TABLE OF RIGHT DIAGONALS
GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part VB)
Square of Squares Tables
Andrew Bremner's article on squares of squares included the 3x3 square:
Bremner's square
3732 | 2892 | 5652 |
360721 | 4252 | 232 |
2052 | 5272 | 222121 |
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 bn+1/bn or cn+1/cn 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 bn or cn 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 VA. The initial simple tuple (−1,1,1) is the only tuple stands on its own. Our fifth example is then (−1, 5741, 8119).
Table T
an | bn | cn |
−1 | 1 | 1 |
−1 | 5 | 7 |
−1 | 29 | 41 |
−1 | 169 | 239 |
−1 | 985 | 1393 |
−1 | 5741 | 8119 |
−1 | 33461 | 47321 |
Construction of two Tables of Right Diagonal Tuples
- 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.
- To generate table I we take the tuple (−1,5741,8119) and add 2 to each entry in the tuple to produce
Table I with +1 entries in the first column.
- 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 | 5743 | 8121 |
1 | 5743+e | 8121+g |
- 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 + 5743)2 +
(gn + 8121)2
− 3(en +5743)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.
- This number f when added to the square of
each member in the tuple (1,b,c) generates
(f + 1)2 +
(f + en
+ 5743)2 + f + gn + 8121)2
− 3(f + en +5743)2
producing the resulting tuple in table II. This is the desired tuple obeying the rule
a2 + b2 +
c2 − 3b2 = 0.
Table I
1 | 5743 | 8121 |
1 | 5743+e | 8121+g |
|
| ⇒ |
Table II
1 | 5743 | 8121 |
1 + f | 5743+e + f |
8121+g + f |
|
|
- The Δs are calculated, the difference in Table 2 between columns 2 or 3, and the results placed in the last column.
- Note that the third column in Table II is identical to column 2 but shifted up 29 rows. In addition, only one magic square with six squares was found between
n = 0 to 40.
- The final tables produced after the algebra is performed are shown below:
|
|
Table I
1 | 5743 | 8121 |
1 | 5859 | 8353 |
1 | 5975 | 8585 |
1 | 6091 | 8817 |
1 | 6207 | 9049 |
1 | 6323 | 9281 |
1 | 6439 | 9513 |
1 | 6555 | 9745 |
… | … | … |
1 | 8295 | 13225 |
|
|
f = S/d
−2 |
166 |
342 |
526 |
718 |
918 |
1126 |
1342 |
… |
5542 |
|
|
Table II
−1 | 5741 | 8119 |
167 | 6025 | 8519 |
343 | 6317 | 8927 |
527 | 6617 | 9343 |
719 | 6925 | 9767 |
919 | 7241 | 10199 |
1127 | 7565 | 10639 |
1343 | 7897 | 1087 |
… | … | … |
5543 | 13837 | 18767 |
|
|
Δ
32959082 |
36272736 |
39786840 |
43506960 |
47438664 |
51587520 |
55959096 |
60558960 |
… |
160737720 |
|
To obtain e, g, f
and d the algebraic calculations are performed as follows:
- The condition we set is g = 2e
- Generate the equation: (12 + (en + 5743)2
+ (gn + 8121)2
− 3(en + 5743)2 (a)
- Add f to the numbers in the previous equation:
(f + 1)2 +
(f + en + 5743)2 +
(f + gn + 8121)2
− 3(f + en + 5743)2
(b)
- Expand the equation in order to combine and eliminate terms:
(f2 + 2f + 1) +
(f2 + 2enf +
11486f + e2n2 +
11486en + 32982049)
+ (f2 + 2gnf +
16242f + g2n2
+ 16242gn + 65950641) +
(−3f2 − 6en
f − 34458f − 3e2n2
− 34458en − 98946147) = 0 (c)
-
−6728f + (2gn
f −4en f)
+ (g2n2
− 2e2n2) + (16242gn
− 22972en) − 13456 = 0 (d)
- Move f to the other side of the equation and
since g = 2e then
6728f = (4e2n2
−2e2n2) +
(32484en − 22972en) − 13456
(e)
6728f = 2e2n2 +
9512en − 13456 (f)
- At this point the divisor d is equal to the coefficent of f,
i.e. d = 6728.
For 4 to divide the right side of
the equation we find the lowest value of e and g
which would satisfy the equation.
e = 116 and g = 232 are those numbers.
- Thus 6728f = 26912n2 + 1103392n − 13456
and (g)
f = 4n2 + 164n − 2 (h)
- Therefore substituting these values for e, f and g into the requisite two equations affords:
for Table I: (12 + (116n + 5743)2
+ (232n + 8121)2
− 3(116n + 5743)2 (i)
for Table II: (4n2 + 164n − 1)2 +
(4n2 + 280n + 5741)2 +
(4n2 + 396n + 8119)2
− 3(4n2 + 280n + 5741)2
(j)
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 is an example of a magic square of order number n = 22 produced from the tuple (5543, 13837, 18767).
The magic sum in this case is 574387797.
Magic square A
710692 | -4828615393 | 187672 |
-450713903 | 138372 | 699292 |
55432 | 721912 | -4667877623 |
This concludes Part VB. To continue to Part VIA.
Go back to homepage.
Copyright © 2011 by Eddie N Gutierrez. E-Mail: edguti144@outlook.com