When are eight integers entirely determined by their pairwise sums? Alice picks 8 numbers, which are not necessarily different. Once she has picked them, she writes out the addition of all the pairs on a piece of paper, which she gives to Basil. Basil wins if he can guess correctly the original n numbers, which Alice chose. Can Basil be certain that he will win?
After a lot of trial and error I found the case where Alice picks the numbers $1,5,7,9,12,14,16,20$ which have the same pairwise sums as the numbers $2,4,6, 10,11,15,17,19$. However the trial and error method is extremely laborious and tedious. Is there a more mathematical approach, which can immediately give you the solution?

Hint: if the collections $(a_1, \dots, a_k)$ and $(b_1, \dots, b_k)$ have identical pairwise sums, then the collections $(a_1, \dots, a_k, b_1+m, \dots, b_k+m)$ and $(b_1, \dots, b_k, a_1+m, \dots, a_k+m)$ also have identical pairwise sums. (The number $m$ has to be such that $a_i \neq b_j \pm m$ for all $i,j$, so that the numbers in each collection would be different.)
