TABLE OF RIGHT DIAGONALS
GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part IB)
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 (2052,4252,5652) 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 IA. The initial simple tuple (−1,1,1) is the only tuple stands on its own. Our first example is then (−1,5,7).
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 (−1,5,7) 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.
- 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 + 5)2 +
(gn + 7)2
− 3(en +5)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
+ 5)2 + f + gn + 7)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 II
1 | 7 | 9 |
1 + f | 7+e + f |
9+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 two rows.
- The final tables produced after the algebra is performed are shown below:
n
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
|
|
Table I
1 | 7 | 9 |
1 | 9 | 13 |
1 | 11 | 17 |
1 | 13 | 21 |
1 | 15 | 25 |
1 | 17 | 29 |
1 | 19 | 33 |
1 | 21 | 37 |
1 | 23 | 41 |
1 | 25 | 45 |
1 | 27 | 49 |
1 | 29 | 53 |
1 | 31 | 37 |
1 | 33 | 61 |
|
|
f = S/d
−2 |
1 |
6 |
13 |
22 |
33 |
46 |
61 |
78 |
97 |
118 |
141 |
166 |
193 |
|
|
Table II
−1 | 5 | 7 |
2 | 10 | 14 |
7 | 17 | 23 |
14 | 26 | 34 |
23 | 37 | 47 |
34 | 50 | 62 |
47 | 65 | 79 |
62 | 82 | 98 |
79 | 101 | 119 |
98 | 122 | 142 |
119 | 145 | 167 |
142 | 170 | 194 |
167 | 197 | 223 |
194 | 226 | 3254 |
|
|
Δ
24 |
96 |
240 |
480 |
840 |
1344 |
2016 |
2880 |
3960 |
5280 |
6864 |
8756 |
10920 |
13440 |
|
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 + 7)2
+ (gn + 9)2
− 3(en + 7)2 (a)
- Add f to the numbers in the previous equation:
(f + 1)2 +
(f + en + 7)2 +
(f + gn + 9)2
− 3(f + en + 7)2
(b)
- Expand the equation in order to combine and eliminate terms:
(f2 + 2f + 1) +
(f2 + 2enf +
14f + e2n2 +
14en + 49)
+ (f2 + 2gnf +
18f + g2n2
+ 18gn + 81) +
(−3f2 − 6en
f − 42f − 3e2n2
− 42en − 147) = 0 (c)
-
−8f + (2gn
f −4en f)
+ (g2n2
− 2e2n2) + (18gn
− 28en) − 16 = 0 (d)
- Move f to the other side of the equation and
since g = 2e then
8f = (4e2n2
−2e2n2) +
(36en − 20en) − 16
(e)
8f = 2e2n2 +
8en − 16 (f)
- At this point the divisor d is equal to the coefficent of f,
i.e. d = 8.
For 8 to divide the right side of
the equation we find the lowest value of e and g
which would satisfy the equation.
e = 2 and g = 4 are those numbers.
- Thus 8f = 8n2 + 16n − 16
and (g)
f = n2 + 2n − 2 (h)
- Therefore substituting these values for e, f and g into the requisite two equations affords:
for Table I: (12 + (2n + 7)2
+ (4n + 9)2
− 3(2n + 7)2 (i)
for Table II: (n2 + 2n − 1)2 +
(n2 + 4n + 5)2 +
(n2 + 6n + 7)2
− 3(n2 + 4n + 5)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 example of a magic square of order number n = 9 produced from the tuple (98, 122, 142).
Squares B and C are magic squares of order n = 13 produced from the tuple (194, 226, 254).
The magic sums in this case are 44652 and 153228, respectively.
Magic square A
732 | 19159 | 1422 |
29719 | 1222 | 72 |
982 | 1032 | 24439 |
|
| |
Magic square B
1162 | 75256 | 2542 |
102136 | 2262 | 42 |
1942 | 1642 | 88696 |
|
| |
Magic square C
1482 | 66808 | 2542 |
93688 | 2262 | 922 |
1942 | 1882 | 80248 |
|
This concludes Part IB. To continue to Part II.
Go back to homepage.
Copyright © 2011 by Eddie N Gutierrez. E-Mail: edguti144@outlook.com