TABLE OF RIGHT DIAGONALS
GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part IVA)
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.
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
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. 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.
Table I
1 | 985 | 1393 |
1 | 985+e | 1393+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 + 29)2 +
(gn + 41)2
- 3(en +29)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
+ 29)2 + f + gn + 41)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 I
1 | 985 | 1393 |
1 | 985+e | 1393+g |
|
| ⇒ |
Table II
1 | 985 | 1393 |
1 + f | 985+e + f |
1393+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:
n
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
|
|
Table I
1 | 985 | 1393 |
1 | 1009 | 1441 |
1 | 1033 | 1489 |
1 | 1054 | 1537 |
1 | 1081 | 1585 |
1 | 1105 | 1633 |
1 | 1129 | 1681 |
1 | 1153 | 1729 |
1 | 1177 | 1777 |
1 | 1201 | 1825 |
1 | 1225 | 1873 |
1 | 1249 | 1921 |
1 | 1273 | 1969 |
|
|
f = S/d
0 |
35 |
72 |
111 |
152 |
195 |
240 |
287 |
336 |
387 |
440 |
495 |
552 |
|
|
Table II
1 | 985 | 1393 |
36 | 1044 | 1476 |
73 | 1105 | 1561 |
112 | 1168 | 1648 |
153 | 1233 | 1737 |
196 | 1300 | 1828 |
241 | 1369 | 1921 |
288 | 1440 | 2016 |
337 | 1513 | 2113 |
388 | 1588 | 2212 |
441 | 1665 | 2313 |
496 | 1744 | 2416 |
553 | 1825 | 2521 |
|
|
Δ
970224 |
1088640 |
1215696 |
1351680 |
1496880 |
1651584 |
1816080 |
1990656 |
2175600 |
2371200 |
2577744 |
2795520 |
3024816 |
|
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 + 985)2
+ (gn + 1393)2
- 3(en +985)2 (a)
- Add f to the numbers in the previous equation:
(f + 1)2 +
(f + en + 985)2 +
(f + gn + 1393)2
- 3(f + en +985)2
(b)
- Expand the equation in order to combine and eliminate terms:
(f2 + 2f + 1) +
(f2 + 2enf +
1970f + e2n2 +
1970en + 970225)
+ (f2 + 2gnf +
2786f + g2n2
+ 2786gn + 1940449) +
(−3f2 − 6en
f −
5910f − 3e2n2
− 5910en - 85683) = 0 (c)
-
-1152f + (2gnf - 4en
f)
+ (g2n2
-2e2n2) + (2786gn
- 3940en) = 0 (d)
- Move f to the other side of the equation and
since g = 2e then
1152f = (4e2n2
-2e2n2) +
(5572en - 3940en)
(e)
1152f = 2e2n2 +
1632en (f)
- At this point the divisor d is equal to the coefficent of f, i.e. d = 1152.
For 1152 to divide the right side of
the equation we find the lowest value of e and g
which would satisfy the equation.
e = 24 and g = 48 are those numbers.
- Thus 1152f = 1152n2 + 39168n
and (g)
f = n2 + 34n (h)
- Therefore substituting these values for e, f and g into the requisite two equations affords:
for Table I: (12 + (24n + 985)2
+ (48n + 1393)2
- 3(24n + 985)2 (i)
for Table II: (n2 + 34n + 1)2 +
(n2 + 58n> + 985)2 +
(n2 + 82n + 1393)2
- 3(n2 + 58n + 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.
Three squares that can be formed from order number n = 1. (There may be more.) These magic squares (A) and (B) are
produced from the tuple (1103, 1537, 1873). A third square formed from the two other tuples from magic squares (A) and (B) as shown
in magic square (C).
Magic square A
13322 | -682992 | 18762 |
1494208 | 10442 | 8282 |
362 | 16922 | 405648 |
|
| |
Magic square B
20342 | -3056389 | 14762 |
-879109 | 10442 | 14762 |
362 | 22862 | -1967749 |
|
| |
Magic square C
20342 | -1677348 | 16922 |
499932 | 13322 | 17462 |
8282 | 22862 | -588708 |
|
Three squares that can be formed from order numbers n = 3, 4 or 11, respectively. (There may be more).
Magic square D
19522 | -2433536 | 16482 |
269824 | 11682 | 15682 |
1162 | 22722 | -1081856 |
|
| |
Magic square E
45392 | -19058823 | 17372 |
-16065063 | 12332 | 43712 |
1532 | 47012 | -17561943 |
|
| |
Magic square F
18642 | -186944 | 24162 |
5404096 | 17442 | 8242 |
4962 | 25042 | 2608576 |
|
This concludes Part IVA. To continue to Part IVB which treats tuples of the type (−1,b,c).
To continue to Part V.
Go back to homepage.
Copyright © 2011 by Eddie N Gutierrez. E-Mail: edguti144@outlook.com