张益唐在孪生素数猜想上做出了什么突破性贡献？
What breakthrough did Yitang Zhang make on the twin prime conjecture?
孪生素数猜想认为存在无穷多对相差为 2 的素数（如 11 和 13）。2013 年，张益唐证明存在某个有限常数，使得存在无穷多对素数之差不超过该常数（他给出的界是 7000 万），首次将「无穷多对」与「有限间隔」联系起来。随后 Polymath 协作项目把该上界大幅缩小到 246。
The twin prime conjecture asserts infinitely many prime pairs differing by 2 (e.g., 11 and 13). In 2013 Yitang Zhang proved that there is some finite bound such that infinitely many prime pairs differ by no more than it—he gave 70 million—the first result linking 'infinitely many' with 'bounded gaps.' The Polymath project later reduced the bound to 246.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-03
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《数学年刊》（Annals of Mathematics，Zhang 2014）；PolyMath 协作项目
Annals of Mathematics (Zhang 2014); Polymath Project
独立核验依据 · Independent verification: 核验依据 1 ARCHIVE 标识 arxiv-1409.8361：The "bounded gaps between pri
#孪生素数 #张益唐 #有界间隔 #素数分布
佩尔方程是什么？
What is Pell's equation?
形如 x² − D y² = 1（D 为非平方正整数）的二元二次丢番图方程。它有无穷多组整数解，可由连分数展开求出最小解再递推。
x²−D y²=1 with nonsquare D has infinitely many integer solutions, found via continued fractions of √D.
难度 专业 · 适合 高中(15-18)
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #丢番图
费马大定理说的是什么？谁最终证明了它？
What does Fermat's Last Theorem state, and who finally proved it?
费马大定理断言：当整数 n > 2 时，方程 xⁿ + yⁿ = zⁿ 没有正整数解。皮埃尔·德·费马在 1637 年页边写下这一猜想并声称有「奇妙证明」，但空白太小写不下。安德鲁·怀尔斯在 1994 年借助椭圆曲线与模形式（谷山-志村猜想）最终完成证明。
Fermat's Last Theorem states that for integer n > 2, the equation xⁿ + yⁿ = zⁿ has no positive integer solutions. Pierre de Fermat noted the conjecture in 1637 with a claim of a 'marvelous proof' too long for the margin. Andrew Wiles completed the proof in 1994 using elliptic curves and modular forms (the Taniyama–Shimura conjecture).
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-03
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《数学年刊》（Annals of Mathematics，Wiles 1995）；参考：Simon Singh《费马大定理》
Annals of Mathematics (Wiles 1995); Simon Singh, Fermat's Last Theorem
独立核验依据 · Independent verification: 核验依据 1 OPENLIBRARY 标识 /works/OL31155W：Fermat's Last Theorem（首版
#费马大定理 #怀尔斯 #椭圆曲线 #数论
什么是费马大定理？
What is Fermat's Last Theorem?
费马大定理断言：当整数 n > 2 时，方程 aⁿ + bⁿ = cⁿ 没有正整数解。这个猜想由皮埃尔·德·费马在 1637 年提出，直到 1994 年才由安德鲁·怀尔斯证明。
Fermat's Last Theorem states that for any integer n > 2, the equation aⁿ + bⁿ = cⁿ has no positive integer solutions. Pierre de Fermat stated it in 1637, and it was finally proved by Andrew Wiles in 1994.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#费马大定理 #数论 #证明
偶完全数有什么统一公式？目前找到的完全数都是偶数吗？
What is the unified formula for even perfect numbers, and are all known perfect numbers even?
欧几里得-欧拉定理：当且仅当2^p−1是梅森素数时，2^(p−1)(2^p−1)是偶完全数。至今未发现奇完全数，但其存在性仍是未解问题。
The Euclid–Euler theorem: 2^(p-1)(2^p-1) is an even perfect number exactly when 2^p-1 is a Mersenne prime. No odd perfect number is known, and their existence remains open.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#完全数 #梅森素数 #数论
费马大定理为什么困扰了数学家三百多年？
Why did Fermat's Last Theorem baffle mathematicians for over 350 years?
