Let $m, n \in \mathbb{N}$. Is there $A, B \subset \{0, \ldots , mn\}$ with $|A|=n+1$, $|B|=m+1$ such that $a+b \neq a'+b'$ for any distinct $(a,b), (a',b') \in A \times B$?
I know the answer is no for the following restricted version.
Let the elements of $A$ (resp. $B$) to be $0=a_0 \leq \cdots \leq a_n$ (resp. $0=b_0 \leq \cdots \leq b_m$). If $a_i-a_{i-1} \leq m$ for all $i$ and $b_j -b_{j-1} \leq n$ for all $j$, then the answer is no.
This is verified by pigeon hole principle. WLOG, assume that $a_n \geq b_m$. Define $\phi:\{0, \ldots, m\} \rightarrow \{0, \ldots, n\}$ so that $\phi(j)$ is the least index $i$ satisfying $a_i \geq b_j$. Then, $a_{\phi(j)}-b_j \in \{0, \ldots, m-1\}$ for all $j$. By pigeon hole principle, there exists $j$ and $j'$ such that $a_{\phi(j)}-b_j = a_{\phi(j')}-b_{j'}$.
But this proof uses $a_i-a_{i-1} \leq m$ for all $i$, so cannot be generalized. I would like to know if there is a counterexample or not.