TABLE OF RIGHT DIAGONALS GENERAL METHOD

GENERATION OF RIGHT DIAGONALS FOR MAGIC SQUARE OF SQUARES (Part IVE)

Picture of a square

Square of Squares Tables

Andrew Bremner's article on squares of squares included the 3x3 square:

Bremmer's square
5824621272
942113222
972822742

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.

We will show a general method for generating the squares for a right diagonal of a magic square. Beginning with the the tuple (1,b1, c1) we can generate the tuple (a, b, c) which when squared gives the diagonal numbers. Initially either b1 or c1 will be equal to ± k where k is any natural number 1,2,3,4.... Again the end result is that a12 + b12 + c123b120 = S is converted to a2 + b2 + c23b2 = 0 which is a necessary condition for the square to be magic.

To summarize the tuples of Table II below will be used as entries into a right diagonal of a magic square. Knowing the difference (b2a2) or (c2b2) will give us a value Δ which can be used to produce other entries into the magic square. To date only one magic square containing 7 entries has been found. Most other squares will contain 6 entries.

As to the reason for the picture of a square, the entries to the square occur as three tuples,viz, (a,b,c), (l,m,n) and (x,y,z) showing their connectivity. In addition, six or more of these entries are present as their squares.

Generation of Tables where b1 = ±1 and c1 = ±1

  1. The object of this exercise is to generate a Table I with a set of tuples that obey the rule: a12 + b12 + c123b120 and convert these tuples into a second set of tuples (Table II) that obey the rule: a2 + b2 + c23b2 = 0.
  2. In addition, we need to know two numbers e and g where g = 2e which when added to the b1 and c1 numbers of Table I, produce the next line of numbers (n + 1) in the next row of Table I. The number a1 will always be 1.
  3. Two other numbers f and d are calculated using the equation
    f = [2e2n2 + (4c1 − 4b1) en +(1 − 2b12 + c12)] / {2(2b1 − c1 − 1)}
    where n is the line number of the tables. f can also be generated directly from Table II from S/d. However, the value of d is equal to the denominator of the general equation above.
  4. Finally Δs are calculated by taking the difference in Table II between (b2a2) or (c2b2), and the results placed under the Δ column. Both differences must be the same.
  5. As an example we begin with the tuple (1,1,1), where a1 = 1, b1 = 1 b1 = 1 and c1 = 1 and use the equation to generate f. Substituting the values for a1, b1 and c1 into the above equation above would set d = 0 and division by zero is undefined.
  6. So that the only two examples allowed are the tuple (1,−1,1), where a1 = 1, b1 = −1 and c1 = 1 and the tuple (1,1,−1), where a1 = 1, b1 = 1 and c1 = −1. Using these two tuples each individual f can then be calculated.
    For b1 = −1 the result is:
  7. f = [2e2n2 + (4 + 4)en + 0] / 2× (−4) = [2e2n2 − 8en ] / (−8)
    Setting e = 2 and g = 4 affords f = −n2 − 2n
    Substituting for f in (b) gives
    a = (−n2 −2n + 1)
    b = (−n2 − 1)
    c = (−n2 + 2n + 1)
  8. Substituting the appropriate n into the equations for a, b, and c produces Table II below. Using a computer program and the requisite calculations produced the tables below. As can be seen taking the value of f from the middle table and adding to a1, b1, c1, produced a, b, c, respectively of Table II.

      
    n
    0
    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    Table I
    a1 b1 c1
    1-11
    115
    139
    1513
    1717
    1921
    11125
    11329
    11533
    11737
    11941
    12145
    12349
      
    f = S/d
    0
    -3
    -8
    -15
    -24
    -35
    -48
    -63
    -80
    -99
    -120
    -143
    -168
    Table II
    a b c
    1-11
    -2-22
    -7-51
    -14-10-2
    -23-17-7
    -34-26-14
    -47-37-23
    -62-50-34
    -79-65-47
    -98-82-62
    -119-101-79
    -142-122-98
    -167-145-119
      
    Δ
    0
    0
    -24
    -96
    -240
    -480
    -840
    -1344
    -2016
    -2880
    -3960
    -5280
    -6864
  9. For c1 = −1 the result is as follows:
    f = [2e2n2 + (−4 − 4)en + 0] / 2× (2) = [2e2n2 − 8en ] / (4)
    Setting e = 2 and g = 4 affords f = 2n2 − 4n
    Substituting for f in (b) gives
    a = (2n2 −4n + 1)
    b = (2n2 −2n + 1)
    c = (2n2 − 1)
  10. Substituting the appropriate n into the equations for a, b, and c produces Table II below. Using a computer program and the requisite calculations produced the tables below. As can be seen taking the value of f from the middle table and adding to a1, b1, c1, produced a, b, c, respectively of Table II.

      
    n
    0
    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    Table I
    a1 b1 c1
    11-1
    133
    157
    1711
    1915
    11119
    11323
    11527
    11731
    11935
    12139
    12343
    12547
      
    f = S/d
    0
    -2
    0
    6
    16
    30
    48
    70
    96
    126
    160
    198
    240
    Table II
    a b c
    11-1
    -111
    157
    71317
    172531
    314149
    496171
    718597
    97113127
    127145161
    161181199
    199221241
    241265287
      
    Δ
    0
    0
    24
    120
    336
    720
    1320
    2184
    3360
    4896
    6840
    9240
    12144


  1. Two examples for b1 = −1 are A and B having the right diagonal tuple (23, 37, 47) and (98, 122, 142) as their squares. The magic sum, Sm, for these cases are 4107 and 44652 and the n's are 6 and 11, respectively. And the other two examples for c1 = −1 are C and D having the right diagonal tuple (97, 113, 127) and (1151, 1201, 1249) as their squares. The magic sum, Sm, for these cases are 38307 and 4327203 and the n's are 8 and 25, respectively.
Magic square A
2921057472
273737212
2324121897
  
Magic square B
732191591422
29719122272
982103224439
  
Magic square C
582188141272
25534113222
97282222174
  
Magic square D
3502264470212492
287990212012702
1151249022762302

This concludes Part IVE. To go back to Part IVD. To continue to Part VE shows the javascript methods used to calculate the tables.
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Copyright © 2012 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com