A **magic square** contains positive consecutive integers starting from 1 such that the rows, columns, and diagonals all add to the same value.

A **heterosquare** contains positive consecutive integers starting from 1 such that the rows, columns, and diagonals all add to *different* values.

If the sums resulting from a heterosquare form a consecutive sequence, the heterosquare is also called an **antimagic square**. (In the example above, if the sums of 9 and 24 were 16 and 17 instead, it would be a antimagic square.)

**Prove that there are no 3 by 3 antimagic squares.**

THE TEXT OF THE PROOF IS HERE.

Yes, we have a proof! Thanks to: Mike Harding, Steven Jim, Marcus Luebke, Stefan Van der Waal, and Steven Yuan. (I believe that's everyone who contributed - please chime in if I missed someone.)

Now, there still are quite possibly improvements to be made, so even if you haven't contributed yet, feel free to check the proof at the wiki to see if any of the steps can be condensed. It's particularly useful to do so in this case because there was already a brute-force computer proof of the above fact; therefore it would be nice to have the most elegant proof possible.

Fairly soon (starting in two weeks) I will put the text down in a LaTeX file, add an introduction and citations, and work on getting it published in an actual journal.

Open Problem #3 will start January 8th.

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## Comments

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TopNewestThe first part of the proof can be condensed.

We all know that \(2(b+d+f+h)+3(a+c+g+i)+4e=8n+28\) or \(a+c+g+i+2e=8n-62\). Also, the minimum and maximum values of \((c+g+e)+(a+i+e)\) are \(2n+1\) and \(2n+15\), respectively, which means \(2n+1 \le 8n-62 \le 2n+15\) or \(11 \le n \le 12\) (as \(n\) must be an integer.

That's it.

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