TABLE OF RIGHT DIAGONALS

GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF COMPLEX SQUARES (Part IIB)

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 Ti below. The first number in each tuple all a start with +√7i 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 + √2)2, i.e., 5.8284271247...

Furthermore, only seven of the tuples will be used in Table Ti in which all a start with +√7i (the rest are listed as previously shown. The initial simple tuple is (√7i,1,3). Our first example is then (√7i,1,3).

Table Ti
anbncn
√7i13
√7i35
√7i913
√7i1927
√7i5375
√7i111157
√7i309437

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
    √7i13
    √7i1+e3+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 (√7i)2 + (en + 1)2 + (gn + 3)2 − 3(en +1)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 + √7i)2 + (f + en + 1)2 + f + gn + 3)2 − 3(f + en +1)2 producing the resulting tuple in table II. This is the desired tuple obeying the rule a2 + b2 + c23b2 = 0.
  6. Table I
    √7i13
    √7i1+e3+g
    Table II
    √7i13
    √7i + f1+e + f 3+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
√7i13
√7i511
√7i919
√7i1327
f = S/d
0
2(-2±2√7i)
6(-2±2√7i)
12(-2±2√7i)
Table II
√7i13
-4±5√7i1±4√7i7±4√7i
-12±13√7i-3±12√7i7±12√7i
-24±25√7i-11±24√7i3±24√7i
Table III
-719
-159±40√7i-111±8√7i-63±56√7i
-1039 ∓ 312√7i-999 ∓ 72√7i-959±168√7i
-3799 ∓ 1200√7i-3911 ∓ 528√7i-4023±144√7i
Δ
8
48±48√7i
40±240√7i
-112±672√7i

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 −. For the unsquared tuples, the first has the tuple (-4 ± 5√7i, 1 ± 4√7i, 7 ± 4√7i), the second (-12 ± 13√7i, -3 ± 12√7i, 7 ± 12√7i) and the third (-24 ± 25√7i, -11 ± 24√7i, 3 ± 24√7i).
  2. Magic square A
    36 ± 32√7i-306 ∓ 216√7i-63 ± 56√7i
    -210 ± 32√7i-111 ± 8√7i-12 ∓ 16√7i
    -159 ∓ 40√7i 84 ± 80√7i-258 ∓16√7i
      
    Magic square A (Unsquared)
    (8 ± 2√7i)2-306 ∓ 216√7i(7 ± 4√7i)2
    -210 ± 32√7i(1 ± 4√7i)2(4 ∓ 2√7i)2
    (-4 ± 5√7)284 ± 80√7i-258 ∓16√7i
    Magic square B
    -19 ± 252√7i-2019 ∓ 636√7i-959 ± 168√7i
    -1939 ∓ 156√7i-999 ∓ 72√7i-59 ± 12√7i
    -1039 ∓ 312√7i 21 ± 492√7i-1979 ∓ 396√7i
      
    Magic square B (Unsquared)
    (18 ± 7√7i)2-2019 ∓ 636√7i(7 ± 12√7i)2
    -1939 ∓ 156√7i(-3 ± 12√7i)2(2 ± 3√7i)2
    (-12 ± 13√7)221 ± 492√7i-1979 ∓ 396√7i
    Magic square C
    217 ± 840√7i-7927 ∓ 2568√7i-4023 ± 144√7i
    -8151 ∓ 1224√7i-3911 ∓ 528√7i329 ± 168√7i
    -3799 ∓ 1200√7i 105 ± 1512√7i-8039 ∓ 1896√7i
      
    Magic square C (Unsquared)
    (35 ± 12√7i)2-7927 ∓ 2568√7i(3 ± 24√7i)2
    -8151 ∓ 1224√7i(-11 ± 24√7i)2(21 ± 4√7i)2
    (-24 ± 25√7)2105 ± 1512√7i-8039 ∓ 1896√7i
  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√7i) × (a + b√7i) = a2 − 7b2 + 2ab√7i
    Division = (a + b√7i) / (a + b√7i)
    Square root of (a + b√7i) :
    r = sqrt(a2 − 7b2)
    r1 = x + y√7i
    r2 = −x − y√7i
    where,
    x = sqrt((r − a) / 2) and y = b / 2x

This concludes Part IIB.
Go back to homepage.


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