Durango Bill's
Magic Squares
Magic Squares exist for many different sizes and may have various qualifications on the numbers they contain. We will look at simple 3x3 Magic Squares and take a brief look at the “3x3 Magic Square of Squares” problem.
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| 8 | 1 | 6 |
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| 3 | 5 | 7 |
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| 4 | 9 | 2 |
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The Magic Square above illustrates the properties of a Magic Square. Each cell contains a unique integer number greater than 0. For this example, the numbers used are consecutive integers, but this does not have to hold true for all Magic Squares.
Each row, column, and diagonal in a Magic Square will sum to the same result (in this case “15”). In any 3x3 Magic Square this “Magic Sum” will be 3 times whatever the value is in the center cell.
The example above uses a center value of “5”. It is the only Magic Square solution if we require the center number to be 5. The lowest possible value in the center cell for a 3x3 Magic Square is 5.
We might ask the question: “What happens if the center number can be something larger – for example “6”?
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| 10 | 1 | 7 | | 9 | 1 | 8 | | 9 | 2 | 7 |
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| 3 | 6 | 9 | | 5 | 6 | 7 | | 4 | 6 | 8 |
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| 5 | 11 | 2 | | 4 | 11 | 3 | | 5 | 10 | 3 |
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The 3 diagrams above show the 3 possible solutions if the value of the center cell is increased to 6. The “Magic Sum” for each row, col. and diagonal has increased to 3 x 6 = 18.
If you have a solution for a 3x3 Magic Square and the center cell has some value “N”, you can always generate a solution for a center value of “N + 1” by simply adding “1” to the value in all 9 cells. For that matter, you can always add any arbitrary constant to all 9 cells and get a solution with larger values.
The 2 sections above show the solutions for center cell values of 5 and 6. What happens if we try “7” for the value in the center cell?
12 1 8 11 2 8 10 2 9 10 3 8
3 7 11 4 7 10 6 7 8 5 7 9
6 13 2 6 12 3 5 12 4 6 11 4
If the center cell is 7, then there are 4 Magic Square solutions. In each of these solutions, the “Magic Sum” is 3 x 7 = 21.
14 1 9 13 1 10 13 2 9 12 1 11
3 8 13 5 8 11 4 8 12 7 8 9
7 15 2 6 15 3 7 14 3 5 15 4
12 3 9 11 3 10 11 4 9
5 8 11 7 8 9 6 8 10
7 13 4 6 13 5 7 12 5
If the center cell is equal to 8, we get the 7 solutions shown above. In all of the examples given above and in the table below, we are only counting “primitive” solutions. You can swap the outside rows and/or columns, rotate the Magic Square, generate mirror images using the diagonals, etc. to generate trivial variations of these; but the examples given above show all the “primitive” solutions.
16 1 10 15 1 11 15 2 10 14 1 12
3 9 15 5 9 13 4 9 14 7 9 11
8 17 2 7 17 3 8 16 3 6 17 4
14 2 11 14 3 10 13 2 12 13 4 10
6 9 12 5 9 13 8 9 10 6 9 12
7 16 4 8 15 4 6 16 5 8 14 5
12 4 11 12 5 10
8 9 10 7 9 11
7 14 6 8 13 6
If we extend the process again we get the 10 solutions shown above that have a center value of “9”. It can be noted that the number of solutions that have a center value of “N+1” is equal to the number of solutions that have a center value of “N” plus the number of new solutions that have a single digit of “1” in the middle of the top row.
Computer programs can generate solutions much easier than humans can. The following table shows the number of primitive solutions that exist given the value in the center cell.
Center Magic Number of
Cell Sum Solutions
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5 15 1
6 18 3
7 21 4
8 24 7
9 27 10
10 30 13
11 33 17
12 36 22
13 39 26
14 42 32
15 45 38
16 48 44
17 51 51
18 54 59
19 57 66
20 60 75
25 75 124
30 90 187
40 120 348
50 150 560
75 225 1,308
100 300 2,368
200 600 9,735
500 1,500 61,835
1,000 3,000 248,668
2,000 6,000 997,335
5,000 15,000 6,243,335
10,000 30,000 24,986,668
20,000 60,000 99,973,335
50,000 150,000 624,933,335
100,000 300,000 2,499,866,668
200,000 600,000 9,999,733,335
500,000 1,500,000 62,499,333,335
1,000,000 3,000,000 249,998,666,668
10,000,000 30,000,000 24,999,986,666,668
100,000,000 300,000,000 2,499,999,866,666,668
1,000,000,000 3,000,000,000 249,999,998,666,666,668
3,999,999,999 11,999,999,997 3,999,999,992,666,666,670
4,000,000,000, 12,000,000,000 3,999,999,994,666,666,668
By the time you get to 1,000,000, there are recognizable patterns in the number of solutions. If you are only interested in how many Magic Squares exist for a given center cell value, and not interested in generating the actual Magic Squares; then the following recursive equation will get you there.
Center Number of Magic
Cell Value Squares “F[N}”
5 1
6 3
N+1 = 2 * F[N] – F[N-1] + G[MOD(N, 6)]
where G[MOD(N,6)] is from the following table:
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MOD(N,6) | 0 | 1 | 2 | 3 | 4 | 5 |
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G[MOD(N,6)] | -1 | 2 | 0 | 0 | 1 | 1 |
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For example, if you know how many Magic Squares exist for center cell values of “5” and “6”, then the number of Magic Squares for the next center cell value (in this case: N + 1 = 7) can be calculated as follows:
1) N = 6 Last center cell value where we have known information
2) F[N] = 3 Known (Number of solutions with center cell value of 6)
3) F[N-1] = 1 Known (Number of solutions with center cell value of 5)
4) G[MOD(N,6)] = -1 MOD(N,6) is zero.
