【作者】 高磊；

【导师】 秦厚荣；

【作者基本信息】 南京大学 ， 基础数学， 2016， 博士

【摘要】 给定非负整数n和正定整系数三元二次型Q(x,y,z),我们称方程Q{x,y,z) = n整数解的个数为Q表n的表示数,记为RQ(n)。本文,我们对三元二次型：x2+y2 + z2, x2+y2 + 2z2, x2 + y2 + 3z2, x2 + y2 + 4z2, x2+y2 + 5z2, x2 + y2 + 6z2, x2 + y2 + 8z2, x2+2y2 + 3z2, x2 + 2y2 + 4z2, x2 + 2y2 + 6z2, x2 + 3y2 + 6z2, x2 + 4y2+4z2, x2 + 4y2 + 8z2, 2x2+2y2+3z2, 2x2+3y2+3z2, x2+2y2+2z2, x2+3y2+3z2, x2 + 5y2+5z2, x2+6y2 + 6z2, 2x2+3y2+6z2, x2+y2+2z2+yz, x2+y2+7z2, x2+y2+11z2, x2+y2 + 13z2, x2+y2 + 19z2, x2+3y2 + 5z2逐一进行讨论,揭示它们的表示数与对应虚二次域类数的关系。其中最后五个二次型,因为它们的类数大于1,我们需要考虑其genus内所有类的二次型表示数,进而建立它们的线性组合与对应虚二次域类数的关系。下面我们例举我们得到的若干结果。假设p是一个除去有限个例外值的奇素数,令Q=x2+y2 + 2z2 + yz,则我们有：类似的,令Q1=x2+3y2 + 5z2, Q2 = x2 + 2y2 + 8z2-2yz,则我们有这里h(d)表示虚二次域Q((?)d)的类数。我们在文中还会给出一些“对偶”的结果,比如对Q的表示数,我们有需要提到的是,二次型x2+y2+3z2的情形是由孙智宏教授提出的一个猜想,这个猜想在最近被郭-彭-秦[3]证明。本文我们将揭示上述这一现象广泛地存在于一般表示数与类数之间。我们首先主要对裴定一得到解析公式的二十个对应尖形式空间为零的对角型正定整系数三元二次型进行讨论,得到类似的若干关系式。进一步的,我们对更多对应尖形式空间不为零的三元二次型(且未必为对角型)加以讨论,建立其解析公式,得到类似上述的关系式,并给出证明。我们还特别对x2+py2+qz2型(p,q是奇素数)的三元二次型进行了深入的研究,通过计算模形式尖点处的值,结合genus中其他代表元,建立其表示数与虚二次域类数的公式。在本文的最后一章,我们还对Cooper和Lam提出的关于表示数的一系列猜想做了进一步的探讨,证明了b=1,c=21时猜想成立。精确的说,我们证明了RQ(n2)= 4H(1,21,n)。这里其中ep表示p模n的指数。对于猜想中其他没有解决的某些情形,比如b=3,c=10时,由于对应二次型的类数为1,且对应爱森斯坦级数空间的基与b=1,c=21有类似的形式,故我们有希望用相同的方法来给予证实。不过,本文我们没有给出具体的证明。更多还原

【Abstract】 Let n be an integer, and Q(x, y, z) a positive definite ternary quadratic form with integral coefficients. The representation number of n by Q is the number of integral solutions (x, y, z) of the equation Q(x,y, z) = n. We denote it by RQ(n). In this paper, we will study case by case for ternary quadratic forms: x2 + y2 + z2, x2 + y2 + 2z2, x2 + y2 + 3z2, x2 + y2 + 4z2, x2 + y2 + 5z2, x2 + y2 + 6z2, x2 + y2 + 8z2, x2 + 2y2 4- 3z2, x2 + 2y2 + 4z2, x2 + 2y2 + 6z2, x2 + 3y2 + 6z2, x2 + Ay2 + 4z2, x2 + 4y2 + 8z2, 2x2 + 2y2 + 3z2, 2x2 + 3y2 + 3z2, x2 + 2y2 + 2z2, x2 + 3y2 + 3z2,x2 + 5y2 + 5z2,x2 + 6y2 + 6z2,2x2 + 3y2 + 6z2,x2 + y2 + 2z2 + yz,x2+y2+7z2, x2+y2 +11z2, x2+y2 + 13z2, x2+y2 + 19z2, x2+3y2+5z2 and find the relations between their representation numbers and the class numbers of corresponding imaginary quadratic fields. Since the class numbers of ternary quadratic forms in last five cases are greater than 1, we need to associate with other representatives in the genus to establish the relations. We will formulate following results as examples.We assume that p is an odd prime. Let Q = x2 + y2 + 2z2 + yz. Then we have Similarly, let Q1 = x2 + 3y2 + 5z2, Q2 = x2 + 2y2 + 8z2 - 2yz. Then we have Where h(d) stands for the class numbers of Q((?)).We will give some "dual" results. For example, we haveThe result about ternary quadratic form x2 + y2 + 3z2 was first conjectured by Z.H.Sun and was verified by Guo-Peng-Qin in[3].In this paper,we will show that the obove phenomenon occurs in a much wider situation. We will first establish a series of similar relations for those diagonal positive ternary forms that the corresponding cusp form spaces are trivial and in which cases Pei has got the analytic formulas for the representation numbers. Moreover, we will study for those ternary quadratic forms whose corresponding cusp form spaces are not trivial and establish the analytic formulas and then get the similar relations as bove.For the ternary quadratic forms x2+py2+qz2, where p and q are odd primes, we obtain a formula which relates the representation numbers of these ternary quadratic forms and the class numbers of the corresponding imaginary quadratic fields.In the last chapter, we will verify one case of the conjectures raised by Cooper and Lam on the representation numbers of n2= x2+by2+cz2. More preciesely, we will prove that RQ(n2)= 4H(1,21,n), where and ep is the non-negative integer such that pep||n.We may verify more cases of the conjectures. For example, if b= 3, c= 10, we may verify it by similar way since the class number of the corresponding ternary quadratic form is equal to 1, and the basis of the corresponding space of Eisenstein series is known.更多还原