Taiwanese Journal of Mathematics


Zhi-Hong Sun

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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 $r=0,1,2,3$,  $$ \begin{array}{ll} & \displaystyle p^{\frac{q-1}4} \equiv \Big(\frac ab\Big)^r\pmod q\\ &  \displaystyle\Leftrightarrow \sum_{k=0}^{[p/4]}\binom{4k}{2k}\Big(\frac{a^2}{16q}\Big)^k\equiv(-1)^{\frac{p^2-1}8a+\frac{p-1}2\cdot \frac{q-1}4}\Big(\frac pq\Big) \Big(\frac ab\Big)^r\pmod p,  \end{array} $$ 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 $m=17,18,20,32,52,80,272$.

Article information

Taiwanese J. Math., Volume 19, Number 3 (2015), 803-818.

First available in Project Euclid: 4 July 2017

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Zentralblatt MATH identifier

Primary: 11A07: Congruences; primitive roots; residue systems
Secondary: 11A15: Power residues, reciprocity 11B39: Fibonacci and Lucas numbers and polynomials and generalizations 11B65: Binomial coefficients; factorials; $q$-identities [See also 05A10, 05A30] 11E25: Sums of squares and representations by other particular quadratic forms

congruence Lucas sequence quartic residue quartic reciprocity binary quadratic form


Sun, Zhi-Hong. QUARTIC RESIDUES AND SUMS INVOLVING $\binom{4k}{2k}$. Taiwanese J. Math. 19 (2015), no. 3, 803--818. doi:10.11650/tjm.19.2015.4132. https://projecteuclid.org/euclid.twjm/1499133663

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