So I've messing around with magic squares and something occured to me: Assume we have a nxn grid of numbers which respects the sum conditions of a magic square as in it has the appropriate column, diagonal and row sums. Is it guaranteed that it is a magic square or do I also need to check whether it contains the appropriate numbers? Basically does a non-magic square grid that respects the summing conditions of a magic square grid exist? If such a grid exist, does it also exist if all the numbers in the grid are guaranteed to be positive? I was thinking about this and I haven't figured out a way to prove any of these statements wrong or true. Any help is much appreciated!


If I understand what you are asking, you want to know if any arrangement of numbers in a square, where the rows, columns, and diagonals all sum to the same amount, that is not a magic square. If this is the case, then the only other requirement is that the numbers in the square all be distinct. However, there is a simple counterexample: $$ \begin{array}{ccc} 1&1&1\\1&1&1\\1&1&1 \end{array} $$ Here, every row, column, and diagonal sums to 3, but it is not a magic square because not every number is distinct (in fact, none of them are!).

  • $\begingroup$ Maybe this is not clear from my question but when I say the summing conditions must be respected I am also talking about the magic constant. Which means that everything must add up to 45 in this case. $\endgroup$ – Veritas Nov 28 '14 at 19:10
  • $\begingroup$ A 3x3 square has a magic constant of 15, so you could create a 3x3 square of 5's, which is still not a magic square for the same reason as above. $\endgroup$ – KSmarts Nov 28 '14 at 19:13
  • $\begingroup$ ^ this should suffice. I got caught up in trying to come up with a formal proof and didn't think of a good counterexample. Thanks for the fast answer. $\endgroup$ – Veritas Nov 28 '14 at 19:15

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