费马 1637 年在书边写下“xⁿ+yⁿ=zⁿ 在 n>2 时无正整数解”，并声称有巧妙证明但未留下。欧拉解决了 n=3 的情形，n=4 可用费马的无穷递降法处理，此后每代数学家只攻克了个别指数。直到 1994 年怀尔斯（在泰勒协助下补正）借助椭圆曲线与模形式的联系完成证明，所用方法远超原题本身。
In 1637 Fermat wrote in a book margin that xⁿ+yⁿ=zⁿ has no positive integer solutions for n>2, claiming a marvelous proof he never recorded. Euler handled n=3; the n=4 case follows from Fermat's own infinite descent; each generation conquered one exponent at a time. Only in 1994 did Wiles—with Taylor's help to patch a gap—finish the proof using a link between elliptic curves and modular forms, methods far beyond the original problem.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#费马大定理 #椭圆曲线 #怀尔斯
刘维尔数是什么？
What is a Liouville number?
刘维尔数是能被有理数以异常快的速度逼近的实数，例如 Σ 10^{-n!}。刘维尔1844年构造出它，给出了历史上第一个被证明的超越数。
A Liouville number is approximated by rationals extraordinarily fast, e.g. Σ 10^{-n!}. Liouville constructed one in 1844, the first number proved transcendental.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #超越数 #逼近
费马小定理是什么？
What is Fermat's little theorem?
费马小定理说：若p是质数且a不是p的倍数，则 a^(p-1) ≡ 1 (mod p)。它是现代公钥密码学（如RSA）的重要基础。
Fermat's little theorem: if p is prime and a is not divisible by p, then a^(p-1) ≡ 1 (mod p); it underlies modern cryptography.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #模运算
模形式是什么？
What are modular forms?
模形式是定义在复上半平面、在特定变换群下具有高度对称性的全纯函数，常带权 k。它们在证明费马大定理（经由谷山-志村猜想）和现代数论中起着核心作用。
Modular forms are highly symmetric holomorphic functions on the upper half-plane; they were central to Wiles's proof of Fermat's Last Theorem via the Taniyama–Shimura conjecture.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-03
来源 · Source
《数学年刊》（Annals of Mathematics，Wiles 1995）；参考：Simon Singh《费马大定理》
Annals of Mathematics (Wiles 1995); Simon Singh, Fermat's Last Theorem
独立核验依据 · Independent verification: 核验依据 1 OPENLIBRARY 标识 /works/OL31155W：Fermat's Last Theorem（首版
#数论 #模形式
费马小定理说 a^p≡a (mod p)，它为什么总是成立？
Fermat's little theorem says a^p≡a (mod p) — why does it always hold?
先用二项式定理：对 (a+1)^p 展开，中间所有二项式系数 C(p,k)（1≤k≤p-1）都含因子 p，因为 p 是质数，分子里的 p 不会被 1 到 p-1 约掉，所以 (a+1)^p≡a^p+1 (mod p)。对 a 从 0 开始归纳，结合 1^p≡1，就得 a^p≡a (mod p)。p 不整除 a 时两边除以 a 得 a^(p-1)≡1，这是 RSA 解密正确性的根基。
Expand (a+1)^p by the binomial theorem. Every middle coefficient C(p,k) for 1≤k≤p-1 is divisible by p, since p is prime and the factor p in the numerator is not canceled by any k in the denominator; hence (a+1)^p≡a^p+1 (mod p). Inducting on a from 1^p≡1 gives a^p≡a (mod p). When p does not divide a, divide by a to get a^(p-1)≡1, the cornerstone of RSA decryption.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-03
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《什么是数学》(柯朗)（数论章）
What Is Mathematics? (Courant), number theory chapter
#费马小定理 #二项式定理 #数论证明
黎曼猜想到底在猜什么？它和质数的分布有什么关系？
What exactly does the Riemann hypothesis claim, and how does it connect to the distribution of primes?
黎曼把欧拉的乘积公式 ζ(s)=Σ 1/n^s=Π(1-p^(-s))^(-1) 延拓到整个复平面，发现 ζ 的非平凡零点与质数分布密切相关。素数定理 π(x)~Li(x) 1900 年前后被证明，但它只给出主项；零点位置决定误差。黎曼猜想断言：所有非平凡零点都落在实部为 1/2 的临界线上。若成立，π(x) 与 Li(x) 的误差可降到约 √x·log x。它被列为千禧年七大难题之一，至今未解，且其成立与否都会对质数分布的精细结构产生巨大影响。
Riemann extended Euler's product formula ζ(s)=Σ 1/n^s=Π(1-p^(-s))^(-1) to the whole complex plane and showed the non-trivial zeros of ζ govern the distribution of primes. The prime number theorem π(x)~Li(x) was proved around 1900, but it gives only the leading term; the zeros dictate the error. The Riemann hypothesis asserts that every non-trivial zero lies on the critical line Re(s)=1/2. If true, the error between π(x) and Li(x) drops to about √x·log x. One of the seven Millennium Prize problems, it remains unsolved, and either outcome would profoundly shape the fine structure of prime distribution.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 来源待核验 · Source pending review
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出处备注：著录含书名，但缺版次 / 出版社 / 年份中的一部分，需编辑补全后才能做独立比对。
Edwards《Riemann's Zeta Function》；Conrey《The Riemann Hypothesis》（2003，Notices of the AMS）
Edwards, Riemann's Zeta Function; Conrey, The Riemann Hypothesis (2003, Notices of the AMS)
独立核验依据 · Independent verification: 核验依据 1 OPENLIBRARY 标识 /works/OL2434709W：Riemann's zeta function
#黎曼猜想 #黎曼ζ函数 #素数定理
怎样用扩展欧几里得算法求出模逆元？
How does the extended Euclidean algorithm compute modular inverses?
