Interestingly enough, the 2-at-a-time sum problem for n=4 isn't necessarily unique! For example, {1, 2, 3, 6} and {0, 3, 4, 5} produce the same sum set, {3, 4, 5, 7, 8, 9}, when the sums are taken 2 at a time.
and there's an easy bootstrapping to produce such an example for n = 2^k+1 , starting from an example for n = 2^k . this is basically part (a) of the proposed theorem. it appears that the bootstrapping can be arranged to get the starting multisets to be actual sets (no repeats). i do not see if it is possible to also get the same for the 2-at-a-time sums multiset by the bootstrapping method i have in mind. perhaps another method can accomplish that. part (c*) of the problem, which is starred, meaning its solution is unknown, asks if there are 3 distinct multisets that have the same 2-at-a-time sums multiset.
I have a book (at home, alas) that talks about this problem. Basically, for n a power of 2, the numbers are not necessarily determined by the sums taken two at a time. If n is not a power of 2, then the numbers are uniquely determined. I can dig up the reference later, if anyone is interested.
count me interested! mike