Sam Northshield 用一句话就说明了质数的无穷性. 这个很精彩的证明是这样的:

Proof.  If the set of primes is finite, then

$0\lt \prod_p \sin\left(\frac \pi p\right)= \prod_p \sin\left(\frac{\pi\Big(1+2\prod\limits_{p’}p’\Big)}p\right)=0. \qquad \Box$

$\sin\left(\frac{\pi\Big(1+2\prod\limits_{p’}p’\Big)}q\right)=\sin k\pi=0.$

### References

1. Sam Northshield, A One-Line Proof of the Infinitude of Primes, The American Mathematical Monthly, Vol. 122, No. 5 (May 2015), p. 466
2. Michael Hardy and Catherine Woodgold, Prime Simplicity,  the Mathematical Intelligencer, Volume 31, Issue 4, December 2009,  44-52
3. Harold Edwards, Contradict or Construct?,  the Mathematical Intelligencer, Volume 32, Issue 1, March 2010, p.3
4. Harold Edwards, Essays in Constructive Mathematics,  Springer, 2010

Yitang Zhang is giving the last invited talk at ICM 2014, “Small gaps between primes and primes in arithmetic progressions to large moduli”.

Yitang Zhang is giving the last invited talk 1

Yitang Zhang stepped onto the main stage of mathematics last year with the announced of his achievement that is hailed as “a landmark  theorem in the distribution of prime numbers”.

Yitang Zhang is giving the last invited talk 2

Yitang Zhang is giving the last invited talk 3

7 月他在母校北京大学的北京国际数学研究中心 (BICMR) 有一个系列的学术报告: Distribution of Prime Numbers and the Riemann Zeta Function I, II, III. 这个报告分三场, 原定时间是 July 8, 10, 15,  2014 16:00-17:00, 地点是镜春园 78 号院的 77201 室.

BICMR 官网上这个报告的 Abstract 是这么写的:

The distribution of prime numbers is one of the most important subjects in number theory.

There are many interesting problems in this field. It may not be difficult to understand the problems themselves, but the solutions are extremely difficult.

In this series of talks we will describe the application of certain analytic tools to the distribution of prime numbers. In particular, the role played by the Riemann zeta function will be discussed. We will also describe some early and current researches on the Riemann Hypothesis.

These talks are open to everyone in the major of mathematics, including undergraduate students.

Yitang Zhang at BICMR :Distribution of Prime Numbers and the Riemann Zeta Function

8 日下午 4 点, 田刚现身. 因为人比较多, 改为在镜春园82号甲乙丙楼的中心报告厅进行. 主持人刘若川是 1999 年的 IMO 金牌(他本来也是 1998 年中国国家队的队员).

$\zeta(2k)=\sum_{n=1}^\infty\frac1{n^{2k}}=(-1)^{k+1}\frac{(2\pi)^{2k}B_{2k}}{2(2k)!}$

10 日下午 4 点的第二场, 依旧在镜春园82号甲乙丙楼的中心报告厅. 不过, 15 日的一场会在镜春园 78 号院的 77201 室, 16:30 开始.

15 日下午 4:30 的最后一场, 要深入一点. 田刚坐在教室最后一排, 刘若川, 许晨阳坐在教室左边的走廊.张大师谈到有 Goldston, Pintz and Yildirim 的工作, 说他自己最大的贡献是把 $$c$$ 改进为 $$\dfrac14+\dfrac1{1168}$$.

Primes of the Form

Primes of the Form $$x^2+ny^2$$: Fermat, Class Field Theory, and Complex Multiplication
David A. Cox,  John Wiley & Sons Inc, 2nd Revised edition

## 质数 $$k$$-tuples 猜想

$\mathcal H=(h_1,h_2,\dotsc,h_{k_0}),$

Hardy-Littlewood prime tuples conjecture  如果 $$k_0$$-tuples $$\mathcal H$$ 是允许的, 那么, 存在无穷多个正整数 $$n$$, 使得 $$n+\mathcal H$$ 全部由质数组成.

## Second Hardy–Littlewood conjecture

1923 年, Hardy 和 Littlewood 发表了一篇论文[1]. 这篇长达 $$70$$ 页, 已经是数论史上的经典, 的论文提出, 对任意整数 $$m,n\geqslant2$$,

$\pi(m+n)\leqslant\pi(m)+\pi(n).$

## 质数 tuples 猜想与第二 Hardy–Littlewood 猜想不能同时成立

1974 年, Ian Richards 和他的博士研究生 Douglas Hensley 指出[2], Hardy-Littlewood 的两个猜想, 是不相容的. 也就是说, 这两个猜想, 至少有一个是不成立的.

$\pi(n+3159)-\pi(n)=447\gt\pi(3159)=446.$

### Annotations

1. 第一部分, 对于质数 tuples 猜想的介绍, 参考了 Tao 的一篇 blog.

### References

1. G. H. Hardy and J. E. Littlewood, On some problems of “partitio numerorum III: On the expression of a number as a sum of primes. Acta Math, 1923, 44: 1–70.
2. D. Hensley and I. Richards, Primes in intervals. Acta Arith. 25 (1974), pp. 375-391.
3. conjectures concerning primes; a discussion of the use of computers in attacking a theoretical problem. Bulletin of the American Mathematical Society 80:3 (1974), pp. 419-438.

#### 证明 1

• a=1. 此时 $$1+(d+2)d=(d+1)^2$$ 是合数;
• $$a\geqslant2$$. 此时 $$a+ad=a(d+1)$$ 是合数.

#### 证明 2

$$(m+1)!+2,(m+1)!+3,\dotsc,(m+1)!+m+1$$ 是 $$m$$ 个连续合数.

#### 证明 3

$a, a+d, a+2d, \dotsc, a+(m-1)d$

#### 证明 4

$f(x)=\sum\limits_{i=0}^ma_ix^i,$

#### 证明 6

$m_i|\left(a+i\right), i=1, 2, \dotsc, n.$

[证明 6 更新于 北京时间 2015 年 6 月 24 日]