Here's a fun puzzle for anyone who wants to have a go. It's not amazingly difficult or anything. I'm sure you maths super-geniuses will do it in 5 minutes :)

Take the set of integers 1 to 12 and divide them into two groups, such that the sum of the cubes of the integers in each group is equal. Then, transfer as few integers as possible between the two groups to yield two new groups in which the sum of the squares of each integer is equal. Then, transfer as few integers as possible between the two groups to yield two even newer groups in which the sum of the integers in each group is equal.

List the two groups in all three cases and, in the latter two cases, note which integers you moved to create them.

wtf! bump!

I was seriously close to getting the squares, but I was 3 off and it was 3 am and I wanted to go to sleep.  I promise to work on this tomorrow T.T

Hint: There is only one possible solution.

More hints in black (better use the classic forum color scheme!):

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There is only 1 way to divide the numbers to 2 groups so that the sum of cubes is equal.
Can you tell, before choosing the numbers, what that sum of cubes should be?
Knowing that number, can you think of an easy way to choose the numbers?

Start by placing 12 in one of the groups and think where 11 should be; continue from there.
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Hmm, that's not how I did it at all... not sure what you're getting at :)

Care to post your solution/reasoning?

Maybe if noone else gets it. I don't like the way I did it. Your way is probably better.

My way feels too "brute-force"-y.

Nah, mine was too.

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Knowing that there is only one solution helps. I found one solution quickly, but I couldn't convince myself there was only one solution without enumerating possibilities a long time and that got boring.
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