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.

(In reply to lower square in progress... by Tristan)

Hmph.  I was brute forcing this using pairs summing up to 90.  After a few failed tries, I thought of a proof of the lack of possibilities there.

Take the two opposite columns.  Their total sum must be 270, because of the three pairs it includes.  Therefore, each row and column must add to 135.  Then, the middle number must be 45, which isn't a prime.

Similarly, the sum of opposite numbers can't be any multiple of ten.

Pushing this further, if the sum of opposite numbers is x, then the sum of each column must be 1.5 * x, and the middle number .5 x.  Therefore, the sum of opposite sides must be double of a prime, and the sum of each row triple of a prime.

I hope this helps others if they happen to be searching the same way I am.

Posted by Tristan on 2004-03-09 19:32:03