WILLIES MISTAKE PEOPLE PLEASE CORRECT HIS MISTAKE

The set (1,2,3,4) can be partioned into two subsets (1,4) and (2,3) of the same size. 1+4=2+3
a)find the next whole number n, above 4, for which the set (1,2,...n) can be paritioned into two subsets S and T of the same size, with the sum of S and T the same.

b)find all partition in a) with the additional property that the sum of the squares of the numbers in S equals the sum of the squares of the numbers in T.

Terri says she can partition the set (1,2...16) into subsets S and T of the same size so that:
- the sum of the numbers in S equals the sum of T
-the sum of the squares of the number in S equals the sum of the squares of the numbers in T.
-the sum of the cubes of the numbers in S equals the sum of the cubes of the numbers in T.

c) show terri is right.

Willy says she can partition the set (1,2...8) into subsets S and T, not necessarily the same size so that:
- the sum of the numbers in S equals the sum of T
-the sum of the squares of the number in S equals the sum of the squares of the numbers in T.
-the sum of the cubes of the numbers in S equals the sum of the cubes of the numbers in T.

d)show why you dont belive willy

a) 1,..,8 T={1,2,7,8} S={3,4,5,6}
EDIT: Not sure about the rest but I guess it involves the Prouhet-Tarry-Escott problem in some way:
http://mathworld.wolfram.com/Prouhet-Tarry-EscottProblem.html

willy's mistake, not willies.

b)
8 + 5 + 3 + 2 = 7 + 6 + 4 + 1 = 18
64 + 25 + 9 + 4 = 49 + 36 + 16 + 1 = 102

1 + 2 + 3 + 4 + + 13 + 14 + 15 + 16 = 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 68



c) Sum of the cubes from 1 to 8 = 1296
Hence each partition must contain 648.
So any solution must have two odd cubes in each partition. (Can't be zero and four.)

8^3 = 512
This means the other cubes in the set must sum to 136.
By inspection, there is no combination of other cubes that sum to 136.
Hence there is no solution.

I would be very surprised if this could be done. A certain corollary from the above mention theorem states how the numbers from 0 to (2^n) - 1 can be divided in two even groups such that the sum of their powers up to n-1 is equal.

I did this puzzle. It's on this year's "maths challenge for young Australians"

a. 8
b.
c. {1,4,6,7,10,11,13,16} & {2,3,5,8,9,12,14,15} I think. Can't remember it.
d. The only set {1,4,6,7} & {2,3,5,8} can't stand the cube bit.