Find “0” in the MOD(N,6) table above and use the “-1”
underneath it.
F[N+1] = 2 * F[N] – F[N-1] + G[MOD(N,6)]
F[7] = 2 * 3 - 1 + (-1)
F[7] = 4
Repeat the above as needed for F[8], F[9], etc.
Here is a solution with the value in the center cell equal to 1,000,000.
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| 1,999,998 | 1 | 1,000,001 |
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| 3 | 1,000,000 | 1,999,997 |
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| 999,999 | 1,999,999 | 2 |
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Magic Squares Using Prime Numbers
There are many variations of magic squares that impose restrictions on the numbers that are used in the individual cells. One variation is to restrict entrees to just prime numbers. There appear to be an infinite number of 3x3 magic squares that use just prime numbers.
67 1 43 101 5 71 101 29 83
13 37 61 29 59 89 53 71 89
31 73 7 47 113 17 59 113 41
109 7 103 149 11 107
67 73 79 47 89 131
43 139 37 71 167 29
The 5 diagrams above show the 5 possible magic squares composed of prime numbers with the value in center square less than 100. As the value in the center square is allowed to increase, many more prime-number magic squares are possible. If the value in the center square can be any prime number under 1,000, there are 474 possible solutions.
If the value in the center square can be any prime number under 1,000,000, then there are 1,044,538,640 possible solutions. The last prime number under 1,000,000 is 999,983. There are 33,542 possible 3x3 prime-number magic squares with a value of 999,983 in the center square. Here’s one of them.
1,009,373 982,343 1,008,233
998,843 999,983 1,001,123
991,733 1,017,623 990,593
As can be seen above, it’s also possible to restrict prime number solutions to a specific last digit. Here are a few examples.
491 41 311 883 13 673
101 281 461 313 523 733
251 521 71 373 1033 163
577 7 337 569 59 449
67 307 547 239 359 479
277 607 37 269 659 149
Or even the last several digits:
77,011,111 6,511,111 53,911,111 61,033,333 10,333,333 41,533,333
22,711,111 45,811,111 68,911,111 18,133,333 37,633,333 57,133,333
37,711,111 85,111,111 14,611,111 33,733,333 64,933,333 14,233,333
109,477,777 75,277,777 104,077,777 116,099,999 58,499,999 110,699,999
90,877,777 96,277,777 101,677,777 89,699,999 95,099,999 100,499,999
88,477,777 117,277,777 83,077,777 79,499,999 131,699,999 74,099,999
Magic Square of Squares
While the above magic squares use simple integers to fill the cells, a much more difficult variation is to fill the cells with perfect squares. Thus integers (A, B, C, D, E, F, G, H, I) have to be found such that the following becomes a Magic Square.
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| A^2 | B^2 | C^2 |
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| D^2 | E^2 | F^2 |
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| G^2 | H^2 | I^2 |
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1) A^2 + B^2 + C^2 = Sum
2) D^2 + E^2 + F^2 = Sum
3) G^2 + H^2 + I^2 = Sum
4) A^2 + D^2 + G^2 = Sum
5) B^2 + E^2 + H^2 = Sum
6) C^2 + F^2 + I^2 = Sum
7) A^2 + E^2 + I^2 = Sum
8) C^2 + E^2 + G^2 = Sum
Sum = 3(E^2)
(Add equations 5, 7, and 8 and then subtract equations 1 and 3).
For more information, please see the following article in Scientific American Magazine.
“Can You Solve a Puzzle Unsolved Since 1996?”
“A challenge that the famous puzzler Martin Gardner put forward long ago remains open” http://www.scientificamerican.com/article/can-you-solve-a-puzzle-unsolved-since-1996/
Alternately, please see Christian Boyer’s web page about the “Magic Square of Squares” problem.
http://www.multimagie.com/English/SquaresOfSquares.htm
This is an unsolved problem in mathematics as no one has found a solution, but no one has proved that a solution is not possible.
The author gave this one a crack, but the only result that I got was that if a solution exists, then E2 has to be greater than 3,600,000,000,000,000. (3.60E15)
A solution at this range (or greater) does not look very promising. If we check equations 1 - 8 above, we note that there are 4 equations that have a term of E2. Thus there have to be 4 triplets where the sum of their squares is equal to some single number – the “Magic Sum” - which is equal to 3(E2).
If E = 60,000,000, then 3(E2) is equal to 1.08E16. If we check how many ways that numbers near 1.08E16 can be split into the sum of 3 squares, where the middle term is E2, we find that a significant number don’t have the required 4 triplets. For example, for the 10,000 values of E from E = 59,990,001 to E = 60,000,000, only 3,314 values for E have at least 4 qualifying triplets, and 2,373 of these had just the minimum of 4. Sample runs for other large E values showed similar ratios. (For that matter, if there aren’t at least 5 triplets, there cannot be a solution. http://mathpages.com/home/kmath417.htm )
Of the candidates that do have 5 or more triplets (about 9% of the candidates for “E”), they then face astronomical odds of matching up the row and column sums. It appears highly unlikely that a solution exists for the Magic Square of Squares problem.
Magic squares – all possible powers
>= 2
Another variation that was tried was to try to form a magic square allowing all possible powers >= 2.
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| A^R | B^S | C^T |
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| D^U | E^V | F^W |
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| G^X | H^Y | I^Z |
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In the diagram above, the letters A thru I can be any positive integer >= 1. The associated powers (letters R thru Z) can be any integer power >= 2. In practice, R thru Z can be restricted to just prime numbers which is a great help for computer run time.
A search was made using all center cell values ( E^V ) up to 1.0E14. No solutions were found.
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