TABLE OF RIGHT DIAGONALS

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

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 − 2 and placed into table T below. The first number in each tuple all a start with +√2 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 +√2. The initial simple tuple is (√2,1,0). Our first example is then (√2,3,4).

Table T
√210
√234
√21724
√299140
√2577816
√233634756
√21960127720

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
    √234
    √23+e4+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 (√2)2 + (en + 3)2 + (gn + 4)2 − 3(en +3)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 + √2)2 + (f + en + 3)2 + f + gn + 4)2 − 3(f + en +3)2 producing the resulting tuple in table II. This is the desired tuple obeying the rule a2 + b2 + c23b2 = 0.
  6. Table I
    √234
    √23+e4+g
    Table II
    √234
    √2 + f3+e + f 4+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
√234
√258
√2712
√2916
f = S/d
0
4(2±√2)
12(2±√2)
24(2±√2)
Table II
√234
8±5√213±4√216±4√2
24±13√231±12√236±12√2
48±25√257±24√264±24√2
Table III
2916
114±80√2201±104√2288±128√2
914±624√21249±744√21584±864√2
3554±2400√24401±2736√25248±3072√2
Δ
7
87±4√2
335±120√2
847±336√2

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 examples are listed on for each of the latter three entries in table III where the radix part can be either + or −. The first has the tuple (8 ± 5√2, 13 ± 4√2, 16 ± 4√2), the second (24 ± 13√2, 31 ± 12√2, 36 ± 12√2) and the third (48 ± 25√2, 57 ± 24√2, 64 ± 24√2)
  2. Magic square A
    904 ± 576√2-589 ± 392√2288 ± 128√2
    -415 ± 344√2201 ± 104√2817 ± 552√2
    114 ± 80√2 991 ± 600√2-502 ± 368√2
      
    Magic square A (Unsquared)
    (16 ± 18√2)2-589 ± 392√2(16 ± 4√2)2
    -415 ± 344√2(13 ± 4√2)2(23 ± 12√2)2
    (8 ± 5√2)2991 ± 600√2-502 ± 368√2
    Magic square B
    792 ± 432√21371 ± 936√21584 ± 864√2
    2041 ± 1176√21249 ± 744√2457 ± 312√2
    914 ± 624√2 1127 ± 552√21706 ± 1056√2
      
    Magic square B (Unsquared)
    (12 ± 8√2)21371 ± 936√2(36 ± 12√2)2
    2041 ± 1176√2(31 ± 12√2)2(13 ± 12√2)2
    (24 ± 13√2)21127 ± 552√21706 ± 1056√2
    Magic square C
    1793 ± 840√26162 ± 4296√25248 ± 3072√2
    7856 ± 4968√24401 ± 2736√2946 ± 504√2
    3554 ± 2400√2 2640 ± 1176√27009 ± 4632√2
      
    Magic square C (Unsquared)
    (15 ± 28√2)26162 ± 4296√2(64 ± 24√2)2
    7856 ± 4968√2(57 ± 24√2)2(13 ± 12√2)2
    (48 ± 25√2)22640 ± 1176√27009 ± 4632√2
  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√2) × (a + b√2) = a2 + 2b2 + 2ab√2
    Division = (a + b√2) / (a + b√2)
    Square root of (a + b√2) :
    r = sqrt(a2 − 2b2)
    r1 = x + y√2
    r2 = −x − y√2
    where,
    x = sqrt((a ± r) / 2) and y = b / 2x
    and where r may take either a + or − value.

This concludes Part IA.
Part IIIA continues using (√2, 99, 140) from Table T above.
Part IB continues using complex square root numbers.
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Copyright © 2011 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com