partition the set

I need some help on this question please.

The set {1, 2, 3, 4} can be partitioned into two sets {1, 4} and {2, 3} of the same size. Notice that 1+4 = 2+3.

Question
Jane says she can partition the set {1, 2, .........16} into two subsets S and T of the same size so that:
- the sum of the numbers in S equals the sum of the numbers in T
- ther sum of the squares of the numbers 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
Is Jane right or wrong? Clearly explain your answer.

I would appreciate any help.

The hint suggests that you pair 1 with 16, 2 with 15, and so on. This will ensure that the sum of the numbers in each partition is equal. We'll keep this in mind as we partition the set in such a way that the sum of squares is equal.

1² + 16² = 257
2² + 15² = 229
3² + 14² = 205
4² + 13² = 185
5² + 12² = 169
6² + 11² = 157
7² + 10² = 149
8² + 9² = 145

Note that the sum of all squares is 1496, so the sum of squares for each partition must be 748. Start with 1 and 16 in subset S. Since the sum of their squares is 257, the sum of the squares of the remaining numbers in S must be 491. Using the above table as reference, we see that the least sum of squares for any two pairs is 294. This from 491 leaves 197, so we know that the pairs with squares totaling more than 197 cannot be placed in subset S with 1 and 16. They must be placed in subset T.

Now we have subset T with 2, 15, 3, and 14. The sum of their squares is 434, leaving 314 as the sum of the remaining squares. This is obviously 169 + 145, so we complete subset T with 5, 12, 8, and 9. The remaining numbers belong in subset S, and the sum of squares of the numbers in each subset is 748.

Calculating the sum of the cubes for the numbers in each subset confirms this solution, as the sum is 9248 for each subset.

S = {1, 4, 6, 7, 10, 11, 13, 16}
T = {2, 3, 5, 8, 9, 12, 14, 15}