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Even-Numbered Magic Squares


Date: 04/30/2001 at 13:23:58
From: David
Subject: Magic square

Dear Dr. Math,

I believe your method of constructing magic squares (see 
http://mathforum.org/dr.math/problems/magic_squares_needed.html   , 
Magic Square Puzzle) is good for odd integers, but does not work for, 
say, n = 4. Do you have a different way that works for even numbers?

David


Date: 05/11/2001 at 10:01:44
From: Doctor Toughy
Subject: Re: Magic square

Hi David,

Thanks for writing to Dr. Math.

I found the following information in the Dr. Math archive by searching 
on the keywords 'magic square':

  How to Construct Magic Squares - Allan Adler
  http://mathforum.org/alejandre/magic.square/adler/adler4.html   

This will give you the following 4x4 magic square and directions for 
how to create it:

         1  15  14   4       
        12   6   7   9       
         8  10  11   5       
        13   3   2  16       

Each row and column adds up to 34.  In order to create 6x6, 8x8, and 
larger magic squares, subtract 8 1/2 from each of the above numbers to 
create a new 4x4 square where each row and column now add up to zero:

         1  15  14   4         -7 1/2    7 1/2   6 1/2   -4 1/2
        12   6   7   9          3 1/2   -2 1/2  -1 1/2      1/2
         8  10  11   5           -1/2   -1 1/2  -2 1/2   -3 1/2
        13   3   2  16          4 1/2   -5 1/2  -6 1/2    7 1/2

In creating our 6x6 we will use the numbers -17 1/2 through -8 1/2  
and 8 1/2 through 17 1/2. Except for the numbers used in the corners, 
each number we use on the edge will have its negative counterpart on 
the opposite end of that row. The corner number will have its opposite 
on the opposite corner.  

So, using the 10 numbers +/- 8 1/2 through +/- 17 1/2, select two 
numbers to determine whether they will work as corner numbers. We'll 
choose +/- 8 1/2 and +/- 9 1/2. This is a random selection, since in 
actuality there are many possible combinations that will work as 
corner numbers.  

When we add these numbers together we can get +/- 1 and +/- 18.  With 
the eight remaining, is it possible to make two sets each with four 
numbers such that one set when added equals +/- 18 and the other 
equals +/- 1?  It turns out that it is possible:

  10 1/2 + 11 1/2 + 13 1/2 - 17 1/2 = + 18
  
  14 1/2 + 15 1/2 - 12 1/2 - 16 1/2 = + 1

Now fill in the four corners with the opposites diagonal to each 
other.  

   8 1/2                   9 1/2


   


  -9 1/2                  -8 1/2

Since the bottom two corners add up to -18, place the four numbers 
that sum to +18 along the bottom in any order. Place their opposites 
across the top. 
 
Since the two left corners add up to -1, place the four numbers that
sum to +1 along that side in any order and the opposites along the 
other side. You're now ready to either create an 8x8 or add 18 1/2 to 
each number to create a 6x6 magic square.
  
In creating an 8x8 magic square we need a total of 28 more numbers 
to fill in the outer edge of our 6x6, making it an 8x8. Since the 
highest/lowest numbers used in our 6x6 were +/- 17 1/2, the numbers 
we will use are +/- 18 1/2 through +/- 31 1/2. We'll try 18 1/2 and 
19 1/2 as our center numbers. Adding the two numbers we get:

       +/- 18 1/2 +/- 19 1/2 = +/-1 and +/- 38

Using the set of numbers +/- 20 1/2 through +/- 31 1/2 we make two 
sets of six numbers each consisting of six numbers which, when added 
together, equal the sums of the corner numbers:

  20 1/2, 21 1/2, 22 1/2, 23 1/2, -24 1/2, -25 1/2 

which, when added together, equal 38, and 

  26 1/2, 28 1/2, 31 1/2, -27 1/2, -29 1/2, -30 1/2 

which, when added together, equal -1

Place +/-18 1/2 in the four corners just as we did earlier.

   18 1/2                      19 1/2






   -19 1/2                    -18 1/2

   
Since the bottom two corner numbers add up to -38, place the numbers 
in the first set, which sum to +38, in the bottom row with their 
opposites in the top. Place the numbers in the second set, which sum 
to -1, on the right side with their opposites on the left, and you 
have an 8x8. 

Hope this helps. If you have any more questions about this or any 
other math topics, please write back.

- Doctor Toughy, The Math Forum
  http://mathforum.org/dr.math/   
    
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