TABLE OF RIGHT DIAGONALS

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

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 − 7 and placed into table T below. The first number in each tuple all a start with +√7 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+2/b_n or c_n+2/c_n converges on (1 + √2)² as the b's or c's get larger. This means that moving down every other row on the table each integer value takes on the previous b_n or c_n multiplied by (1 + √7)², i.e., 5.8284271247...

Furthermore, only seven of the tuples will be used in Table T in which all a start with +√7 (the rest are listed as previously shown. The initial simple tuple is (√7,2,1). Our second example is then (√7,46,65).

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

√7	46	65
√7	46+e	65+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 (√7)² + (en + 46)² + (gn + 65)² − 3(en +46)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 + √7)² + (f + en + 46)² + f + gn + 65)² − 3(f + en +46)² producing the resulting tuple in table II. This is the desired tuple obeying the rule a² + b² + c² − 3b² = 0.

Table I √7 46 65 √7 46+e 65+g

⇒

Table II √7 46 65 √7 + f 46+e + f 65+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 −. For the unsquared tuples, the first has the tuple (108 ± 5√2, 192 ± 4√2, 249 ± 4√2), the second (323 ± 13√2, 446 ± 12√2, 541 ± 12√2) and the third (648 ± 25√2, 808 ± 24√2, 941 ± 24√2).

Magic square A 5041 ± 4416√7 -16226 ∓ 1800√7 62113 ± 1992√7 34048 ∓ 888√7 36976 ± 1536√7 39904 ± 3960√7 11839 ± 1080√7 90178 ± 4872√7 8911 ∓ 1344√7

Magic square A (Unsquared) (23 ± 96√7)2 -16226 ∓ 1800√7 (249 ± 4√7)2 721 ± 264√7 (192 ± 4√7)2 (198 ± 10√7)2 (108 ± 5√7)7 90178 ± 4872√7 8911 ∓ 1344√7

Magic square B 95593 ± 2472√7 210490 ± 16656√7 293689 ± 12984√7 398020 ± 21216√7 199924 ± 10704√7 1828 ± 192√7 106159 ± 8424√7 189358 ± 4752√7 304255 ± 18936√7

Magic square B (Unsquared) (309 ± 4√7)2 210490 ± 16656√7 (541 ± 12√7)2 398020 ± 21216√7 (446 ± 16√7)2 (6 ± 16√7)2 (324 ± 13√7)2 189358 ± 4752√7 304255 ± 18936√7

Magic square C 237433 ± 7728√7 843742 ± 63436√7 889513 ±45168√7 1308976 ± 76224√7 656896 ± 38784√7 4826 ± 1344√7 424279 ± 32400√7 470050 ± 14112√7 1076359 ± 69840√7

Magic square C (Unsquared) (21 ± 184√7)2 843742 ± 63436√7 (941 ± 24√7)2 1308976 ± 76224√7 (808 ± 24√7)2 (28 ± 24√7)2 (648 ± 25√7)2 470050 ± 14112√7 1076359 ± 69840√7

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√7) × (a + b√7) = a² + 7b² + 2ab√7
   Division = (a + b√7) / (a + b√7)
   Square root of (a + b√7) :
   r = sqrt(a² − 7b²)
   r1 = x + y√7
   r2 = −x − y√7
   where,
   x = sqrt((a ± r) / 2) and y = b / 2x
   and where r may take either a + or − value.

This concludes Part IIIB.
To go to Part IIB using √7i complex numbers.
Go back to homepage.