Power, Simplicity and Beauty
mathematics
theorem Let \(n\) be a positive integer, then \(n\) can be expressed as the sum of three squares iff it is not of the form \begin{equation}4^a(8b+7)\end{equation} for some \( a,b\in\Bbb Z,a,b\geq 0\). For a …
Sums of three squares Read More定义 定义 1[离散形式] 称 \( p=(p_1,p_2,\dotsc,p_n)\)Majorization(控制) \( q=(q_1,q_2,\dotsc,q_n)\), 记为 \(p \succ …
Majorization inequality and Muirhead’s inequality Read MoreSylvester’s theorem the binomial coefficient \begin{equation}{n\choose k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\end{equation} has a prime divisor \(p\) greater than \(k\). In other words, if \(n\geq 2k\), then the product of \(k\) consecutive integers greater than \(k\) is divisible by a …
Generalizations of Bertrand’s postulate Read More二平方和 引理 \(1\)正整数 \(a, b\) 互质, \((a,b)=1, p>2\) 是质数，并且 \(p\bigm|(a^2+b^2)\), 则 \(p\equiv1\pmod4\). 证明 不难用二次剩余(Quadratic residue)来给出证明： 由 \(p\bigm|(a^2+b^2)\) 得 \begin{equation}a^2\equiv-b^2\pmod p,\end{equation} 故而 \begin{equation}(\frac{a^2}{p})= (\frac{-1}{p})(\frac{b^2}{p}).\end{equation} 这里 \((\frac{n}{p})\) 是 Legendre symbol. 由于 \((a,b)=1\), 因之 \((p,a)=(p,b)=1\), 于是 \begin{equation}(\frac{a^2}{p})=(\frac{b^2}{p})=1\end{equation} 成为事实, …
Sums of two squares Read MoreLet \(\Bbb P\) be the set of all primes, the set \[\{\sin p \mid p\in\Bbb P\} \] is dense in \([-1,1]\). This problem appeared in artofproblemsolving on 22 May 2004, and has been solved in …
\(\sin p\) is dense in \([-1,1]\) Read MoreBertrand’s postulate states that if \(x\geq4\), then there always exists at least one prime \(p\) with\(x<p<2x-2\). A weaker but more elegant formulation is: for every \(x>1\) there is always at least one prime p such that \(x<p<2x\). In …
Ramanujan’s proof of Bertrand’s postulate Read MoreTerence Tao(陶哲轩)\(1\)月\(31\)日, 提交了一篇论文 “Every odd number greater than 1 is the sum of at most five primes“. 这篇文章的主要结果, 正如标题展示的, 每个奇数可以表示为不超过\(5\)个质数之和. 显然, 这个结果和 Goldbach’s conjecture(哥德巴赫猜想)有关, 把奇数情形的哥德巴赫猜想, 即弱哥德巴赫猜想(Goldbach’s weak conjecture)推进了一步, 也改进了 Ramare 的结论: 每个偶数可以表示为不超过\(6\)个质数的和. Tao …
Terence Tao come closer to solving Goldbach’s weak conjecture Read More密率与无限算术级数 Szemerédi’s theorem 任意有正（上）密率的正整数的子集必定包含任意长的算术级数. Van der Waerden’s theorem 把正整数集合任意划分成两个子集, 必有一个子集包含任意长的算术级数. 这里先要说明的是, Szemerédi’s theorem中的子集未必包含无限长的算术级数, Van der Waerden’s theorem 也未必有一个子集包含无限长的算术级数. 看下面的例子: \[ \{1,2,3\}\bigcup \{n: 2^{2i} \leqslant n < 2^{2i+1}, i \in {\Bbb N}\}\] 问题：正整数集合的子集 …
Positive density doesn’t assure infinite arithmetic progression Read More