Products of Differences over Arbitrary Finite Fields

There exists an absolute constant $\delta >0$ such that for all $q$ and all subsets $A\subseteq {\mathbb{F}}_q$ of the finite field with $q$ elements, if $\mathrm{\mid }A\mathrm{\mid }>q^{2\mathrm{/}3-\delta }$, then $$\mathrm{\mid }(A-A)(A-A)\mathrm{\mid }=\mathrm{\mid }\{(a-b)(c-d):a,b,c,d\in A\}\mathrm{\mid }>\frac{q}{2}\mathrm{.}$$ Any $\delta <1\mathrm{/}13,542$ suffices for sufficiently large $q$. This improves the condition $\mathrm{\mid }A\mathrm{\mid }>q^{2\mathrm{/}3}$, due to Bennett, Hart, Iosevich, Pakianathan, and Rudnev, that is typical for such questions. Our proof is based on a qualitatively optimal characterisation of sets $A,X \subseteq $\mathbb{F}$_q$ for which the number of solutions to the equation $$(a_1-a_2)=x(a_3-a_4),a_1,a_2,a_3,a_4\in A,x\in X$$ is nearly maximum. A key ingredient is determining exact algebraic structure of sets $A,X$ for which $\mathrm{\mid }A+XA\mathrm{\mid }$ is nearly minimum, which refines a result of Bourgain and Glibichuk using work of Gill, Helfgott, and Tao. We also prove a stronger statement for $$(A-B)(C-D)=\{(a-b)(c-d):a\in A,b\in B,c\in C,d\in D\}$$ when $A,B,C,D$ are sets in a prime field, generalising a result of Roche-Newton, Rudnev, Shkredov, and the authors.