TABLE OF RIGHT DIAGONALS

GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF COMPLEX 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.

Generation of Diagonal Tuple Sets

As was shown in the web page Generation of Right Diagonals, the first seven tuples of complex squares, i.e., those with real and imaginary coefficients are generated using the formula c2 = 2b2 + 1 and placed into table Ti below. The first number in each tuple of all a start with either +i or −i which employ imaginary 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...

The initial simple tuple (±i,1,1) is the only tuple stands on its own. Our second example is then (±i, 12, 17).

Table Ti
±abc
±i11
±i23
±i1217
±i7099
±i408577
±i23783363
±i1386019601
i80782114243

Construction of 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 an i.
  3. Table I
    ±i1217
    ±i12+e17+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 i2 + (en + 12)2 + (gn + 17)2 − 3(en +12)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 ±i)2 + (f + en + 12)2 + f + gn + 17)2 − 3(f + en +12)2 producing the resulting tuple in table II. This is the desired tuple obeying the rule a2 + b2 + c23b2 = 0.
  6. Table I
    ±i1217
    ±i12+e17+g
    Table II
    ±i1217
    ±i + f12+e + f 17+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. Two Tables II and III are set up. The first takes the values in Table I adds the requisite value from table f to generate Table II. These complex numbers are then squared to generate Table III, which gives the diagonals to be used per magic square.
  9. The Δs are calculated, by subtracting column 2 from column 3 in Table III Squared and the results placed in the last column. For example adding 57 + 24i to −9 + 40i gives 48 + 64i and adding 57 + 24i to that value gives 105 + 88i.
  10. The final tables produced after the algebra is performed are shown below. The divisor d is easily obtained using the algebraic equations below. The value for d = 14 ± 2i and f is found using equations (g), (h), and finally (i).
n
0
1
2
3
4
Table I
±i1217
±i2237
±i3257
±i4277
±i5297
f = S/d
0
28±4i
84±12i
168±24i
280±40i
Table II Unsquared
i1217
28 ± 5i50 ± 4i65 ± 4i
84 ± 13i116 ± 12i141 ± 12i
168 ± 25i210 ± 24i245 ± 24i
280 ± 41i332 ± 40i377 ± 40i
Table III Squared
−1144289
759 ± 2801i2484 ± 400i4209 ± 520i
6887 ± 2184i13312 ± 2784i19737 ± 3384i
27599 ± 8400i43524 ± 10080i59449 ± 11760i
76719 ± 22960i108624 ± 26560i140529 ± 30160i
Δ
5
1725 ± 120i
6425 ± 600i
15925 ± 1680i
31905 ± 3600i

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. The following complex magic squares are exhibited first in square format and in square root format where both forms are equivalent. Only five complex squares with integer values for both the real and imaginary part were found for each of the squares looked at. For Square A (n = 3) the right diagonal tuple is (28 ± 5i, 50 ± 4i, 65 ± 4i) while for Square B (n = 4) the right diagonal tuple is (280 ± 41i, 332 ± 40i, 377 ± 40i) Unsquared. The magic sums in this case are 130572 ± 30240i and 325872 ± 79680i, respectively.
  2. Note that the top + and − go together and the and the bottom − and + also go together.
  3. Magic square A Squared
    860 ± 3648i70263 ± 14832i59449 ± 11760i
    102113 ± 18192i43524 ± 10080i−15065 ± 1968i
    27599 ± 8400i16785 ± 5328i86188 ± 16512i
    Magic square A Unsquared
    (48 ± 38i)270263 ± 14832i(65 ± 4i)2
    102113 ± 18192i(50 ± 4i)2(8 ± 123i)2
    (28 ± 5i)216785 ± 5328i86188 ± 16512i
    Magic square B Squared
    32256 ± 4320i153087 ± 45200i140529 ± 30160i
    216897 ± 52400i)108624 ± 26560i351 ± 720i
    76719 ± 22960i64161 ± 7920i184992 ± 48800i
    Magic square B Unsquared
    (180 ± 12i)2153087 ± 45200i(377 ± 40i)2
    216897 ± 52400i(332 ± 40i)2(24 ± 15i)2
    (280 ± 41i)264161 ± 7920i184992 ± 48800i
  4. We can get six complex squares from n = 4 by using squares that look similar but which are not. For example, (−41 + 280i)2 and (280 + 41i)2 are different numbers. The roots of one number are (−41 + 280i) and (41 − 280i) and of the other (280 ± 41i) and (−280 − 41i).
  5. Magic square C Squared
    108624 ∓ 26560i293967 ± 76080i140529 ± 30160i
    357777 ± 83280i)108624 ± 26560i−140529 ∓ 30160i
    76719 ± 22960i−76719 ∓ 22960i325872 ± 79680i
    Magic square C Unsquared
    (−332 ± 40i)2293967 ± 76080i(377 ± 40i)2
    357777 ± 83280i(332 ± 40i)2(−40 ∓ 377i)2
    (280 ± 41i)2(−41 ± 280i)2325872 ± 79680i
  6. The following equations were used for calculating squares and square roots:
    Multiplication = (a + bi) × (a + bi) = a2 − b2 + 2abi
    Division = (a + bi) / (a + bi)
    Square root of (a + bi) :
    r = sqrt(a2 + b2)
    r1 = x + y
    r2 = −x − y
    where,
    y = sqrt((r − a) / 2) and x = b / 2y

This concludes Part IIA. Part IIIA.
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Copyright © 2011 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com