Constructing x2x^2 for primes p=ax2+by2p=ax^2+by^2

preprint2010arXivOpen access

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)λ(a,b;n) be given by q∏k=1∞(1−qak)3(1−qbk)3=∑n=1∞λ(a,b;n)qnq\prod_{k=1}^\infty (1-q^{ak})^3(1-q^{bk})^3 = \sum_{n=1}^\infty λ(a,b;n)q^n. In the paper, using Jacobi's identity ∏n=1∞(1−qn)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)λ(a,b;n). For example, if 2∤ab2\nmid ab and p∤ab(ab+1)p\nmid ab(ab+1), then (−1)a+b2x+b+12(4ax2−2p)=λ(a,b;((ab+1)p−a−b)/8+1)(-1)^{\frac{a+b}2x+\frac{b+1}2}(4ax^2-2p) = λ(a,b;((ab+1)p-a-b)/8+1). We also give formulas for λ(1,3;n+1),λ(1,7;2n+1)λ(1,3;n+1),λ(1,7;2n+1), λ(3,5;2n+1)λ(3,5;2n+1) and λ(1,15;4n+1)λ(1,15;4n+1).

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