Noordsij Primes of the form x 2 + ny 2
Noordsij Primes of the form x 2 + ny 2
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发表时间：
2015
影响因子：
0.8
通讯作者：
Matthew Bates
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作者：
Matthew Bates
本文将讨论的问题是，哪些素数可以表示为x + ny，其中x，y ∈ Z，对于不同的n ∈ N值。本文的一个目的是利用简单的同余关系对所有这样的素数p进行分类。即，为了找到β1，...，βr ∈ Z和K ∈ Z使得，p = x + ny <$$> p <$β1，.，βr（mod K）对于每个n.（我也希望表明这种关系首先存在！）沿着我希望向读者介绍一些有趣的数学领域。尽管这个问题看起来很简单，但找到答案却远不简单。为了让我们进入状态并出于逻辑原因，我们将从开始（n = 1，2）开始开始;不费吹灰之力，我们将看到这些情况可以用二次互易完全解决。不幸的是，这样的方法通常会失败（事实上，对于所有n 6= 1，2，它都会失败），因此需要一种新的方法。我们考虑Z上二元二次型的一些一般理论。这将有助于我们解决更多的特殊情况，其中大部分是400多年前费马考虑和证明的;为了不被一个400岁的法国人打败，我们发展了一些二次型的亏格理论。这将给出n的许多值的完整解，欧拉所谓的方便数（当一个数的主亏格由一个类组成时，它是方便的，不管这意味着什么！）。然后，我们将考虑有多少方便的数字;最终令人失望的是，只有65个（或者66个？）。为了不被二次型理论在解决我们的问题上的局限性所沮丧，我们继续前进。很明显，要解决这个问题，我们需要一种新的方法，这种新方法就是类域理论。使用希尔伯特类场论，我们得到了无穷多个n的解，但不是所有的！不幸的是，我们得到的解决方案是非建设性的，在这个意义上，我们证明了存在期望的同余关系，但我们不知道他们是什么。要想走得更远，就需要更好地理解类域理论以及模函数和复数乘法的知识。
This essay will address the question of which primes can be represented as x + ny, with x, y ∈ Z, for different values of n ∈ N. One aim of this essay is to classify all such primes p using simple congruence relations. i.e. to find β1, ..., βr ∈ Z and K ∈ Z such that, p = x + ny ⇐⇒ p ≡ β1, ..., βr (mod K) for every n. (I also hope to show that such a relation exists in the first place!) Along the way I hope to introduce the reader to some interesting areas of mathematics. Despite the question’s simple appearance, finding the answer is far from simple. To get us into the mood and for logical reasons, we will begin at the beginning (n = 1, 2); with little effort we will see that these cases can be completely solved using quadratic reciprocity. Unfortunately, such a method fails in general (in fact, it will fail for all n 6= 1, 2), hence a new approach is required. We consider some general theory of quadratic forms in two variables over Z. This will help us to solve a few more special cases, most of which were considered and proved by Fermat over 400 years ago; in order to not be beaten by a 400-year-old Frenchman we develop some genus theory of quadratic forms. This will give a complete solution for many values of n, Euler’s so-called convenient numbers (a number is convenient when its principal genus consists of a single class, whatever that means!). We will then consider how many convenient numbers there are; ultimately coming to the disappointing realisation that there are only 65 (or maybe 66?). Not to be disheartened by the limitations of quadratic form theory in solving our question we press on. It is clear that to solve the problem in general we need a new approach, the new approach being class field theory. Using Hilbert class field theory we derive a solution for infinitely many n, but not all! Unfortunately, the solution we get is non-constructive, in the sense that we prove that there exists desired congruence relations, but we do not know what they are. To go any further one requires a greater understanding of class field theory and knowledge of modular functions and complex multiplication.