TABLE OF RIGHT DIAGONALS

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

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 (205,²425,²565²) appear to have been obtained from elsewhere.. 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 − 1 and placed into table T below. The first number in each tuple all a start with +1 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 −1 instead as of +1 as was shown in Part VIA. The initial simple tuple (−1,1,1) is the only tuple stands on its own. Our fifth example is then (−1, 33461, 47321).

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.
2. To generate table I we take the tuple (−1, 33461, 47321) and add 2 to each entry in the tuple to produce Table I with +1 entries in the first column.
3. We also set a condition for table I. 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.

Table I

1	33463	47323
1	33463+e	47323+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 1² + (en + 33463)² + (gn + 47323)² − 3(en +33463)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 + 1)² + (f + en + 33463)² + f + gn + 47323)² − 3(f + en +33463)² producing the resulting tuple in table II. This is the desired tuple obeying the rule a² + b² + c² − 3b² = 0.

Table I 1 33463 47323 1 33463+e 47323+g

⇒

Table II 1 33463 47323 1 + f 33463+e + f 47323+g + f

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.

Square A and B are examples of magic square of order number n = 18 and n = 25, respectively. They are generated from the produced from the tuples (5687, 42713, 60137) and (8249, 46661, 65471), respectively, having the magic sums 5473201107 and 6531746763, respectively.

This concludes Part VIB.
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