Unique-xor sets Let a set $S$ of nonnegative integers be a "unique-xor" set if all pairs of distinct integers in $S$ have different bitwise xor from all other pairs.
Let $x(n)$ be the least maximum element for any unique-xor set of size $n$.
My question is: Can we prove that $x(n)=\Theta(n^2)$? Or more basically, that $x(n)=O(n^k)$, for some constant $k$?
Has this question been studied?

For instance, the set $[0,1,2,4,...,2^m]$ is a unique-xor set, for any $m$. As another example, $[0,1,2,4,11,12]$ is a unique-xor set.
Of particular interest are unique-xor sets with large size and small maximum value. I believe that for sizes $1,2,3,4,5,6$, smallest possible maximum value for a unique-xor set of that size is $0,1,2,4,8,12$, given by the two constructions above.
A lower bound on $x(n)$ is that a unique-xor set of size $n$ must have maximum value at least $2^{\lceil \log_2 n(n-1)/2 \rceil - 1 }$. To see why, note that the set must have $n(n-1)/2$ distinct xor results. The maximum result must have at least $\lceil \log_2 n(n-1)/2 \rceil$ bits when written in binary. As a consequence, the maximum value must have at least the same number of bits when written in binary.
From the above lower bound, we find that $x(n) = \Omega(n^2)$. Intuitively, this seems like it should be tight, that $x(n) = \Theta(n^2)$, because a maximum value around $n^2$ gives enough space for the xor results to be distinct. However, I have not been able to find a better explicit construction than the exponential construction above.

Note that a unique-xor set is equivalent to a set system with unique symmetric difference, if we map the bits of the numbers to distinct elements. In that case, my question becomes whether there exists such a set system with $n$ sets over $k \log_2 n$ elements, or ideally over $2\log_2 n$ elements. Has this question been studied?
 A: Such sets in additive groups (sets where sums are unique) are called Sidon sets. If we consider the additive group $(\mathbb{F}_2^n,+)$ and associate the group (also a vector space over $\mathbb{F}_2$) with $\mathbb{F}_{2^n}$ via some basis.
I believe the results below show that the answer is $\theta(n^2).$
Sidorenko has shown tight upper and lower bounds on the size of the largest Sidon set, denoted by $\beta(\mathbb{F}_2^n)$ in $\mathbb{F}_2^n:$
$$
\beta(\mathbb{F}_2^n)\leq \sqrt{2^{n+1}-\frac{7}{4}}+\frac{1}{2}, 
$$
and
$$
\beta(\mathbb{F}_2^n)\geq 2^{n/2}, \quad n\textrm{ even}
$$
There is also an old construction of Lindstrom:
Consider all elements of $\mathbb{F}_{2^{2k}}$ of the form $(\mathbf{x},\mathbf{x}^3)$ where we express $\mathbf{x}\in \mathbb{F}_{2^k}$ as a $k$-dimensional vector over $\mathbb{F}_2$ with respect to some basis. It turns out that this forms a Sidon set, and given $(\mathbf{x},\mathbf{x}^3)+(\mathbf{y},\mathbf{y}^3)=(\mathbf{u},\mathbf{v})$, one can solve for $\mathbf{x}$ (and therefore $\mathbf{y}$) by solving a quadratic equation in $\mathbb{F}_2^k$.
