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 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 − 7 and placed into table T below. The first number in each tuple all a start with +√7 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+2/bn or cn+2/cn converges on (1 + √2)2 as the b's or c's get larger. This means that moving down every other row on the table each integer value takes on the previous bn or cn multiplied by (1 + √7)2, i.e., 5.8284271247...

Furthermore, only seven of the tuples will be used in Table T in which all a start with +√7 (the rest are listed as previously shown. The initial simple tuple is (√7,2,1). Our first example is then (√7,2,1).

Table T
√721
√745
√7811
√72231
√74665
√7128181
√7268379

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
    √721
    √72+e1+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 (√7)2 + (en + 2)2 + (gn + 1)2 − 3(en +2)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 + √7)2 + (f + en + 2)2 + f + gn + 1)2 − 3(f + en +2)2 producing the resulting tuple in table II. This is the desired tuple obeying the rule a2 + b2 + c23b2 = 0.
  6. Table I
    √721
    √72+e1+g
    Table II
    √721
    √7 + f2+e + f 1+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
Table I
√721
√745
√769
√7813
f = S/d
0
0
2(6±2√7)
6(6±2√7)
Table II
√721
√745
12±5√718±4√721±4√7
36±13√744±12√749±12√7
Table III
741
71625
319±120√7436±144√7553±168√7
2479±936√72944±1056√73409±1176√7
Δ
-3
9
117±24√7
465±120√7

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. Two examples are listed on for each of the latter three entries in table III where the radix part can be either + or −. For the unsquared tuples, the first has the tuple (12 ± 5√2, 18 ± 4√2, 21 ± 4√2) and the second (36 ± 13√2, 44 ± 12√2, 49 ± 12√2).
  2. Magic square A
    268 ± 48√7487 ± 216√7553 ± 168√7
    721 ± 264√7436 ± 144√7151 ± 24√7
    319 ± 120√7 385 ± 72√7604 ± 240√7
      
    Magic square A (Unsquared)
    (4 ± 6√7)2487 ± 216√7(21 ± 4√7)2
    721 ± 264√7(18 ± 4√7)2(12 ∓ 12√7)2
    (12 ± 5√7)7385 ± 72√7604 ± 240√7
    Magic square B
    497 ± 112√74926 ± 1880√73409 ± 1176√7
    5836 ± 2120√72944 ± 1056√732 ∓ 8√7
    2470 ± 936√7 962 ± 232√75391 ± 2000√7
      
    Magic square B (Unsquared)
    (7 ± 8√7)24926 ± 1880√7(49 ± 12√7)2
    5836 ± 2120√7(44 ± 12√7)2(2 ∓ 8√7)2
    (36 ± 13√7)2962 ± 232√75391 ± 2000√7
  3. The following equations were used for calculating squares and square roots and are described in the page non-complex square root equations:
    Multiplication = (a + b√7) × (a + b√7) = a2 + 7b2 + 2ab√7
    Division = (a + b√7) / (a + b√7)
    Square root of (a + b√7) :
    r = sqrt(a2 − 7b2)
    r1 = x + y√7
    r2 = −x − y√7
    where,
    x = sqrt((a ± r) / 2) and y = b / 2x
    and where r may take either a + or − value.

This concludes Part IIA.
To go to Part IIIB using √7 numbers.
Go back to homepage.


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