Power, Simplicity and Beauty
开篇 Zsigmondy’s theorem states that if \(a>b>0\) and \(n>1\) are positive integers, then \(a^n+b^n\) has at least one prime factor that does not divide \(a^k+b^k\) for all positive integers \(k<n\), with …
Zsigmondy’s theorem Read MoreIn 1949, Leo Moser gave a proof of Bertrand’s postulate in the paper “A theorem of the distribution of primes”(American Mathematical Monthly,624-624,1949,vol56). Maybe this proof is the simplest one among all proofs of Bertrand’s postulate. …
Leo Moser’s proof of Bertrand’s postulate 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 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 More第一个证明是哥德巴赫(Goldbach)给出的，这个证明涉及的所谓 Fermat 数总会使得人们联想起费尔马（Fermat）和高斯（Gauss）！ 我们的目的是指出任意两个 Fermat 数 \(F_n=2^{2^n}+1, n=0,1,2,\dotsc\) 互质. 下述关系式足以 \begin{equation}\prod_{i=0}^{n}F_i=F_{n+1}-2.\end{equation} 下面是Filip Saidak \(2005\) 年的证明, 发表在 “美国数学月刊” (Amer.Math.Monthly) \(2006\)年第\(113\)卷第\(10\)期\(937-938\): 任取不是 \(1\) 的自然数 \(m\), 由于 \(m\) 与 \(m+1\) 互质, 即 \((m,m+1)=1\), 于是 …
Proofs of the infinity of primes 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