Ternary Egyptian fractions with prime denominator

Abstract

For a prime number $p$, let $A_3(p)=|\{m\in \mathbb{N}:\exists m_1,m_2,m_3\in \mathbb{N},\frac{m}{p}=\frac{1}{m_1}+\frac{1}{m_2}+\frac{1}{m_3}\}|$. In 2019 Luca and Pappalardi proved that $x(\log x{)}^3\ll {\sum }_{p\le x}{A}_3(p)\ll x(\log x{)}^5$. We improve the upper bound, showing ${\sum }_{p\le x}{A}_3(p)\ll x(\log x{)}^3(\log \log x{)}^2$.