TABLE OF RIGHT DIAGONALS

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

Square of Squares Tables

Andrew Bremner's article on squares of squares included the 3x3 square:

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 + c² − 3b² = 0 are generated from another set of tuples that (except for the initial set of tuples) obeys the equation a² + b² + c² − 3b² ≠ 0.

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 b_n+1/b_n or c_n+1/c_n 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 b_n or c_n 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).

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 + c² − 3b² ≠ 0
   and convert these tuples into a second set of tuples that obey the rule: a² + b² + c² − 3b² = 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.

Table I

√2	99	140
√2	99+e	140+g

3. 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.
4. 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 + c² − 3b² = 0.

Table I √2 99 140 √2 99+e 140+g

⇒

Table II √2 99 140 √2 + f 99+e + f 140+g + f

5. This explains why when both n and f are both equal to 0 that the first row of both tables are equal.
6. The Δs are calculated, the difference in Table 2 between columns 2 or 3, and the results placed in the last column.
7. The final tables produced after the algebra is performed are shown below:

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)

Magic square A 119081 ± 1468√2 105394 ± 4156√2 287328 ± 4288√2 338848 ± 2383√2 170601 ± 3304√2 2354 ± 480√2 53874 ± 2320√2 235808 ± 2452√2 222121 ± 1399√2

Magic square A (Unsquared) (3 ± 244√2)2 105394 ± 4156√2 (536 ± 4√2)2 338848 ± 2383√2 (413 ± 4√2)2 (48 ± 5√2)2 (232 ± 5√2)2 235808 ± 2452√2 222121 ± 1399√2

Magic square B 523336 ± 57120√2 881387 ∓ 16008√2 1355184 ± 27936√2 1751817 ∓ 6168√2 919969 ± 23016√2 88121 ± 52200√2 484754 ± 18096√2 958551 ± 62040√2 1316602 ∓ 11088√2

Magic square B (Unsquared) (56 ± 510√2)2 881387 ∓ 16008√2 (1164 ± 12√2)2 1751817 ∓ 6168√2 (959 ± 12√2)2 (261 ± 100√2)2 (696 ± 13√2)2 958551 ± 62040√2 1316602 ∓ 11088√2

Magic square C 1243409 ∓ 61464√2 3713826 ± 214440√2 4097728 ± 97152√2 5872640 ± 241992√2 3018321 ± 83376√2 164002 ∓ 75240√2 1938914 ± 69600√2 2322816 ∓ 47688√2 4793233 ± 228216√2

Magic square C (Unsquared) (39 ∓ 788√2)2 6162 ± 4296√2 (2024 ± 24√2)2 5872640 ± 241992√2 (1737 ± 24√2)2 (380 ∓ 99√2)2 (1392 ± 25√2)2 2322816 ∓ 47688√2 4793233 ± 228216√2

2. 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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