Largest Magic Square Ever Known

Date: 09/18/2001 at 03:46:49
From: Jayson Javellana
Subject: Largest Magic Square Ever Known

Dr. Math,

I am eager to know what was the largest magic square ever constructed 
and known in the world of mathematics. What is the easiest way to find 
the sum of each row, column, and main diagonals of magic squares?

Thank you. 
Jayson

Date: 09/18/2001 at 11:35:02
From: Doctor Rob
Subject: Re: Largest Magic Square Ever Known

Thanks for writing to Ask Dr. Math, Jayson.

There are magic squares of every odd order, so there is no maximum.
Here is a 9-by-9 magic square that shows the pattern for the
construction of odd orders (see where the numbers 1, 2, ..., 9 are
placed; then 10, 11, ..., 18; then 19, 20, ..., 27; and so on):

   37 28 19 10  1 73 64 55 46
   47 38 29 20 11  2 74 65 56
   57 48 39 30 21 12  3 75 66
   67 58 49 40 31 22 13  4 86
   77 68 59 50 41 32 23 14  5
    6 78 69 60 51 42 33 24 15
   16  7 79 70 61 52 43 34 25
   26 17  8 80 71 62 53 44 35
   36 27 18  9 81 72 63 54 45

There are also known constructions for large even orders, but they are 
more complicated.

If an n-by-n magic square contains the numbers from 1 to n^2, then
the sum of any row, column, or main diagonal must be n*(n^2+1)/2.
For example, if n = 9, as above, then the sum would be 9*82/2 = 369.

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

Date: 09/18/2001 at 11:48:07
From: Doctor Jubal
Subject: Re: Largest Magic Square Ever Known

Hi Jayson,

Thanks for writing to Dr. Math.

I searched the Dr. Math archives on the keyword:

  magic square (exact phrase)

and found this link

   Finding Magic Squares
   http://mathforum.org/dr.math/problems/holaday5.12.98.html   

which gives instructions for creating a magic square as large as you 
want, so long as the square has an odd number of squares on a side.  
Actually, however, you can create magic squares as large as you want, 
even if they have an even number of squares on a side.

If the number of squares on each side is divisible by four, start off 
by just numbering the squares in order across each row:  (I'll 
illustrate the method for 8x8.)

   1   2   3   4   5   6   7   8

   9  10  11  12  13  14  15  16

  17  18  19  20  21  22  23  24

  25  26  27  28  29  30  31  32

  33  34  35  36  37  38  39  40

  41  42  43  44  45  46  47  48

  49  50  51  52  53  54  55  56

  57  58  59  60  61  62  63  64

Now, divide the square into smaller 4x4 squares, and draw in the 
diagonals of those four-by-four squares. Some of the numbers will be 
crossed out in this process

   \   2   3   /   \   6   7   / 
  
   9   \   /  12  14    \   /  16

  17   /   \  20  21    /   \  24
  
   /  26  27   \   /  30  31   \
  
   \  34  35   /   \  38  39   /
  
  41   \   /  44  45   \   /  48
  
  49   /   \  52  53   /   \  56
  
   /  58  59   \   /   62 63  \

Finally, replace all the numbers that got crossed out with 
n^2 + 1 - a, where n^2 is the number of cells in the square (64 in 
this case) and a is what the number originally was. This gives you

  64   2   3  61  60   6   7  57

   9  55  54  12  14  51  50  16

  17  47  46  20  21  43  42  24

  40  26  27  37  36  30  31  33

  32  34  35  29  28  38  39  25

  41  23  22  44  45  19  18  48

  49  15  14  52  53  11  10  56

   8  58  59   5   4  62  63   1

which is a magic square. This will hold for any square of side n 
divisible by 4.

There is also a way to generate magic squares as large as you want 
with sides of 4n+2 (even, but not divisible by 4), but it's more 
complicated.
  
Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

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