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

Question

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.

Answer 1

There are infinitely many such vectors.

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.