Find a 3x3 magic square that is composed of 9 prime numbers (not the numbers from 1-9) and show how you found it.

(A magic square, as you may already know, is one in which the respective sums of the numbers in all the rows, columns, and both major diagonals all add up to the same number.)
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Since "Magic Square" is a term used outside the scope of this problem, I'm sure you can find an answer on the internet. Please find a solution independently.

Every magic square can be written as:

  [1 1 1]     [0 2 1]     [1 0 2]
a*[1 1 1] + b*[2 1 0] + c*[2 1 0]
  [1 1 1]     [1 0 2]     [0 2 1]

For all the entries to be prime, a cannot be a mult of 2 or 3 and b and c both must be mult of 6.

If we allow 1 in our prime magic square; a=1, b=6, c=30 yields:

  [31 13 19]
  [73 37  1]
  [ 7 61 43]

Posted by Brian Smith on 2004-03-10 11:33:31