each pairwise sum is a perfect square and the sum of all four is a perfect square.
Now that I think of it, with two of my numbers' being equal, there are really only 3+1 = 4 different squares being formed from my solutions.
blackjack
<a href="http://c.casalemedia.com/c?s=51816&f=3&id=0" target="_blank"><img src="http://as.casalemedia.com/s?s=51816&u=http://perplexus.info&f=3&id=0&if=0" width="120" height="600" border="0"></a>
flooble's webmaster puzzle
re(2): A (bad?) guess
