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 3x3 square:
Bremmer's square
| 582 | 462 | 1272 |
| 942 | 1132 | 22 |
| 972 | 822 | 742 |
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 bn+2/bn or cn+2/cn 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 bn or cn 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 first example is then (√7,2,1).
Table T
| √7 | 2 | 1 |
| √7 | 4 | 5 |
| √7 | 8 | 11 |
| √7 | 22 | 31 |
| √7 | 46 | 65 |
| √7 | 128 | 181 |
| √7 | 268 | 379 |
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. 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.
- 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.
- 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 + 2)2 +
(gn + 1)2
− 3(en +2)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 + √7)2 +
(f + en
+ 2)2 + f + gn + 1)2
− 3(f + en +2)2
producing the resulting tuple in table II. This is the desired tuple obeying the rule
a2 + b2 +
c2 − 3b2 = 0.
|
| ⇒ |
Table II
| √7 | 2 | 1 |
| √7 + f | 2+e + f |
1+g + f |
|
|
- This explains why when both n and f are both equal to 0
that the first row of both tables are equal.
- The Δs are calculated, the difference in Table 2 between columns 2 or 3, and the results placed in the last column.
- The final tables produced after the algebra is performed are shown below:
|
|
Table I
| √7 | 2 | 1 |
| √7 | 4 | 5 |
| √7 | 6 | 9 |
| √7 | 8 | 13 |
|
|
f = S/d
| 0 |
| 0 |
| 2(6±2√7) |
| 6(6±2√7) |
|
|
Table II
| √7 | 2 | 1 |
| √7 | 4 | 5 |
| 12±5√7 | 18±4√7 | 21±4√7 |
| 36±13√7 | 44±12√7 | 49±12√7 |
|
Table III
| 7 | 4 | 1 |
| 7 | 16 | 25 |
| 319±120√7 | 436±144√7 | 553±168√7 |
| 2479±936√7 | 2944±1056√7 | 3409±1176√7 |
|
|
Δ
| -3 |
| 9 |
| 117±24√7 |
| 465±120√7 |
|
To obtain e, g, f
and d the algebraic calculations are performed as follows:
- The condition we set is g = 2e
- Generate the equation: ((√7)2 + (en + 2)2
+ (gn + 1)2
− 3(en +3)2 (a)
- Add f to the numbers in the previous equation:
(f ± √7)2 +
(f + en + 2)2 +
(f + gn + 1)2
− 3(f + en +2)2
(b)
- Expand the equation in order to combine and eliminate terms:
(f2 ± 2√7f + 7) +
(f2 + 2enf +
4f + e2n2 +
4en + 4)
+ (f2 + 2gnf +
2f + g2n2
+ 2gn + 1) +
(−3f2 − 6enf
− 12f − 3e2n2
− 12en − 12) = 0 (c)
-
(±2√7f − 6f)
+ (2gnf −
4enf)
+ (g2n2
−2e2n2) + (2gn
− 8en) = 0 (d)
- Move f to the other side of the equation and
since g = 2e then
6f ∓ 2√7f =
(4e2n2
−2e2n2) +
(4en − 8en)
(e)
6f ∓ 2√7f =
2e2n2 −
4en (f)
- At this point the divisor d is equal to the coefficient of f,
i.e. d = 6 ∓ 2√7.
For 6 ∓ 2√7 to divide the right side of
the equation we find the lowest value of e and g
which would satisfy the equation.
To obtain e and g we must first multiply by the factor
(6 ± 2√7) ⁄ (6 ± 2√7) which gives (g) followed by
(h).
And where the + of the first factor in the denominator is multiplied
by the − of the second factor in the denominator and vice versa.
- f = (2e2n2 −
4en) ⁄ (6 ∓ 2√7) × (6 ± 2√7) ⁄ (6 ± 2√7)
(g)
f = [(2e2n2 −
4en) × (6 ± 2√7)] ⁄ 8 (h)
- At this point setting e = 2 and therefore, g = 4 affords
f = (n2 − n)(6 ± 2√7)
(i)
- Therefore substituting these values for e, f and g into the requisite two equations affords:
for Table I: ((√7)2 + (2n + 2)2
+ (4n + 1)2
− 3(2n + 2)2 (j)
for Table II:
((n2 − n)(6 ± 2√7)
± √7)2 +
((n2 − n)(6 ± 2√7) + 2n + 2)2
+ ((n2 − n)(6 ± 2√7) + 4n + 1)2
− 3((n2 − n)(6 ± 2√7) + 2n + 2)2
(k)
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.
- Two 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
(12 ± 5√2, 18 ± 4√2, 21 ± 4√2) and the second
(36 ± 13√2, 44 ± 12√2, 49 ± 12√2).
Magic square A
| 268 ± 48√7 | 487 ± 216√7 | 553 ± 168√7 |
| 721 ± 264√7 | 436 ± 144√7 | 151 ± 24√7 |
| 319 ± 120√7 | 385 ± 72√7 | 604 ± 240√7 |
|
| |
Magic square A (Unsquared)
| (4 ± 6√7)2 | 487 ± 216√7 | (21 ± 4√7)2 |
| 721 ± 264√7 | (18 ± 4√7)2 | (12 ∓ 12√7)2 |
| (12 ± 5√7)7 | 385 ± 72√7 | 604 ± 240√7 |
|
Magic square B
| 497 ± 112√7 | 4926 ± 1880√7 | 3409 ± 1176√7 |
| 5836 ± 2120√7 | 2944 ± 1056√7 | 32 ∓ 8√7 |
| 2470 ± 936√7 | 962 ± 232√7 | 5391 ± 2000√7 |
|
| |
Magic square B (Unsquared)
| (7 ± 8√7)2 | 4926 ± 1880√7 | (49 ± 12√7)2 |
| 5836 ± 2120√7 | (44 ± 12√7)2 | (2 ∓ 8√7)2 |
| (36 ± 13√7)2 | 962 ± 232√7 | 5391 ± 2000√7 |
|
- 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 IIA.
To go to Part IIIB using √7 numbers.
Go back to homepage.
Copyright © 2011 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com
