TABLE OF RIGHT DIAGONALS

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

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 (√2,99,140).

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
    √299140
    √299+e140+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 + 99)2 + (gn + 140)2 − 3(en + 99)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 + 99)2 + f + gn + 140)2 − 3(f + en + 99)2 producing the resulting tuple in table II. This is the desired tuple obeying the rule a2 + b2 + c23b2 = 0.
  6. Table I
    √299140
    √299+e140+g
    Table II
    √299140
    √2 + f99+e + f 140+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
√299140
√2181304
√2263468
√2345632
f = S/d
0
2(116±2√2)
6(116±2√2)
12(116±2√2)
Table II
√299140
232±5√2413±4√2536±4√2
696±13√2959±12√21164±12√2
1392±25√21737±24√22024±24√2
Table III (Table II Squared)
2980119600
53874±2320√2170601±3304√2287328±4288√2
484754±18096√2919969±23016√21355184±27936√2
1938914±69600√23018321±83376√24097728±97152√2
Δ
9799
116727±984√2
435215±4920√2
1079407±13776√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 (232 ± 5√2, 413 ± 4√2, 536 ± 4√2), the second (696 ± 13√2, 959 ± 12√2, 1164 ± 12√2) and the third (1392 ± 25√2, 1737 ± 24√2, 2024 ± 24√2)
  2. Magic square A
    119081 ± 1468√2105394 ± 4156√2287328 ± 4288√2
    338848 ± 2383√2170601 ± 3304√22354 ± 480√2
    53874 ± 2320√2 235808 ± 2452√2222121 ± 1399√2
      
    Magic square A (Unsquared)
    (3 ± 244√2)2105394 ± 4156√2(536 ± 4√2)2
    338848 ± 2383√2(413 ± 4√2)2(48 ± 5√2)2
    (232 ± 5√2)2235808 ± 2452√2222121 ± 1399√2
    Magic square B
    523336 ± 57120√2881387 ∓ 16008√21355184 ± 27936√2
    1751817 ∓ 6168√2919969 ± 23016√288121 ± 52200√2
    484754 ± 18096√2 958551 ± 62040√21316602 ∓ 11088√2
      
    Magic square B (Unsquared)
    (56 ± 510√2)2881387 ∓ 16008√2(1164 ± 12√2)2
    1751817 ∓ 6168√2(959 ± 12√2)2(261 ± 100√2)2
    (696 ± 13√2)2958551 ± 62040√21316602 ∓ 11088√2
    Magic square C
    1243409 ∓ 61464√23713826 ± 214440√24097728 ± 97152√2
    5872640 ± 241992√23018321 ± 83376√2164002 ∓ 75240√2
    1938914 ± 69600√2 2322816 ∓ 47688√24793233 ± 228216√2
      
    Magic square C (Unsquared)
    (39 ∓ 788√2)26162 ± 4296√2(2024 ± 24√2)2
    5872640 ± 241992√2(1737 ± 24√2)2(380 ∓ 99√2)2
    (1392 ± 25√2)22322816 ∓ 47688√24793233 ± 228216√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 IIIA.
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Copyright © 2011 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com