TABLE OF RIGHT DIAGONALS
GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part IVB)
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 IVA. The initial simple tuple (−1,1,1) is the only tuple stands on its own. Our fourth example is then (−1,985,1393).
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 (−1,985,1393) 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 | 987 | 1395 |
1 | 987+e | 1395+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 + 987)2 +
(gn + 1395)2
− 3(en +987)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
+ 987)2 + f + gn + 1395)2
− 3(f + en +987)2
producing the resulting tuple in table II. This is the desired tuple obeying the rule
a2 + b2 +
c2 − 3b2 = 0.
Table I
1 | 987 | 1395 |
1 | 987+e | 1395+g |
|
| ⇒ |
Table II
1 | 987 | 1395 |
1 + f | 987+e + f |
1395+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 seventeen 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 |
… |
16 |
17 |
|
|
Table I
1 | 987 | 1395 |
1 | 1021 | 1463 |
1 | 1055 | 1531 |
1 | 1089 | 1599 |
1 | 1123 | 1667 |
1 | 1157 | 1735 |
1 | 1191 | 1803 |
1 | 1225 | 1871 |
1 | 1259 | 1939 |
1 | 1293 | 2007 |
1 | 1327 | 2075 |
1 | 1361 | 2143 |
… | … | … |
1 | 1395 | 2211 |
1 | 1565 | 2551 |
|
|
f = S/d
−2 |
48 |
102 |
160 |
222 |
288 |
358 |
432 |
510 |
592 |
678 |
768 |
… |
1278 |
1392 |
|
|
Table II
−1 | 985 | 1393 |
49 | 1069 | 1511 |
103 | 1157 | 1633 |
161 | 1249 | 1759 |
223 | 1345 | 1889 |
289 | 1445 | 2023 |
359 | 1549 | 2161 |
433 | 1657 | 2303 |
511 | 1769 | 2449 |
593 | 1885 | 2599 |
679 | 2005 | 2753 |
769 | 2129 | 2911 |
… | … | … |
1279 | 2809 | 3741 |
1393 | 2957 | 3943 |
|
|
Δ
970224 |
1140360 |
1328040 |
1534080 |
1759296 |
2004504 |
2270520 |
2558160 |
2868240 |
3201576 |
3558984 |
3941280 |
… |
6254640 |
6803400 |
|
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 + 987)2
+ (gn + 1395)2
− 3(en + 987)2 (a)
- Add f to the numbers in the previous equation:
(f + 1)2 +
(f + en + 987)2 +
(f + gn + 1395)2
− 3(f + en + 987)2
(b)
- Expand the equation in order to combine and eliminate terms:
(f2 + 2f + 1) +
(f2 + 2enf +
1974f + e2n2 +
1974en + 974169)
+ (f2 + 2gnf +
2790f + g2n2
+ 2790gn + 1946025) +
(−3f2 − 6en
f − 5922f − 3e2n2
− 5922en − 2922507) = 0 (c)
-
−1156f + (2gn
f −4en f)
+ (g2n2
− 2e2n2) + (2790gn
− 3948en) − 2312 = 0 (d)
- Move f to the other side of the equation and
since g = 2e then
1156f = (4e2n2
−2e2n2) +
(5580en − 3948en) − 2312
(e)
1156f = 2e2n2 +
1632en − 2312 (f)
- At this point the divisor d is equal to the coefficent of f,
i.e. d = 1156.
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 = 34 and g = 68 are those numbers.
- Thus 1156f = 2312n2 + 55488n − 2312
and (g)
f = 2n2 + 48n − 2 (h)
- Therefore substituting these values for e, f and g into the requisite two equations affords:
for Table I: (12 + (34n + 987)2
+ (68n + 1395)2
− 3(34n + 987)2 (i)
for Table II: (2n2 + 48n − 1)2 +
(2n2 + 82n+ 985)2 +
(2n<2 + 116n + 1393)2
− 3(2n2 + 82n + 985)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 = 4 produced from the tuple (223, 1345, 1889).
while Squares B and C are of the order n = 16 and n = 17 produced from the tuples (1279, 2809, 3761) and
(1393, 2957, 3943), respectively.
The magic sums in this case are 5427075, 23671443 and 26231547, respectively.
Magic square A
38152 | -12695471 | 18892 |
-9176879 | 13452 | 35772 |
2232 | 40392 | -10936175 |
|
| |
Magic square B
46492 | -12086879 | 37612 |
422401 | 28092 | 39192 |
12792 | 52792 | -5832239 |
|
| |
Magic square C
28252 | 2703673 | 39432 |
16310473 | 29572 | 10852 |
13932 | 38452 | 9507073 |
|
This concludes Part IVB. To continue to Part V.
Go back to homepage.
Copyright © 2011 by Eddie N Gutierrez. E-Mail: edguti144@outlook.com