775
161518939
162339103
163159267
163979431
164799595
165619759
166439923
1770912055
f = S/d
0
118
240
366
496
630
768
910
1056
1206
1360
1518
3936
Table II
157418119
11959418401
24161458687
36763538977
49765659271
63167819569
76970019871
911722510177
1057745310487
1207768510801
1361792111119
1519816111441
39371164515991
Δ
32959080
35281320
37702944
40225920
42852216
45583800
48422640
51370704
54429960
57602376
60889920
64294560
120106056

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.


Two squares that can be formed from order number n = 11 and n = 24. (Below n = 11 or between 11 and 24, only magic squares containing 5 squares are formed). These magic squares (A) and (B) are produced from the tuples (1517, 8161, 11441) and (3937, 11645, 15991) as shown below:

Magic square A
360912-1233650999114412
-110506187981612351892
15192369712-1169356439
  
Magic square B
157152-95855231159912
-144356881116452112632
3937219159224250825

This concludes Part VA. To continue to Part VB which treats tuples of the type (−1,b,c).
To continue to Part VIA.
Go back to homepage.


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