thanks R! very informative. in particular, the example answers part (c*) of the proposed problem. my only suggestion is that the text does not make it explicitly clear if you are considering multi-sets (or usual sets). i infer that you are considering multi-sets (i checked that the example works on the level of multi-sets), so perhaps you could state that explicitly. is the usual-set case of any (further) interest? mike

I wonder if the `book at home' is UPINT? Section C5 currently reads as follows:

\usection{C5}{Sums determining members of a set.}

\hGidx{Moser, Leo} Leo Moser asked, and Selfridge, Straus and others largely settled, to what extent the sums of all the pairs of numbers in a set determine the set. They show that if the cardinality is not a power of two, then the members are determined. Suppose that $y_1$, $y_2$, $\ldots$, $y_s$ are the sums $x_i+x_j$ ($i\neq j$)of the numbers $x_1$, $x_2$, $\ldots$, $x_{2^k}$, so that $s=2^{k-1}(2^k-1)$. Are there more than two sets $\{x_i\}$ which give rise to the same set $\{y_j\}$\,? If $k=3$ there may be three such sets, for example. $$\{\pm1,\pm9,\pm15,\pm19\},\quad\{\pm2,\pm6,\pm12,\pm22\}, \quad\{\pm3,\pm7,\pm13,\pm21\}$$ but there can't be more than three. For $k>3$ the problem is open.

The corresponding problem where sums of {\em triples} of elements of a set are given has been settled by Boman \& Linusson. The exceptions are precisely 3, 6, 27, 486. For $n=27$ they give five examples of which the simplest is $\{-4,-1^{10},2^{16}\}$ and its negative, where exponents denote repetitions. For $n=486$ they give $\{-7,-4^{56},-1^{231},2^{176},5^{22}\}$ and its negative.

The corresponding problem for sums of {\em four} distinct elements was settled by Ewell.

\medskip

\small

\Aidx{BakerI} I.~N.~Baker, Solutions of the functional equation $(f(x))^2-f(x^2)=h(x)$, {\it Canad.\ Math.\ Bull.}, {\bf3}(1960) 113--120.

\Aidx{Bolker} E.~Bolker, The finite Radon transform, {\it Contemp.\ Math.}, {\bf63}(1987) 27--50; {\it MR} {\bf88b}:51009.

\Aidx{Boman}\Aidx{ONeil} J.~Boman, E.~Bolker \& P.~O'Neil, The combinatorial Radon transform modulo the symmetric group, {\it Adv.\ Appl.\ Math.}, {\bf12}(1991) 400--411; {\it MR} {\bf92m}:05014.

\Aidx{Linus} Jan Boman \& Svante Linusson, Examples of non-uniqueness for the combinatorial Radon transform modulo the symmetric group, {\it Math.\ Scand.}, {\bf 78}(1996) 207--212; {\it MR} {\bf97g}:05014.

\Aidx{Ewell} John A.~Ewell, On the determination of sets by sets of sums of fixed order, {\it Canad.\ J.\ Math.}, {\bf20}(1968) 596--611.

\Aidx{Gordon}\Aidx{Fraen}\Aidx{Strau} B.~Gordon, A.~S.~Fraenkel \& E.~G.~Straus, On the determination of sets by the sets of sums of a certain order, {\it Pacific J.\ Math.}, {\bf12}(1962) 187--196; {\it MR} {\bf27} \#3576.

\Aidx{Honsb} Ross A.~Honsberger, A gem from combinatorics, {\it Bull.\ ICA}, {\bf1}(1991) 56--58.

\Aidx{Lambek}\Aidx{Moser} J.~Lambek \& L.~Moser, On some two way classifications of the integers, {\it Canad.\ Math.\ Bull.}, {\bf2}(1959) 85--89.

\Aidx{LiuB}\Aidx{gX} B.~Liu \& X.~Zhang, On harmonious labelings of graphs, {\it Ars Combin.}, {\bf36}(1993) 315--326.

L.~Moser, Problem E1248, {\it Amer.\ Math.\ Monthly}, {\bf64}(1957) 507.

\Aidx{Ossow} J.~Ossowski, On a problem of Galvin, {\it Congressus Numerantium}, {\bf96}(1993) 65--74.

\Aidx{RogerD} D.~G.~Rogers, A functional equation: solution to Problem 89-19*, {\it SIAM Review}, {\bf32}(1990) 684--686.

\Aidx{Selfr} J.~L.~Selfridge \& E.~G.~Straus, On the determination of numbers by their sums of a fixed order, {\it Pacific J.\ Math.}, {\bf8}(1958) 847--856; {\it MR} {\bf22} \#4657.

\normalsize

Suggestions for improvement wd be much appreciated.

Best, R.

On Tue, 14 Oct 2008, Michael Reid wrote:

...

...

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