Let $G$ be a finite abelian group and $A$ be a subset of $G \times G$ which is corner--free, meaning that there are no $x, y \in G$ and $d \in G \setminus \{0\}$ such that $(x, y)$, $(x+d, y)$, $(x, y+d) \in A$. We prove that
$|A| \le |G|^2 ... more >>>