以求满足 11x≡1 (mod 30) 的 x 为例。先对 (30,11) 辗转相除：30=2·11+8，11=1·8+3，8=2·3+2，3=1·2+1。再反向回代：1=3−1·2=3−(8−2·3)=3·3−8=(11−8)·3−8=3·11−4·8=3·11−4·(30−2·11)=11·11−4·30，于是 11·11≡1 (mod 30)，逆元是 11。一般地，只要 gcd(a,m)=1，回代总能把逆元表示为 a 与 m 的线性组合。
Take solving 11x≡1 (mod 30). First run Euclid's algorithm on (30,11): 30=2·11+8, 11=1·8+3, 8=2·3+2, 3=1·2+1. Then back-substitute: 1=3−1·2=3−(8−2·3)=3·3−8=(11−8)·3−8=3·11−4·8=3·11−4·(30−2·11)=11·11−4·30. Hence 11·11≡1 (mod 30) and the inverse is 11. In general, whenever gcd(a,m)=1, back-substitution expresses the inverse as a linear combination of a and m.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#扩展欧几里得算法 #模逆元 #辗转相除法
奇完全数存在吗？
Does an odd perfect number exist?
目前已知的完全数（如 6、28、496）都是偶数。是否存在奇完全数仍是未解之谜；已有大量必要条件（如它必须大于 10^1500 且至少有 101 个质因数），但尚无定论。
All known perfect numbers (6, 28, 496, …) are even. Whether an odd perfect number exists is open; many constraints are known (it must exceed 10^1500 with ≥101 prime factors) but none definitive.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-03结论有争议 · Contested
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《什么是数学》(柯朗)（数论章：完全数）
What Is Mathematics? (Courant), number theory (perfect numbers)
#数论 #完全数
狄利克雷算术级数定理是什么？
What is Dirichlet's theorem on arithmetic progressions?
它断言：只要 a 与 d 互质，等差数列 a, a+d, a+2d, … 中就含有无穷多个素数。狄利克雷借助他发明的“L函数”完成了这一证明。
It states that if a and d are coprime, the arithmetic progression a, a+d, a+2d, … contains infinitely many primes. Dirichlet proved it using the L-functions he invented.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #算术级数 #素数
怎样系统生成所有本原勾股数（a²+b²=c²）？
How can we systematically generate all primitive Pythagorean triples (a²+b²=c²)?
取互质且一奇一偶的m>n，则a=m²−n²、b=2mn、c=m²+n²给出本原勾股数，且这样能生成全部本原解。
With coprime m>n of opposite parity, a=m²-n², b=2mn, c=m²+n² gives a primitive triple, and this generates all primitive solutions.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#勾股数 #毕达哥拉斯 #数论
√2 连分数的渐进分数和佩尔方程 x²−2y²=±1 有什么关系？
How are the convergents of √2's continued fraction related to Pell's equation x²−2y²=±1?
