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Constructing $x^2$ for primes $p=ax^2+by^2$

Number Theory 2010-12-20 v1

Abstract

Let aa and bb be positive integers and let pp be an odd prime such that p=ax2+by2p=ax^2+by^2 for some integers xx and yy. Let λ(a,b;n)\lambda(a,b;n) be given by qk=1(1qak)3(1qbk)3=n=1λ(a,b;n)qnq\prod_{k=1}^\infty (1-q^{ak})^3(1-q^{bk})^3 = \sum_{n=1}^\infty \lambda(a,b;n)q^n. In the paper, using Jacobi's identity n=1(1qn)3=k=0(1)k(2k+1)qk(k+1)2\prod_{n=1}^\infty (1-q^n)^3 = \sum_{k=0}^\infty (-1)^k(2k+1)q^{\frac{k(k+1)}2} we construct x2x^2 in terms of λ(a,b;n)\lambda(a,b;n). For example, if 2ab2\nmid ab and pab(ab+1)p\nmid ab(ab+1), then (1)a+b2x+b+12(4ax22p)=λ(a,b;((ab+1)pab)/8+1)(-1)^{\frac{a+b}2x+\frac{b+1}2}(4ax^2-2p) = \lambda(a,b;((ab+1)p-a-b)/8+1). We also give formulas for λ(1,3;n+1),λ(1,7;2n+1)\lambda(1,3;n+1),\lambda(1,7;2n+1), λ(3,5;2n+1)\lambda(3,5;2n+1) and λ(1,15;4n+1)\lambda(1,15;4n+1).

Keywords

Cite

@article{arxiv.1012.3919,
  title  = {Constructing $x^2$ for primes $p=ax^2+by^2$},
  author = {Zhi-Hong Sun},
  journal= {arXiv preprint arXiv:1012.3919},
  year   = {2010}
}

Comments

16 pages

R2 v1 2026-06-21T17:00:36.337Z