Group Action Approaches in Erdos Quotient Set Problem

Let Fq denote the finite field of q elements. For E Fqᵈ, denote the distance set (E) = \\|x-y\|²: = (x₁-y₁) ²+ + (xd-yd) ²: (x, y) E² \. The Erdos quotient set problem was introduced in Iosevich₂019 where it was shown that for even d2 that if |E| Fq² such that |E| >> q^d/2, then (E) (E): = \s{t: s, t (E), t=0\} =Fqᵈ. The proof of the latter result is quite sophisticated and in pham2023group, a simple proof using a group-action approach was obtained for the case of q 3 4 when d=2. In the q 3 4 setting, for each r (Fq) ², pham2023group showed if E Fq, then V (r): = \# \ (a, b, c, d) E²: \|a-b\|²{\|c-d\|² = r \} >> |E|⁴q. In this work we use group action techniques in the q 3 4 setting, for d=2 and improve the results of pham2023group by removing the assumption on r (Fq) ². Specifically we show if d=2 and q 3 4, then for each r Fq^*, V (r) |E|⁴2qif |E| 2q for all r Fq. Finally, we improve the main result of bhowmik2023near using our proof techniques from our quotient set results.