a_507_bc wrote: Let $p>7$ be a prime and let $A$ be subset of $\{0,1, \ldots, p-1\}$ with size at least $\frac{p-1}{2}$. Show that for each integer $r$, there exist $a, b, c, d \in A$, not necessarily distinct, such that $ab-cd \equiv r \pmod p$. There is an $x\in A\setminus\{0\}$. By replacing every element $y$ of $A$ with $yx^{-1}\mod{p}$ and $r$ with $rx^{-2}\pmod{p}$ we can assume that $1\in A$. Let $B=\{ab\pmod{p}\mid a,b\in A\}$. The we have $A\subseteq B$. If $|B|>\frac{p-1}{2}$ we are done. So assume that $|B|=\frac{p-1}{2}$ and thus $|B|=|A|=\frac{p-1}{2}$ and $B=A$. Then for $x\in A\setminus\{0\}$ multiplication with $x$ permutes the elements of $A$. So $A\setminus\{0\}$ is a subgroup of $\mathbb{F}_p^*$. By Lagrange's Theorem $|A\setminus\{0\}|\mid p-1$. For $p>7$ we can not have $\frac{p-3}{2}\mid p-1$. So $0\not\in A$ and $A$ is a subgroup of $\mathbb{F}_p^*\cong\mathbb{Z}/(p-1)\mathbb{Z}$ with $\frac{p-1}{2}$ elements. But the only such subgroup is the subgroup of quadratic residues. So $A=B$ is the set of nonzero quadratic residues. If $r\not\equiv \pm1\pmod{p}$ choose the quadratic residues $ab=\left(\frac{r+1}{2}\right)^2, cd=\left(\frac{r-1}{2}\right)^2$ from $B$. So we just need to show that there are two non-zero quadratic residues with difference $1 \pmod{p}$ for $p>7$. But note that there are $3$ quadratic residues in $\{1,2,3,4\}$ or $4$ quadratic residues in $4,5,6,7,8,9$ or $\frac{p-7}{2}$ quadratic residues in $9,10,\ldots,p-1$. So one of these set contains two quadratic residues with difference $1$.

This problem was proposed by me.