Quadratic-residue distribution conjecture for breaking the square-root barrier

Let $p$ be prime, let $\chi(x)=\left(\frac{x}{p}\right)$ denote the Legendre symbol, and let $\{r_1,\ldots,r_k\}\subseteq\mathbb{Z}_p$. For distinct indices $i,j$, set $a=r_j-r_i$ and

$$I_{r_i,r_j}:=\{r_i-r_\ell:1\leq \ell\leq k,\ \ell\ne i,\ \ell\ne j\},$$

calling $I_{r_i,r_j}$ a one-sided difference set. Quadratic-residue distribution conjecture. There exist constants $0<\alpha<1$, $\nu>1/2$, and a positive integer $m_\alpha$ such that, for any set $\{r_1,\ldots,r_k\}$ in $\mathbb{Z}_p$ with $m_\alpha\leq k\leq p^\nu$, there exist indices $1\leq i<j\leq k$ satisfying

$$\frac{\left|\sum_{\ell\in I_{r_i,r_j}}\chi(\ell)\chi(\ell+a)\right|}{|I_{r_i,r_j}|}<\alpha.$$

This conjecture would provide a route to breaking the square-root barrier for the specified compressed-sensing construction through improved eigenvalue bounds; its resolution is not supplied here.