√2 的连分数是 [1;2,2,2,...]，渐进分数依次为 1/1、3/2、7/5、17/12、41/29……它们恰好就是佩尔方程 x²−2y²=±1 的解：1²−2·1²=−1、3²−2·2²=1、7²−2·5²=−1，正负号交替出现。原因是渐进分数是 √2 的“最佳有理逼近”，其误差通过恒等式 |x²−2y²|=|x+y√2|·|x−y√2| 与判别式直接挂钩：误差小于 1/(2y) 就迫使 |x²−2y²| 取到最小值 1。于是丢番图方程与逼近论这两个看似独立的主题，通过连分数这座桥自然连通。
√2 has continued fraction [1;2,2,2,...], with convergents 1/1, 3/2, 7/5, 17/12, 41/29, …; these are exactly the solutions of Pell's equation x²−2y²=±1: 1²−2·1²=−1, 3²−2·2²=1, 7²−2·5²=−1, with signs alternating. Because convergents are best rational approximations, their error ties to the Pell discriminant through the identity |x²−2y²|=|x+y√2|·|x−y√2|: an error below 1/(2y) forces the discriminant to its minimum absolute value 1. Diophantine equations and approximation theory thus meet naturally through continued fractions.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-03
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《什么是数学》(柯朗)（连分数章）
What Is Mathematics? (Courant), chapter on continued fractions
#连分数 #佩尔方程 #渐进分数
卡塔兰猜想是什么？
What is Catalan's conjecture?
卡塔兰猜想问：除了 8 和 9（即 2³ 与 3²），是否还有两个连续的正整数都是正整数的幂？2002 年米哈伊列斯库证明：没有，这确实是唯一的情况。
Catalan's conjecture asked whether 8 and 9 (2³ and 3²) are the only consecutive perfect powers; Mihăilescu proved in 2002 that they are the only case.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #卡塔兰
黎曼猜想与zeta函数的零点分布有何关系，它为何如此重要？
How does the Riemann hypothesis relate to the zeros of the zeta function, and why is it so important?
黎曼猜想断言黎曼zeta函数所有非平凡零点都位于复平面上实部为二分之一的直线上。若成立，它将极大精化素数分布的误差估计，影响数论众多分支。爱德华兹与康雷的专著讨论了它与素数定理的深刻联系。
The Riemann hypothesis claims that all non-trivial zeros of the Riemann zeta function lie on the line of real part one-half in the complex plane. If true it would sharpen the error term in prime distribution and influence many branches of number theory. Edwards and Conrey discuss its deep link to the Prime Number Theorem.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 来源待核验 · Source pending review
来源 · Source
出处备注：著录含书名，但缺版次 / 出版社 / 年份中的一部分，需编辑补全后才能做独立比对。
Edwards《Riemann's Zeta Function》；Conrey《The Riemann Hypothesis》（2003，Notices of the AMS）
Edwards, Riemann's Zeta Function; Conrey, The Riemann Hypothesis (2003, Notices of the AMS)
独立核验依据 · Independent verification: 核验依据 1 OPENLIBRARY 标识 /works/OL2434709W：Riemann's zeta function
#黎曼猜想 #zeta函数 #非平凡零点
偶完全数和梅森素数之间有什么关系？为什么找完全数等于找梅森素数？
How are even perfect numbers related to Mersenne primes, and why does finding one equal finding the other?
欧几里得证明了“若 2^p-1 是质数则 2^(p-1)(2^p-1) 是完全数”，欧拉 18 世纪证明了逆命题：每个偶完全数都长这样。于是两者一一对应，找偶完全数就是找形如 2^p-1 的质数。GIMPS 项目至今（2026 年）已确认 52 个这样的梅森素数。至于是否存在奇完全数，仍是数论里最古老也最著名的未解问题之一。
Euclid proved that if 2^p-1 is prime then 2^(p-1)(2^p-1) is perfect; Euler proved the converse in the 18th century — every even perfect number has that form. The two are in one-to-one correspondence, so hunting even perfect numbers means hunting primes of the form 2^p-1. As of 2026, the GIMPS project has confirmed 52 such Mersenne primes. Whether any odd perfect number exists remains one of number theory's oldest and best-known open questions.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#梅森素数 #完全数 #欧几里得-欧拉定理
费马大定理是什么？
What is Fermat's Last Theorem?
它断言当整数 n>2 时，方程 x^n + y^n = z^n 不存在非零整数解。怀尔斯于1994年用椭圆曲线与模形式证明。
It states x^n+y^n=z^n has no nonzero integer solutions for n>2; proved by Wiles in 1994 using elliptic curves and modular forms.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #费马
华林问题是什么？
What is Waring's problem?
它问：对任意正整数 k，是否存在 g(k) 使得每个正整数都能表示为至多 g(k) 个 k 次幂之和。华林1770年提出，希尔伯特1909年证明 g(k) 恒存在。
It asks whether, for each k, every positive integer is a sum of at most g(k) k-th powers. Waring posed it in 1770; Hilbert proved g(k) always exists in 1909.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #加法 #幂和
什么是素数定理？
What is the prime number theorem?
