TABLE OF RIGHT DIAGONALS

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

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 − 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 instead as of +1 as was shown in Part IVA. 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
anbncn
−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.
  2. To generate table I we take the tuple (−1,985,1393) and add 2 to each entry in the tuple to produce Table I with +1 entries in the first column.
  3. We also set a condition for table I. 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.
  4. Table I
    19871395
    1987+e1395+g
  5. 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 + 987)2 + (gn + 1395)2 − 3(en +987)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.
  6. This number f when added to the square of each member in the tuple (1,b,c) generates (f + 1)2 + (f + en + 987)2 + f + gn + 1395)2 − 3(f + en +987)2 producing the resulting tuple in table II. This is the desired tuple obeying the rule a2 + b2 + c23b2 = 0.
  7. Table I
    19871395
    1987+e1395+g
    Table II
    19871395
    1 + f987+e + f 1395+g + f
  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 seventeen 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
16
17
Table I
19871395
110211463
110551531
110891599
111231667
111571735
111911803
112251871
112591939
112932007
113272075
113612143
113952211
115652551
f = S/d
−2
48
102
160
222
288
358
432
510
592
678
768
1278
1392
Table II
−19851393
4910691511
10311571633
16112491759
22313451889
28914452023
35915492161
43316572303
51117692449
59318852599
67920052753
76921292911
127928093741
139329573943
Δ
970224
1140360
1328040
1534080
1759296
2004504
2270520
2558160
2868240
3201576
3558984
3941280
6254640
6803400

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. Square A is an example of a magic square of order number n = 4 produced from the tuple (223, 1345, 1889). while Squares B and C are of the order n = 16 and n = 17 produced from the tuples (1279, 2809, 3761) and (1393, 2957, 3943), respectively. The magic sums in this case are 5427075, 23671443 and 26231547, respectively.
Magic square A
38152-1269547118892
-91768791345235772
223240392-10936175
  
Magic square B
46492-1208687937612
4224012809239192
1279252792-5832239
  
Magic square C
28252270367339432
163104732957210852
13932384529507073

This concludes Part IVB. To continue to Part V.
Go back to homepage.


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