# TABLE OF RIGHT DIAGONALS

## Square of Squares Tables

Andrew Bremner's article on squares of squares included the 3x3 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 + c23b2 = 0 are generated from another set of tuples that (except for the initial set of tuples) obeys the equation a2 + b2 + c23b20.

## Generation of Diagonal Tuple Sets

As was shown in the web page Generation of Right Diagonals, the first seven tuples of complex squares, i.e., those with real and imaginary coefficients are generated using the formula c2 = 2b2 + 1 and placed into table Ti below. The first number in each tuple of all a start with either +i or −i which employ imaginary 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...

The initial simple tuple (±i,1,1) is the only tuple stands on its own. Our second example is then (±i, 12, 17).

Table Ti
±abc
±i11
±i23
±i1217
±i7099
±i408577
±i23783363
±i1386019601
i80782114243

## Construction of Tables of Right Diagonal Tuples

1. The object of this exercise is to generate a table with a set of tuples that obey the rule: a2 + b2 + c23b20
and convert these tuples into a second set of tuples that obey the rule: a2 + b2 + c23b2 = 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.
2. 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 an i.
3.  ±i 12 17 ±i 12+e 17+g
4. 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 i2 + (en + 12)2 + (gn + 17)2 − 3(en +12)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.
5. This number f when added to the square of each member in the tuple (1,b,c) generates (f ±i)2 + (f + en + 12)2 + f + gn + 17)2 − 3(f + en +12)2 producing the resulting tuple in table II. This is the desired tuple obeying the rule a2 + b2 + c23b2 = 0.
6.  ±i 12 17 ±i 12+e 17+g
 ±i 12 17 ±i + f 12+e + f 17+g + f
7. This explains why when both n and f are both equal to 0 that the first row of both tables are equal.
8. Two Tables II and III are set up. The first takes the values in Table I adds the requisite value from table f to generate Table II. These complex numbers are then squared to generate Table III, which gives the diagonals to be used per magic square.
9. The Δs are calculated, by subtracting column 2 from column 3 in Table III Squared and the results placed in the last column. For example adding 57 + 24i to −9 + 40i gives 48 + 64i and adding 57 + 24i to that value gives 105 + 88i.
10. The final tables produced after the algebra is performed are shown below. The divisor d is easily obtained using the algebraic equations below. The value for d = 14 ± 2i and f is found using equations (g), (h), and finally (i).
 0 1 2 3 4
 ±i 12 17 ±i 22 37 ±i 32 57 ±i 42 77 ±i 52 97
 0 28±4i 84±12i 168±24i 280±40i
 i 12 17 28 ± 5i 50 ± 4i 65 ± 4i 84 ± 13i 116 ± 12i 141 ± 12i 168 ± 25i 210 ± 24i 245 ± 24i 280 ± 41i 332 ± 40i 377 ± 40i
 −1 144 289 759 ± 2801i 2484 ± 400i 4209 ± 520i 6887 ± 2184i 13312 ± 2784i 19737 ± 3384i 27599 ± 8400i 43524 ± 10080i 59449 ± 11760i 76719 ± 22960i 108624 ± 26560i 140529 ± 30160i
 5 1725 ± 120i 6425 ± 600i 15925 ± 1680i 31905 ± 3600i

To obtain e, g,  f  and d the algebraic calculations are performed as follows:

