Oh, my G**. They had to have recourse to this type of problem, classic and elegant in its solution by Selfridge and Straus, that if two different families of positive integers $(a_1,a_2,\ldots,a_n)$ and $(b_1,b_2,\ldots,b_n)$ have the same family of pairwise sums of pairs of elements, then $n$ must be a power of $2$. (Of course, the result extends to families of real numbers).

The point is that the problem is (slightly) different. Why does uniqueness imply the fact that Bob can find the numbers in one guess?

Because, in case of uniqueness, the numbers are predetermined; the wording slightly induces the idea there is a guess, but in fact the whole crux of the problem is if there is a choice between two, or more, such selections of $n$ numbers. Bob is not asked to show how he finds the numbers; think of him as an infinite intellect genie. Some relevant links to places on AoPS where this is discussed http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849940#849940 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=306646&hilit=Selfridge http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=347130&hilit=Selfridge http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=477523&hilit=Selfridge and the original article by Selfridge & Straus http://msp.org/pjm/1958/8-4/pjm-v8-n4-p17-s.pdf

There exist well defined functions which are not algorithmically computable. The crux of the problem is to explain how Bob can find the numbers in only one guess, as is explicitly stated in the formulation of the problem.

This has nothing to do with algorithmic computability. In the general case, we have a system of $\dfrac {n(n-1)} {2}$ equations in $n$ unknowns, and the theory guarantees it has unique solution, so it is a deterministic process via Linear Algebra methods. I concede, we do not know which variables make up which sum, but there is a finite (factorial) number of ways this assignation can be done, and each can be checked, until the guaranteed unique solution is reached.

mavropnevma wrote: In the general case, we have a system of $\dfrac {n(n-1)} {2}$ equations in $n$ unknowns, and the theory guarantees it has unique solution, so it is a deterministic process via Linear Algebra methods. I completely agree, but this should be mentioned in a solution. However this argument is advanced for JBMO level.

By the way, can an empowered mod add these to the Contests section?

Done.

No need to solve the general problem for a competition of this level. The organizers only asked for simple algorithms for those basic cases. For $n=8$, it is simple to provide a counterexample, using the one they gave for $n=4$. For example, the two sets $\{1,5,7,9,4,6,8,12\}$ and $\{2,4,6,10,3,7,9,11\}$ have the same pairwise sums. For $n=5$, name $a_1$, $a_2$, $a_3$, $a_4$ and $a_5$ the five numbers you're looking for, in increasing order. The lowest of the pairwise sums is $a_1+a_2$, the highest is $a_4+a_5$. Now calculate the sum of all your $10$ numbers, and divide by $4$, you obtain $a_1+a_2+a_3+a_4+a_5$. By subtracting, we know $a_3$. Now, the second highest pairwise sum is $a_1+a_3$, so we now have $a_1$, and the ending is easy. The case $n=6$ is slightly trickier. I use the same notations as before. From the start, we know $a_1+a_2$, $a_1+a_3$, $a_4+a_6$ and $a_5+a_6$. By summing all $15$ numbers and dividing by $5$, we also know the sum of the $a_i$. From this, we deduce the values of $a_3+a_4$, $a_2+a_4$, $a_3+a_5$. By subtracting $a_1+a_2$ and $a_1+a_3$, we get $a_2-a_3$, and similarly we know $a_5-a_4$. By adding those to $a_3+a_4$, we find $a_2+a_5$, and then $a_1+a_6$. Now look at the third lowest pairwise sum. It is either $a_1+a_4$ or $a_2+a_3$, but we have no way of knowing which. However, we have access to $a_1+a_2+a_3+a_4$, so we also have access to the multiset $\{a_1+a_4,a_2+a_3\}$, symmetrically, we also have $\{a_3+a_6,a_4+a_5\}$. There are only two pairwise sums we haven't identified yet, they are necessarily $a_1+a_5$ (for the lowest) and $a_2+a_6$. By subtracting $a_1+a_6$ and $a_2+a_6$, we deduce $a_1-a_2$, and since we know $a_1+a_2$, we have determined $a_1$ and $a_2$. The ending is easy.