It is easy to see (using AC, of course) that there exist two sets $U\subset\mathbb{R}$ and $V\subset\mathbb{R}$ such that any real number $x$ can be represented as unique sum $x=u+v$, where $u\in U$ and $v\in V$.
There are $2^{2^\omega}$ such $(U,V)$ pairs.


The question given by my son: prove existence of pair $(U,V)$ with $U=V$.


In other words, prove the existence of $U\subset\mathbb{R}$ such that for any $x\in\mathbb{R}$ there exists unique pair $\{u,v\}$ with $x=u+v$



Existence of subset of reals such that any real number is unique sum of exactly two elements of the subset
The usual transfinite construction works. Let $\{r\_{\alpha} : \alpha < \mathfrak{c}\}$ list $\mathbb{R}$. Construct $\{U\_{\alpha}: \alpha < \mathfrak{c}\}$ by induction on $\alpha$ such that the following hold.


(a) For every $a, b, c, d \in U\_{\alpha}$, $a + b = c + d \implies \{a, b\} = \{c, d\}$.


(b) There are $a, b \in U\_{\alpha + 1}$ such that $r\_{\alpha} = a + b$.


(c) For limit $\alpha$, $U\_{\alpha} = \bigcup\_{\beta < \alpha} U\_{\beta}$.


Then $U = \bigcup\_{\alpha < \mathfrak{c}} U\_{\alpha}$ is as required. Requirements (a), (c) are trivially satisfied. To ensure (b), at stage $\alpha + 1$, if $r\_{\alpha} \notin U + U$, choose $x$ outside the $\mathbb{Q}$-linear span of $U\_{\alpha} \cup \{r\_{\alpha}\}$ and put $U\_{\alpha+1} = U\_{\alpha} \cup \{x, r\_{\alpha} - x\}$ and note that this does not violate (a).