素数定理描述了质数在自然数中分布的大致规律：小于 x 的质数个数约等于 x / ln x。它说明质数虽然稀疏，但其分布具有可预测的平均趋势。
The prime number theorem describes the rough distribution of primes among the natural numbers: the number of primes less than x is about x divided by ln x. It shows that although primes become sparse, their distribution follows a predictable average trend.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#素数定理 #质数分布 #数论
素数定理说了什么？
What does the Prime Number Theorem say?
素数定理描述不超过 x 的素数个数 π(x) 渐近于 x/ln x，即随着数变大，素数变得越来越稀疏，密度约为 1/ln x。高斯很早就凭直觉猜测了这一规律。
The Prime Number Theorem says π(x) (primes ≤ x) is asymptotic to x/ln x, so primes thin out with density ≈ 1/ln x; Gauss conjectured it from intuition.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #素数分布
梅森素数与卢卡斯-莱默检验是什么？
What are Mersenne primes and the Lucas-Lehmer test?
形如 2^p − 1 的素数叫梅森素数（p 须为质数）。卢卡斯-莱默检验是一种高效判定 2^p − 1 是否为素数的算法，是寻找最大已知素数的主要工具。
Primes of form 2^p−1 are Mersenne primes; the Lucas-Lehmer test efficiently checks them and finds the largest known primes.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #素数
黎曼猜想是什么？为什么它如此重要？
What is the Riemann Hypothesis and why is it so important?
黎曼猜想由波恩哈德·黎曼于 1859 年提出，涉及黎曼 ζ 函数非平凡零点的分布：它猜测这些零点全部位于复平面上实部为 1/2 的直线上。该猜想与素数分布的精确规律紧密相连，是克雷数学研究所公布的七大千禧年难题之一，解决者将获得 100 万美元奖金。
Proposed by Bernhard Riemann in 1859, the hypothesis concerns the zeros of the Riemann zeta function: it conjectures that all its nontrivial zeros lie on the line with real part 1/2 in the complex plane. It is deeply tied to the precise distribution of prime numbers and is one of the seven Millennium Prize Problems of the Clay Mathematics Institute, carrying a one-million-dollar award.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#黎曼猜想 #ζ函数 #素数分布 #千禧年难题
RSA 里为什么“先加密再解密”能还原出原文？背后是哪条数论定理？
Why does encrypting then decrypting in RSA recover the original message, and which number-theory theorem is behind it?
设 n=pq，φ(n)=(p-1)(q-1)。选 e 与 φ(n) 互素，求 d 使 ed≡1 (mod φ(n))。加密 c=m^e (mod n)，解密 m'=c^d=m^(ed) (mod n)。由欧拉定理 a^φ(n)≡1 (mod n)（a 与 n 互素时），而 ed-1 是 φ(n) 的倍数，所以 m^(ed)≡m (mod n)，还原原文。安全性来自：不知道 p、q 就难以算出 φ(n)，也就求不出 d。
Let n=pq and φ(n)=(p-1)(q-1). Choose e coprime to φ(n), then find d with ed≡1 (mod φ(n)). Encryption is c=m^e (mod n) and decryption m'=c^d=m^(ed) (mod n). Euler's theorem says a^φ(n)≡1 (mod n) whenever gcd(a,n)=1, and since ed-1 is a multiple of φ(n), we get m^(ed)≡m (mod n) — the original text returns. Security rests on the fact that without p and q you cannot compute φ(n), hence cannot find d.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#RSA原理 #欧拉定理 #模运算
素数定理描述了素数分布的什么规律？
What pattern of prime distribution does the Prime Number Theorem describe?
素数定理指出，小于或等于 x 的素数个数 π(x) 近似等于 x / ln x（或其更精确的积分形式 li(x)），即素数在自然数中逐渐变得稀疏，但遵循可预测的密度。该定理在 1896 年由阿达马和德拉瓦莱-普桑各自独立证明，核心工具是复分析中的 ζ 函数。
The Prime Number Theorem states that the number of primes up to x, denoted π(x), is asymptotically x / ln x (more precisely li(x))—primes thin out but in a predictable way. It was proved independently in 1896 by Hadamard and de la Vallée Poussin, with the Riemann zeta function as the central tool.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-04
来源 · Source
美国数学学会（AMS）数学综述；G.H. Hardy 与 E.M. Wright《数论导引》
American Mathematical Society (AMS); Hardy & Wright, An Introduction to the Theory of Numbers
独立核验依据 · Independent verification: 核验依据 1 ISBN 9783540643326 (1938)
#素数定理 #素数分布 #阿达马 #复分析
费马小定理是什么，它在密码学中有什么用？
What is Fermat's little theorem, and how is it used in cryptography?
