TABLE OF RIGHT DIAGONALS

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

Square of Squares Tables

Andrew Bremner's article on squares of squares included the first 3x3 square having 7 squares:

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. The initial simple tuple (1,1,1) is the only tuple stands on its own. Our second example is then (1,29,41).

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

1	29	41
1	29+e	41+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 1² + (en + 29)² + (gn + 41)² − 3(en +29)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 + 1)² + (f + en + 29)² + f + gn + 41)² − 3(f + en +5)² producing the resulting tuple in table II. This is the desired tuple obeying the rule a² + b² + c² − 3b² = 0.

Table I 1 29 41 1 29+e 41+g

⇒

Table II 1 29 41 1 + f 29+e + f 41+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. Note that the third column in Table II is identical to column 2 but shifted up four rows.
8. 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. Again as was shown previously Bremner's square contains 7 squares. The three diagonal squares are divisible by 5 and thus correspond to the tuple (41,85,113) of table II. Two other examples magic square (A) and (B) produced from the tuple (953, 1013, 1217) of order n = 28 and (1, 29, 41) of order n = 0, are also shown.

This concludes Part II. To continue to Part IIB which treats tuples of the type (−1,b,c). To continue to Part III.
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