Quartic congruences and eta products

Abstract

Let $a_{15}(n),a_{20}(n)$ and $a_{24}(n)$ be defined by

$$\begin{aligned} q\prod _{k=1}^{\infty }(1-q^{k})(1-q^{3k})(1-q^{5k})(1-q^{15k})= & {} \sum \limits _{n=1}^{\infty }a_{15}(n)q^n,\\ q\prod _{k=1}^{\infty }(1-q^{2k})^2(1-q^{10k})^2= & {} \sum \limits _{n=1}^{\infty }a_{20}(n)q^n,\\ q\prod _{k=1}^{\infty }(1-q^{2k})(1-q^{4k})(1-q^{6k})(1-q^{12k})= & {} \sum \limits _{n=1}^{\infty }a_{24}(n)q^n \quad (|q|<1), \end{aligned}$$

and let $p>3$ be a prime. In this paper, for $p\equiv 3\ (\textrm{mod}\ 4)$ we reveal the connection between $a_{20}(p)$ and residue-counts of $x^4-4x^2+4x$ modulo p as x runs over $0,1,\ldots ,p-1$, and the connection between $a_{24}(p)$ and residue-counts of $x^3+c/x$ modulo p as x runs over $1,2,\ldots ,p-1$, where c is an integer not divisible by p. We also deduce the congruences for $a_{15}(p),a_{24}(p)$ modulo 16 and $a_{20}(p)$ modulo 4, and pose some analogous conjectures.