TABLE OF RIGHT DIAGONALS GENERAL METHOD

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

Square of Squares Tables

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

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.

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 + c₁² − 3b₁² ≠ 0 = S is converted to a² + b² + c² − 3b² = 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 (b² − a²) or (c² − b²) 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 + c₁² − 3b₁² ≠ 0 and convert these tuples into a second set of tuples (Table II) that obey the rule: a² + b² + c² − 3b² = 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 (b² − a²) or (c² − b²), 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:
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)

7. 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 -1 1 1 1 5 1 3 9 1 5 13 1 7 17 1 9 21 1 11 25 1 13 29 1 15 33 1 17 37 1 19 41 1 21 45 1 23 49

   f = S/d 0 -3 -8 -15 -24 -35 -48 -63 -80 -99 -120 -143 -168

   Table II a b c 1 -1 1 -2 -2 2 -7 -5 1 -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

8. 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)
   
9. 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 1 -1 1 3 3 1 5 7 1 7 11 1 9 15 1 11 19 1 13 23 1 15 27 1 17 31 1 19 35 1 21 39 1 23 43 1 25 47

   f = S/d 0 -2 0 6 16 30 48 70 96 126 160 198 240

   Table II a b c 1 1 -1 -1 1 1 1 5 7 7 13 17 17 25 31 31 41 49 49 61 71 71 85 97 97 113 127 127 145 161 161 181 199 199 221 241 241 265 287

   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, S_m, 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, S_m, for these cases are 38307 and 4327203 and the n's are 8 and 25, respectively.

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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