Jacobsthal sums, Legendre polynomials and binary quadratic forms
Zhi-Hong Sun
Abstract
Open-access reader
Zhi-Hong Sun
Abstract
Open-access reader
Let $p>3$ be a prime and $m,n\in\Bbb Z$ with $p\nmid mn$. Built on the work of Morton, in the paper we prove the uniform congruence: $$&\sum_{x=0}^{p-1}\Big(\frac{x^3+mx+n}p\Big) \equiv {-(-3m)^{\frac{p-1}4} \sum_{k=0}^{p-1}\binom{-\frac 1{12}}k\binom{-\frac 5{12}}k (\frac{4m^3+27n^2}{4m^3})^k\pmod p&\t{if $4\mid p-1$,} \frac{2m}{9n}(\frac{-3m}p)(-3m)^{\frac{p+1}4} \sum_{k=0}^{p-1}\binom{-\frac 1{12}}k\binom{-\frac 5{12}}k (\frac{4m^3+27n^2}{4m^3})^k\pmod p&\text{if $4\mid p-3$,}$$ where $(\frac ap)$ is the Legendre symbol. We also establish many congruences for $x\pmod p$, where $x$ is given by $p=x^2+dy^2$ or $4p=x^2+dy^2$, and pose some conjectures on supercongruences modulo $p^2$ concerning binary quadratic forms.
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Let $p>3$ be a prime and $m,n\in\Bbb Z$ with $p\nmid mn$. Built on the work of Morton, in the paper we prove the uniform congruence: $$&\sum_{x=0}^{p-1}\Big(\frac{x^3+mx+n}p\Big) \equiv {-(-3m)^{\frac{p-1}4} \sum_{k=0}^{p-1}\binom{-\frac 1{12}}k\binom{-\frac 5{12}}k (\frac{4m^3+27n^2}{4m^3})^k\pmod p&\t{if $4\mid p-1$,} \frac{2m}{9n}(\frac{-3m}p)(-3m)^{\frac{p+1}4} \sum_{k=0}^{p-1}\binom{-\frac 1{12}}k\binom{-\frac 5{12}}k (\frac{4m^3+27n^2}{4m^3})^k\pmod p&\text{if $4\mid p-3$,}$$ where $(\frac ap)$ is the Legendre symbol. We also establish many congruences for $x\pmod p$, where $x$ is given by $p=x^2+dy^2$ or $4p=x^2+dy^2$, and pose some conjectures on supercongruences modulo $p^2$ concerning binary quadratic forms.
Key concepts: Legendre symbol, Congruence relation, Combinatorics, Legendre polynomials, Quadratic residue, Binary quadratic form, Modulo, Binary number