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Let $\Lambda :=\{\lambda_1, \dots, \lambda_n\}$ be a set of $n$ distinct real numbers.

For a given $p \in \mathbb N$, consider further the set

$$I_p := \{ \{i_1, i_2, \dots, i_p\} : i_j \in \{1, \dots,n\} \text{ for all } j=1, \dots, p\}.$$

It is an easy and well-known fact that $|I_p| = \binom{n+p-1}{p}$. For example, if $n=3$ and $p=2$, then

$$ I_2 = \{ \{1,1\}, \{1,2\}, \{1,3\}, \{2,2\}, \{2,3\}, \{3,3\} \}.$$

Now for a given $p \in \mathbb N$ we write down all $\binom{n+p-1}{p}$ sums $$\lambda_{i_1} + \lambda_{i_2} + \dots + \lambda_{i_p}. $$

I was asking myself if there was a (rather) simple condition on the set $\Lambda$ such that one can guarantee that for all $p \in \mathbb N$ all the $\binom{n+p-1}{p}$ sums will be distinct, i.e. a condition on $\Lambda$ such that

$$ \lambda_1, \dots, \lambda_n \text{ pairwise distinct and also}$$ $$ \lambda_1 + \lambda_1, \lambda_1 + \lambda_2, \dots, \lambda_1+\lambda_n, \lambda_2 + \lambda_2, \dots \text{ pairwise distinct and also...}$$ $$ \text{etc.}$$

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If you choose the numbers to be algebraically independent, then this guarantees this condition. This is very strong: even linearly independent over $\mathbb Q$ should be enough.

The set of $\sqrt{p}$ where $p$ runs over the primes, is such a set, for example.

Also, the set $t,t^2,t^3,\dots,t^k$ is such a set where $t$ is your favourite non-algebraic number, for example $\pi$ or $e$.

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    $\begingroup$ In fact, if we want the condition to hold for all $p$, then linear independence over $\mathbb{Q}$ is necessary and sufficient. $\endgroup$ – Richard Stanley Aug 16 '14 at 21:50
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The condition equivalent to the statement that for all $p$, the sums are distinct is that the space of rational dependencies does not intersect the space $W = (1,1,...,1)^\perp$ of $n$-tuples of coefficients with sum $0$ except at the origin. Since $W$ has dimension $n-1$, any space of dimension $2$ or higher intersects $W$ nontrivially, but it is possible and in some sense typical for a $1$-dimensional space of rational dependencies to intersect $W$ only at the origin.

For example, consider $\lbrace 1, \sqrt{2}, \sqrt{2}-1 \rbrace.$ The rational dependencies are the multiples of $(1,-1,1)$, which has trivial intersection with $W$.

That two multisets of size $p$ have the same sum means

$a_1\lambda_1 + ... + a_n \lambda_n = b_1 \lambda_1 + ... + b_n \lambda_n$

with $\sum a_i = p = \sum b_i$. This means $\sum (a_i-b_i) = 0$, so $(a_1-b_1,...,a_n-b_n)$ is a rational dependency perpendicular to $(1,...,1)$. Conversely, if there is a nontrivial rational dependency perpendicular to $(1,...,1)$ then we can separate the positive and negative coefficients and clear denominators to find multisets of the same size with the same sum.


Anthony Quas commented on Linear dependency of real numbers with integer coefficients adding up to zero that this is equivalent to saying that the differences $\lbrace \lambda_1 - \lambda_2, \lambda_2 - \lambda_3, ... , \lambda_{n-1}-\lambda_n \rbrace$ are linearly independent over $\mathbb Q$.

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    $\begingroup$ In other words, $(\lambda_1,1),\dots,(\lambda_n,1)$ are $\mathbb Q$-linearly independent. $\endgroup$ – Emil Jeřábek Aug 17 '14 at 10:29

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