若p是素数、a不被p整除，则a^(p−1) ≡ 1 (mod p)。它既可做素性测试基础，也是RSA中指数运算与解密正确性的关键依据。
If p is prime and p∤a, then a^(p-1) ≡ 1 (mod p). It underlies primality tests and is key to RSA's exponentiation and decryption correctness.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#费马小定理 #模幂 #密码学基础
哈代–拉马努金定理说，一个典型正整数的不同素因子个数大约服从什么分布？
What distribution does the Hardy–Ramanujan theorem assign to the number of distinct prime factors of a typical integer?
哈代与拉马努金在 1917 年证明：对绝大多数不大于 x 的整数 n，其不同素因子个数 ω(n) 接近 log log x，且偏差的标准阶约为 √(log log x)。这意味着 ω(n) 具有“正态阶”log log x，即整数的素因子个数增长得极慢却呈类正态分布。
Hardy and Ramanujan proved in 1917 that for almost all integers n ≤ x, the number of distinct prime factors ω(n) is about log log x, with fluctuations of order √(log log x). Thus ω(n) has the 'normal order' log log x—its typical value grows astonishingly slowly yet follows an approximate normal distribution.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-04 · 2 条独立证据
来源 · Source
Hardy, G. H. 与 Wright, E. M., 《数论导引》(第6版), Oxford University Press
Hardy, G. H. & Wright, E. M., 'An Introduction to the Theory of Numbers' (6th ed.), Oxford University Press
独立核验依据 · Independent verification: 核验依据 1 ISBN 9783540643326 (1938) · 核验依据 2 ISBN 9783540643326 (2008)
#哈代-拉马努金定理 #素因子个数 #正态阶
朗兰兹纲领是什么？
What is the Langlands program?
朗兰兹纲领是一组宏伟的猜想，试图把数论中的伽罗瓦表示与自守形式/表示联系起来，被称为“数学的大统一理论”。它把看似不同的领域（数论、分析、几何）编织成一张网。
The Langlands program is a web of conjectures connecting Galois representations (number theory) with automorphic forms, often called a 'grand unified theory' of mathematics.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #朗兰兹
维费里希素数（Wieferich prime）是什么，它为什么与费马大定理的第一个情形有关？
What is a Wieferich prime, and why is it tied to the first case of Fermat's Last Theorem?
若素数 p 满足 2^(p−1) ≡ 1 (mod p²)，则称 p 为维费里希素数；目前只知 1093 与 3511 两个。库默尔证明：只要指数 p 不是维费里希素数，费马大定理在指数 p 下的“第一情形”就成立，因此这类素数成了该定理最棘手的例外。
A prime p is Wieferich if 2^(p−1) ≡ 1 (mod p²); only 1093 and 3511 are known. Kummer showed that if the exponent p is not a Wieferich prime, the first case of Fermat's Last Theorem holds for that p, so such primes mark the hardest exceptional cases of the theorem.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
Ribenboim, P., 《更大的素数小书》, Springer
Ribenboim, P., 'The Little Book of Bigger Primes', Springer
独立核验依据 · Independent verification: 核验依据 1 OPENLIBRARY 标识 /works/OL18252048W：The little book of big
#维费里希素数 #费马大定理 #同余
ABC 猜想是什么？
What is the abc conjecture?
对互质正整数 a,b,c 满足 a+b=c，令 rad(abc) 为它们所有不同质因数的乘积。ABC 猜想说对任意 ε>0，c 通常远小于 rad(abc)^(1+ε)。它与费马大定理、许多丢番图方程都有深刻联系。
For coprime a,b,c with a+b=c, let rad(abc) be the product of their distinct prime factors. The abc conjecture says c is usually much smaller than rad(abc)^(1+ε), linking to many Diophantine results.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #ABC猜想
二次域 Q(√d) 里的“代数整数”和理想类群，如何把费马大定理与数的几何联系起来？
How do the algebraic integers and ideal class groups of the quadratic field Q(√d) link Fermat's Last Theorem to the geometry of numbers?
