Set of integers with a particular additive property I have a set of integers $S = \{a_1,a_2,\ldots,a_n\}$. Let $K = a_1+a_2+\ldots+a_n$.
Consider the space of all $n$-tuples whose values are taken from the set $S$. 
For example $(a_1,a_2,\ldots,a_n)$ is one $n$-tuple and its sum is $K$, $(a_1,a_1,a_2,a_2,\ldots,a_k)$ is another. 
I need $S$ such that the only $n$-tuple that sums to $K$ among the space of all $n$-tuples is $(a_1,a_2,\ldots,a_n)$. 
An example that wont work is, if $S = \{1,4,7\}$, $K = 12 (= 1+4+7)$. $\{1,4,7\}$ and $\{4,4,4\}$ both sum to $12$. Similarly if we take $S = \{1,2,3,4,5,6,7\}$, I can think of 2 $7$-tuples that add to $28$.
I tried the set $S = \{1, 2, 4, 8, \ldots, 2^n\}$ and it seems to work. Can anyone give me an example that is polynomial instead of exponential in $n$. Would the set $S = \{1, 4, 9, 16, \ldots, n^2\}$ or $\{1, 8, 27,\ldots,n^3\}$ work? Thanks for the comments and suggestions.
 A: The set of squares will not work in general. Any notrivial solution to $a^2+b^2=c^2+d^2$ allows us to drop $a^2$ and $b^2$ and instead have $c^2$ and $d^2$ twice. And such solutions are easy to find since the equation is equivalent to $(a-d)(a+d)=a^2-d^2=c^2-b^2=(c-b)(c+b)$. For example $1\cdot 15 = 3\cdot 5$, which leads to $a=8, b=1, c=4, d=7$. Indeed, $8^2+1^2=65=4^2+7^2$. There is also the smaller solution $5^2+5^2=1^2+7^2$. Thus already $S=\{1,4,9,16,25,36,49\}$ does not have the desired property.
And this reminds us of Ramanujan and Hardy's taxi: $1729 = 10^3+9^3=12^3+1^3$. So the set $S=\{1,8,27,\ldots, 12^3\}$ of cubes won't work either. 
Here's why exponential growth is necessary:
Let $S=\{a_1,\ldots, a_n\}$. There are $2^n$ subsets of $\{1,\ldots, n\}$ be a "good" set. If $A,B\subseteq\{1,\ldots, n\}$ are distict subsets of the same size $k$ and with $\sum_{i\in A}a_i=\sum_{i\in B}a_i$, the set $S$ is "bad" as this allows us to replace summands coming from $A$ in $a_1+\ldots +a_n$ with summands coming from $B$. Since $\sum_{i\in A}a_i$ is between $0$ and $ka_n$ and there are $n\choose k$ subsets of size $k$, we conclude that $ka_n\ge {n\choose k}$. If we pick $k=\lfloor n/2\rfloor$, then using Stirling's approximation we see that ${n\choose k}\approx\frac{2^n\sqrt {2}}{\sqrt{\pi n}}$ and obtain
$$ a_n\gtrsim\frac{2^{n+1}\sqrt 2}{n\sqrt{\pi n}}.$$
