Index lifting theorem in elementary number theory
index lifting theorem in elementary number theory 初等数论中的指数提升定理
Index lifting theorem in elementary number theory Read Morenumber theory
Number theory
Fermat’s theorem on sums of two squares: History
Fermat 的平方和定理： 素数 \(p\equiv1\pmod4\)，则 \(p\) 能表成两个整数 \( a, b\) 的平方和 \(p=a^2+b^2\). 是很精彩的定理，在堆垒数论很经典，我们很感兴趣。今天先来谈一点关于它的历史。 在历史上，最早考虑把正整数（不仅仅是素数）表示成两个正整数的平方和的可能性的问题的数学家是 Albert Girard. 他的论文发表在 1625 年。前面刚刚提到的Fermat 平方和定理有时候也称为 Girard 定理。至于 Fermat, 他在1640年12月25日给 Marin Mersenne 的一封信对这个定理给出了一个详尽的描述，同时也定出了把 \(p\) 的幂表成两个整数的平方和有多少种方法。 Albert Girard 小传 …
Fermat’s theorem on sums of two squares: History Read More\(\{\sqrt n\}\) is linearly independent over the rationals
集合 \[\{\sqrt n|n\in\Bbb N\; \text{is Square-free integer}\}\] 在有理数域上线性无关. 这其实是非常古老的问题, 早已经有很一般的结果. 先厘清无平方因子整数Square-free integer这个概念: \(1\) 到底是不是无平方因子整数? wiki 给出的定义是: 不被不是 \(1\) 的完全平方整除的整数称为无平方因子整数. 因此, \(1\) 算无平方因子正整数. 鉴于此, 我们认为: 无平方因子整数定义为”不被质数的平方整除的整数”更为恰当. 下面的证明来自 Iurie Boreico 的文章 Linear Independence of Radicals. …
\(\{\sqrt n\}\) is linearly independent over the rationals Read MoreAn Elementary Proof on dense of \(\{n^k\alpha\}\)
\(\alpha\) 是无理数, 则 \(\{n^k\alpha\}\)\((k=1, 2, \dotsc)\) 在区间 \((0, 1)\) 稠密． 这个证明原创应该是 Tao 的, 但 Tao 只证明了一个特殊情况, 并且指出这个手段可以采用来证明更一般的结果(本文结论), 但他没有详细写出. {nkα}An Elementary Proof on dense of
An Elementary Proof on dense of \(\{n^k\alpha\}\) Read MoreA one sentence Proof of the Infinitude of Primes
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 \[0\lt \prod_p \sin\left(\frac \pi p\right)= \prod_p \sin\left(\frac{\pi\Big(1+2\prod\limits_{p’}p’\Big)}p\right)=0. \qquad \Box\] 有更短的数学证明吗?应该没有! 这个证明正确吗?好像不是一目了然啊! …
A one sentence Proof of the Infinitude of Primes Read MoreThe elementary proof of the fact that there are infinite prime quadratic residues modulo \(p\)
\(p\) is a odd prime, then there are infinite primes \(q\) such that \(q\) is a quadratic residue modulo \(p\); there are infinite primes \(q\) such that \(q\) is a …
The elementary proof of the fact that there are infinite prime quadratic residues modulo \(p\) Read MoreIntegers represented by \(a^3+b^3+c^3-3abc\)
Which integers can be expressed as \(a^3+b^3+c^3-3abc\)? \(a\), \(b\), \(c\in\Bbb Z\). \[(a\pm1)^3+a^3+a^3-3(a\pm1)a^2=3a\pm1\] \[(a-1)^3+a^3+(a+1)^3-3a(a+1)(a-1)=9a\] \[2(a^3+b^3+c^3-3abc)=3(a+b+c)(a^2+b^2+c^2)-(a+b+c)^3\] If \(3\mid(a^3+b^3+c^3-3abc)\), then \(3\mid(a+b+c)^3\), \(3\mid(a+b+c)\). so \(9\mid(a^3+b^3+c^3-3abc)\). All \(n\) such that \(3\nmid n\) or \(9\mid n\).
Integers represented by \(a^3+b^3+c^3-3abc\) Read MoreA Hungarian Olympiad number theory problem related to Hasse principle
Let \(f(x)=x^n+a_{n-1}x^{n-1} +\dotsb+a_1x+a_0\) be a polynomial with integer coefficients, and let \(d_1\),\(\dotsc\), \(d_n\) be pairwise distinct integers. Suppose that for infinitely many prime numbers \(p\) there exists an integer \(k_p\) …
A Hungarian Olympiad number theory problem related to Hasse principle Read More