Question

How to divide `1^2, 2^2, ..., 64^2` set into 4 subsets with equal sum and 16 numbers in each?

Answer 1

Here's one method...

Explanation:

Start with `4` empty sums and add squares to them from the list, trying to keep the sums as even as possible:

`{ (1^2+8^2 = 65), (2^2+7^2 = 53), (3^2+6^2 = 45), (4^2+5^2 = 41) :}`

Do the same with the next group of `8` squares:

`{ (9^2+16^2 = 337), (10^2+15^2 = 325), (11^2+14^2 = 317), (12^2+13^2 = 313) :}`

Notice that the differences of the sums are `4, 7, 12`

We can combine these two sets of sums to smooth out the differences in each group of two sums by transpositions to get:

`{ (1^2+8^2+10^2+15^2 = 390), (2^2+7^2+9^2+16^2 = 390), (3^2+6^2+12^2+13^2 = 358), (4^2+5^2+11^2+14^2 = 358) :}`

Use the same permutation with the next `16` squares to find:

`{ (17^2+24^2+26^2+31^2 = 2502), (18^2+23^2+25^2+32^2 = 2502), (19^2+22^2+28^2+29^2 = 2470), (20^2+21^2+27^2+30^2 = 2470) :}`

We can now add these four sums in reverse order to the previous ones to cancel out the difference `32` as follows:

`{ (1^2+8^2+10^2+15^2+20^2+21^2+27^2+30^2 = 2860), (2^2+7^2+9^2+16^2+19^2+22^2+28^2+29^2 = 2860), (3^2+6^2+12^2+13^2+18^2+23^2+25^2+32^2 = 2860), (4^2+5^2+11^2+14^2+17^2+24^2+26^2+31^2 = 2860) :}`

Having arrived at a permutation that works for `32` squares, repeat it with the next `32` squares to get a solution for `64` squares.