# find all the vectors such that the pairwise differences of whose entries are all different.

Given a natural number $$n$$, I want to find all vectors in $$\mathbb{R}^n$$ such that

(1) The sum of the entries is $$0$$

(2) Pairwise differences of the entries are all different.

Let $$\mathbf{a}=(a_1,a_2,\dots,a_n)$$ be such an example. Then $$\sum_i a_i=0$$ and $$a_i-a_j$$ ($$i\neq j$$) are all different.

How to find and describe the set of all such vectors? It seems to be a nonlinear problem in the sense that the statement $$A-B\neq 0$$ cannot be part of a linear system.

Hint Given any such vector $$\bf a$$, it's immediate that $$\lambda {\bf a}$$ is also such a vector for any $$\lambda \in \Bbb R - \{ 0 \}$$, so to prove the claim it suffices to give a single example.
Now, for any vector $${\bf b}$$,
1. the differences of the entries of the vector $${\bf b} - \mu {\bf 1}$$, where $${\bf 1} := (1, \ldots, 1)$$, are $$(b_j - \mu) - (b_i - \mu) = b_j - b_i$$, namely, just the differences of $${\bf b}$$, and
2. the sum of the entries of $${\bf b} - \mu {\bf 1}$$ is $${\bf 1}^{\perp} ({\bf b} - \mu {\bf 1}) = {\bf 1}^{\perp} {\bf b} - \mu n$$.
In particular, if we set $$\mu = \frac{1}{n} {\bf 1}^{\perp} {\bf b}$$, then $${\bf b} - \mu {\bf 1}$$ has the same differences of entries as $$\bf b$$ but the sum of its entries is zero and so satisfies (1). Thus, it's enough to find a $$\bf b$$ vector satisfying (2).
For any $$n$$, the differences of the entries of the vector $${\bf b} := (1, 2, 4, \ldots, 2^{n - 1})$$ are all different.