Quartic residues and sums involving $\binom{4k}{2k}$

TL;DR

This paper explores the relationship between quartic residues and specific binomial sum expressions, providing new congruences and connections in number theory involving primes, quadratic forms, and p-adic integers.

Contribution

It establishes novel congruences linking quartic residues with binomial sums and extends these results to particular cases involving specific values of m.

Findings

Derived congruences connecting quartic residues and binomial sums.

Established conditions for prime representations involving sums and Legendre symbols.

Provided explicit congruences for sums with specific parameters m=17,18,20,32,52,80,272.

Abstract

Let $p$ be an odd prime and let $m\not\equiv 0\pmod p$ be a rational p-adic integer. In this paper we reveal the connection between quartic residues and the sum $\sum_{k=0}^{[p/4]}\binom{4k}{2k}\frac 1{m^k}$, where $[x]$ is the greatest integer not exceeding $x$. Let $q$ be a prime of the form $4k+1$ and so $q=a^2+b^2$ with $a,b\in\Bbb Z$. When $p\nmid ab(a^2-b^2)q$, we show that for $r=0,1,2,3$, $p^{\frac{q-1}4}\equiv (\frac ab)^r\pmod q$ if and only if $$\sum_{k=0}^{[p/4]}\binom{4k}{2k}\Big(\frac{a^2}{16q}\Big)^k\equiv (-1)^{\frac{p^2-1}8a+\frac{p-1}2\cdot \frac{q-1}4}\Big(\frac pq\Big) \Big(\frac ab\Big)^r\pmod p,$$ where $(\frac pq)$ is the Legendre symbol. We also establish congruences for $\sum_{k=0}^{[p/4]}\binom{4k}{2k}\frac 1{m^k}\pmod p$ in the cases $m=17,18,20,32,52,80,272$.