Quartic residues and sums involving $\left(\genfrac{}{}{0ex}{}{4k}{2k}\right)$

Abstract

Let $p$ be an odd prime and let $m\not\equiv 0(\mathrm{mod}p)$ be a rational p-adic integer. In this paper we reveal the connection between quartic residues and the sum ${\sum }_{k=0}^{[p/4]}\left(\genfrac{}{}{0ex}{}{4k}{2k}\right)\frac{1}{m^k}$, 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 $a,b\in \mathbb{Z}$. When $p\nmid ab(a^2-b^2)q$, we show that for $r=0,1,2,3$, $p^{\frac{q-1}{4}}\equiv (\frac{a}{b}{)}^r(\mathrm{mod}q)$ if and only if $$\sum _{k=0}^{[p/4]}\left(\genfrac{}{}{0ex}{}{4k}{2k}\right)(\frac{a^2}{16q}{)}^k\equiv (-1{)}^{\frac{p^2-1}{8}a+\frac{p-1}{2}\cdot \frac{q-1}{4}}\left(\frac{p}{q}\right)(\frac{a}{b}{)}^r(\mathrm{mod}p),$$ where $(\frac{p}{q})$ is the Legendre symbol. We also establish congruences for ${\sum }_{k=0}^{[p/4]}\left(\genfrac{}{}{0ex}{}{4k}{2k}\right)\frac{1}{m^k}(\mathrm{mod}p)$ in the cases $m=17,18,20,32,52,80,272$.

Section 3 is new