在二次域 Q(√d) 中，整数环的“理想”弥补了普通整数分解可能失效的缺陷，其理想类群衡量分解唯一性偏离的程度（类数为 1 时唯一分解成立）。库默尔用分圆域的理想分解证明费马大定理的“第一情形”，而类数公式又通过圆法与椭圆函数把类群大小与“数的几何”（格点）联系起来。
In the quadratic field Q(√d), the ring of algebraic integers may fail unique factorization, and its ideal class group measures how far it deviates (class number 1 means unique factorization holds). Kummer used ideal factorization in cyclotomic fields to prove the first case of Fermat's Last Theorem, while class-number formulas tie the size of the class group to the geometry of numbers via lattices and elliptic functions.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
Cox, D. A., 《形如 x²+ny² 的素数》, Wiley
Cox, D. A., 'Primes of the Form x²+ny²', Wiley
独立核验依据 · Independent verification: 核验依据 1 OPENLIBRARY 标识 /works/OL1849155W：Primes of the Form x2 +
#二次域 #代数整数 #理想类群 #类数
二次互反律陈述了什么内容？高斯为什么称它为“黄金定理”？
What does the law of quadratic reciprocity state, and why did Gauss call it his golden theorem?
互异奇素数 p、q 满足 (q/p)(p/q)=(−1)^{((p−1)/2)((q−1)/2)}：“q 是否为模 p 平方剩余”反向依赖“p 是否为模 q 剩余”，仅当两者 ≡3 (mod 4) 时翻转。高斯引理以格点数的奇偶揭示此对称机制；配合两补充律可判定一切二次同余可解性，又是高次互反律原型，堪称“黄金定理”。
For distinct odd primes p, q the law reads (q/p)(p/q) = (−1)^{((p−1)/2)((q−1)/2)}: whether q is a square modulo p generally depends reciprocally on whether p is a square modulo q, the relation flipping only when both primes are 3 mod 4. Gauss's lemma exposes the mechanism through the parity of lattice points inside a rectangle. Combined with the two supplementary laws it decides solvability of every quadratic congruence, and as the prototype of higher reciprocity laws Gauss named it his aureum theorema.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-04
来源 · Source
K. Ireland 与 M. Rosen《A Classical Introduction to Modern Number Theory》，第 2 版，GTM 84，Springer，1990，第 5 章
Ireland & Rosen, A Classical Introduction to Modern Number Theory, 2nd ed., GTM 84, Springer, 1990, Ch. 5
独立核验依据 · Independent verification: 核验依据 1 ISBN 9781475721041 (1990)
#数论 #二次互反律 #勒让德符号
什么是完全数？请举例。
What is a perfect number? Give an example.
完全数是等于其所有真因数（不含自身）之和的数。最小的完全数是6（1+2+3=6），下一个是28（1+2+4+7+14=28）。
A perfect number equals the sum of its proper divisors; the smallest is 6 (1+2+3) and the next is 28.
难度 进阶 · 适合 小学高段(9-12)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #完全数
什么是阿姆斯特朗数（水仙花数），最小的几个是哪些？
What is an Armstrong (narcissistic) number, and what are the smallest ones?
一个n位数等于其各位数字的n次方之和，就是阿姆斯特朗数，如153=1³+5³+3³。三位的水仙花数只有153、370、371、407四个。
An n-digit number equal to the sum of its digits each raised to the nth power is Armstrong (narcissistic); e.g., 153=1³+5³+3³. The three-digit ones are exactly 153, 370, 371, 407.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 来源待核验 · Source pending review
来源 · Source
出处备注：本题原挂出处与答案内容不符，已撤回；具体书目待补，暂不展示来源。
本题原挂出处与答案内容不符，已撤回；具体书目待补。
Previous citation withdrawn as unsupported for this answer; a specific source is pending.
#阿姆斯特朗数 #水仙花数 #数字性质
怎样生成所有的基本勾股数？
How do you generate all primitive Pythagorean triples?
所有满足 a²+b²=c² 且两两互素的基本勾股数都能由欧几里得公式得到：取互素且一奇一偶的 m>n，令 a=m²−n²、b=2mn、c=m²+n²。例如 m=2,n=1 得到 3,4,5；m=3,n=2 得到 5,12,13。每个基本三元组恰好这样出现一次，不会有遗漏。
Every primitive triple with a²+b²=c² and pairwise coprime entries comes from Euclid's formula: take coprime m>n of opposite parity, then a=m²−n², b=2mn, c=m²+n². For example m=2,n=1 gives 3,4,5, and m=3,n=2 gives 5,12,13. Each primitive triple arises exactly once in this way, with no omissions.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-10-03
来源 · Source
《什么是数学》(柯朗)（数论章）
What Is Mathematics? (Courant), number theory chapter
#勾股数 #丢番图方程 #欧几里得公式
什么是梅森素数，目前最大的已知素数多来自它吗？
What is a Mersenne prime, and do the largest known primes come from them?
