TABLE OF RIGHT DIAGONALS

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

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 complex 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 +√2i 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 +√2i. The initial simple tuple is (√2i,1,2) and thus is the first example.

Table T
√2i12
√2i710
√2i41 58
√2i239338
√2i13931970
√2i811911482
√2i4732166922

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
    √2i12
    √2i1+e2+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 (√2i)2 + (en + 1)2 + (gn + 2)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 + √2i)2 + (f + en + 1)2 + f + gn + 2)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
    √2i12
    √2i1+e2+g
    Table II
    √2i12
    √2i + f1+e + f 2+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
√2i12
√2i36
√2i510
√2i714
f = S/d
0
4(√2i)
12(√2i)
24(√2i)
Table II
√2i12
5√2i3+4√2i6+4√2i
13√2i5+12√2i10+12√2i
25√2i7+24√2i14+24√2i
Table III
√2i12
-5069−72√2i4+48√2i
-338-263+120√2i-188+240√2i
-1250-1103+336√2i-956+672√2i
Δ
3
27+24√2i
75+120√2i
147+336√2i

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 (±5√2, 3 ± 4√2, 6 ± 4√2), the second (±13√2, 5 ± 12√2, 10 ± 12√2) and the third (±25√2, 7 ± 24√2, 14 ± 24√2)
  2. Magic square A
    -92 ± 336√2i19 ∓ 322√2i4 ± 48√2i
    73 ∓ 264√2i-23 ± 24√2i-119 ± 312√2i
    -50 -65 ± 360√2i46 ∓ 288√2i
      
    Magic square A (Unsquared)
    (14 ± 12√2i)219 ∓ 312√2i(2 ± 4√2i)2
    73 ∓ 264√2i(3 ± 4√2i)2(13 ± 12√2i)2
    ( ±5√2i)2-65 ± 360√2i46 ∓ 288√2i
    Magic square B
    28 ∓ 24√2i949 ± 144√2i-188 ± 240√2i
    1099 ± 384√2i-263 ± 120√2i-47 ∓ 144√2i
    -338103 ± 96√2i1024 ± 264√2i
      
    Magic square B (Unsquared)
    (6 ∓ 2√2i)2949 ± 144√2i(10 ± 12√2i)2
    1099 ± 384√2i(5 ± 12√2i)2(9 ∓ 8√2i)2
    ( ±13√2i)2103 ± 996√2i1024 ± 264√2i
    Magic square C
    124 ± 360√2i4141 ∓ 24√2i956 ± 672√2i
    4435 ± 648√2i-1103 ± 336√2i-23 ± 24√2i
    -1250√2i 271 ± 696√2i4288 ± 312√2i
      
    Magic square C (Unsquared)
    (18 ± 10√2i)24141 ∓ 24√2i(14 ± 24√2i) 2
    4435 ± 648√2i(7 ± 24√2i)2(3 ± 4√2i)2
    (±25√2i)2271 ± 696√2i4228 ± 312√2i
  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√2i) × (a + √2i) = a2 − 2b2 + 2ab√2i
    Division = (a + b√2i) / (a + b√2i)
    Square root of (a + b√2i) :
    r = sqrt(a2 − 2b2)
    r1 = x + y√2i
    r2 = −x − y√2i
    where,
    x = sqrt((r − a) / 2) and y = b / 2x

This concludes Part IB.
Part IIA continues using the √7 in complex square root numbers.
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Copyright © 2011 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com