TABLE OF RIGHT DIAGONALS

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

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 (√2,3,4).

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	3	4
√2	3+e	4+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 + 3)² + (gn + 4)² − 3(en +3)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 + 3)² + f + gn + 4)² − 3(f + en +3)² producing the resulting tuple in table II. This is the desired tuple obeying the rule a² + b² + c² − 3b² = 0.

Table I √2 3 4 √2 3+e 4+g

⇒

Table II √2 3 4 √2 + f 3+e + f 4+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 (8 ± 5√2, 13 ± 4√2, 16 ± 4√2), the second (24 ± 13√2, 31 ± 12√2, 36 ± 12√2) and the third (48 ± 25√2, 57 ± 24√2, 64 ± 24√2)

Magic square A 904 ± 576√2 -589 ± 392√2 288 ± 128√2 -415 ± 344√2 201 ± 104√2 817 ± 552√2 114 ± 80√2 991 ± 600√2 -502 ± 368√2

Magic square A (Unsquared) (16 ± 18√2)2 -589 ± 392√2 (16 ± 4√2)2 -415 ± 344√2 (13 ± 4√2)2 (23 ± 12√2)2 (8 ± 5√2)2 991 ± 600√2 -502 ± 368√2

Magic square B 792 ± 432√2 1371 ± 936√2 1584 ± 864√2 2041 ± 1176√2 1249 ± 744√2 457 ± 312√2 914 ± 624√2 1127 ± 552√2 1706 ± 1056√2

Magic square B (Unsquared) (12 ± 8√2)2 1371 ± 936√2 (36 ± 12√2)2 2041 ± 1176√2 (31 ± 12√2)2 (13 ± 12√2)2 (24 ± 13√2)2 1127 ± 552√2 1706 ± 1056√2

Magic square C 1793 ± 840√2 6162 ± 4296√2 5248 ± 3072√2 7856 ± 4968√2 4401 ± 2736√2 946 ± 504√2 3554 ± 2400√2 2640 ± 1176√2 7009 ± 4632√2

Magic square C (Unsquared) (15 ± 28√2)2 6162 ± 4296√2 (64 ± 24√2)2 7856 ± 4968√2 (57 ± 24√2)2 (13 ± 12√2)2 (48 ± 25√2)2 2640 ± 1176√2 7009 ± 4632√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 IA.
Part IIIA continues using (√2, 99, 140) from Table T above.
Part IB continues using complex square root numbers.
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