• The condition we set is g = 2e
• Generate the equation: i)2 + (en + 12)2 + (gn + 17)2 − 3(en +12)2 (a)
• Add  f  to the numbers in the previous equation:
(f ±i)2 + (f + en + 12)2 + (f + gn + 17)2 − 3(f + en +12)2 (b)
• Expand the equation in order to combine and eliminate terms:
(f2 ± 2if  − 1) + (f2 + 2enf + 24f + e2n2 + 24en + 144)
+ (f2 + 2gnf + 34f + g2n2 + 34gn + 289) + (−3f2 − 6en f − 72f − 3e2n2 − 72en − 432) = 0 (c)
• (±2i − 14)f + (2gnf − 4enf) + (g2n2 −2e2n2) + (34gn − 48en) = 0 (d)
• Move f to the other side of the equation and since g = 2e then
(14 ± 2i)f = (4e2n2 −2e2n2) + (68en − 48en) (e)
(14 ± 2i)f = 2e2n2 + 48en (f)
• At this point the divisor d is equal to the coefficent of f, i.e. d = 14 ± 2i.
For 14 ± 2i 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
(14 + 2i) ⁄ (14 + 2i) 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 ⁄ (14 ± 2i) × (14 ∓ 2i) ⁄ (14 ∓ 2i) (g)
f = [(2e2n2 + 4en) × (14 ± 2i)] ⁄ 8 (h)
• At this point setting e = 10 and therefore, g = 20 affords
f = (n2 + n)(14 ± 2i) (i)
• Therefore substituting these values for e, f and g into the requisite two equations affords:
for Table I:  i2 + (10n + 2)2 + (20n + 3)2 − 3(10n + 2)2 (j)

for Tables II and III:  [(n2 + n)(14 ± 2i) + i]2 + [(n2 + n)(14 ± 2i) + 10n + 12]2 +
[(n2 + n)(14 ± 2i) + 20n + 17]2 − 3[(n2 + n)(14 ± 2i) + 10n + 12]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.

1. The following complex magic squares are exhibited first in square format and in square root format where both forms are equivalent. Only five complex squares with integer values for both the real and imaginary part were found for each of the squares looked at. For Square A (n = 3) the right diagonal tuple is (28 ± 5i, 50 ± 4i, 65 ± 4i) while for Square B (n = 4) the right diagonal tuple is (280 ± 41i, 332 ± 40i, 377 ± 40i) Unsquared. The magic sums in this case are 130572 ± 30240i and 325872 ± 79680i, respectively.
2. Note that the top + and − go together and the and the bottom − and + also go together.
3.  860 ± 3648i 70263 ± 14832i 59449 ± 11760i 102113 ± 18192i 43524 ± 10080i −15065 ± 1968i 27599 ± 8400i 16785 ± 5328i 86188 ± 16512i
 (48 ± 38i)2 70263 ± 14832i (65 ± 4i)2 102113 ± 18192i (50 ± 4i)2 (8 ± 123i)2 (28 ± 5i)2 16785 ± 5328i 86188 ± 16512i
 32256 ± 4320i 153087 ± 45200i 140529 ± 30160i 216897 ± 52400i) 108624 ± 26560i 351 ± 720i 76719 ± 22960i 64161 ± 7920i 184992 ± 48800i
 (180 ± 12i)2 153087 ± 45200i (377 ± 40i)2 216897 ± 52400i (332 ± 40i)2 (24 ± 15i)2 (280 ± 41i)2 64161 ± 7920i 184992 ± 48800i
4. We can get six complex squares from n = 4 by using squares that look similar but which are not. For example, (−41 + 280i)2 and (280 + 41i)2 are different numbers. The roots of one number are (−41 + 280i) and (41 − 280i) and of the other (280 ± 41i) and (−280 − 41i).
5.  108624 ∓ 26560i 293967 ± 76080i 140529 ± 30160i 357777 ± 83280i) 108624 ± 26560i −140529 ∓ 30160i 76719 ± 22960i −76719 ∓ 22960i 325872 ± 79680i
 (−332 ± 40i)2 293967 ± 76080i (377 ± 40i)2 357777 ± 83280i (332 ± 40i)2 (−40 ∓ 377i)2 (280 ± 41i)2 (−41 ± 280i)2 325872 ± 79680i
6. The following equations were used for calculating squares and square roots:
Multiplication = (a + bi) × (a + bi) = a2 − b2 + 2abi
Division = (a + bi) / (a + bi)
Square root of (a + bi) :
r = sqrt(a2 + b2)
r1 = x + y
r2 = −x − y
where,
y = sqrt((r − a) / 2) and x = b / 2y

This concludes Part IIA. Part IIIA.
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