形如2^p−1（p为素数）的素数叫梅森素数。目前人类发现的最大素数几乎都是梅森素数，得益于专门的分布式搜索（GIMPS）。
A Mersenne prime has the form 2^p-1 with p prime. The largest known primes are almost all Mersenne primes, found via distributed projects like GIMPS.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-10-04
来源 · Source
K. H. Rosen《Elementary Number Theory and Its Applications》（第6版），Pearson，2011，§6.2
K. H. Rosen, Elementary Number Theory and Its Applications, 6th ed., Pearson, 2011, §6.2
独立核验依据 · Independent verification: 核验依据 1 ISBN 9780321717757 (2011)
#梅森素数 #2^p-1 #大素数
模 p 的离散对数问题为什么被认为困难，它又支撑了哪些现代密码？
Why is the discrete logarithm problem modulo a prime considered hard, and which modern cryptosystems does it underpin?
给定素数 p、生成元 g 与 h = gˣ mod p，求指数 x 就是离散对数问题；目前没有已知的高效经典算法，最佳通用方法（指标计算法）随 p 的位数超多项式增长。它直接支撑了 Diffie–Hellman 密钥交换与 ElGamal 加密等公钥密码体制。
Given a prime p, generator g, and h = gˣ mod p, recovering the exponent x is the discrete logarithm problem; no efficient classical algorithm is known, and the best general methods (index calculus) grow faster than any polynomial in the bit-length. It underpins public-key systems such as Diffie–Hellman key exchange and ElGamal encryption.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-03
来源 · Source
Hoffstein, J., Pipher, J., Silverman, J. H., 《数学密码学导引》(第2版), Springer
Hoffstein, J., Pipher, J., Silverman, J. H., 'An Introduction to Mathematical Cryptography' (2nd ed.), Springer
独立核验依据 · Independent verification: 核验依据 1 OPENLIBRARY 标识 /works/OL13764896W：An introduction to mat
#离散对数 #Diffie-Hellman #公钥密码
网络传输中，公钥和私钥是怎么做到“别人能锁、只有你能开”的？
In secure communication, how do a public key and a private key make it possible for anyone to lock but only you to unlock?
RSA 用两个很大且保密的质数 p、q 相乘得到 n=pq，公开 n 和加密指数 e 作为公钥，任何人用它把消息加密成密文；而只有知道 p、q 的人才能算出私钥 d 把密文解回原文。攻击者想破解就必须把 n 分解成 p、q，而分解超大合数目前没有快速算法，这正是 RSA 安全的根基。
RSA multiplies two large secret primes p, q to get n=pq. The public key, consisting of n and an exponent e, lets anyone encrypt a message into ciphertext, while only someone who knows p and q can derive the private key d and decrypt. An attacker must factor n back into p and q to break it, and no fast factoring algorithm for huge composites is known — that difficulty is the root of RSA's security.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#RSA #公钥加密 #质因数分解
什么是完全数？为什么说 6 和 28 是“完美的”？
What is a perfect number, and why are 6 and 28 called perfect?
一个正整数如果等于它所有真因数（小于它本身的正因数）之和，就称为完全数。6 的真因数是 1、2、3，而 1+2+3=6；28 的真因数是 1、2、4、7、14，加起来也正好是 28。古希腊人把完全数看得十分神圣，欧几里得还证明了：只要 2^p-1 是质数，那么 2^(p-1)(2^p-1) 就是偶完全数。
A positive integer is perfect when it equals the sum of its proper divisors (its positive divisors less than itself). For 6 these are 1, 2, 3, and 1+2+3=6; for 28 they are 1, 2, 4, 7, 14, which also sum to 28. The ancient Greeks held perfect numbers in near-mystical regard, and Euclid proved that whenever 2^p-1 is prime, the number 2^(p-1)(2^p-1) is an even perfect number.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-10-03
来源 · Source
《什么是数学》(柯朗)（数论章：完全数）
What Is Mathematics? (Courant), number theory (perfect numbers)
#完全数 #真因数 #欧几里得
中国剩余定理是什么？
What is the Chinese Remainder Theorem?
它说明若多个模数两两互质，那么同余方程组 x ≡ a_i (mod m_i) 在模其乘积下有唯一解。它用于日历换算和加密。
If moduli are pairwise coprime, the system x≡a_i (mod m_i) has a unique solution modulo their product; used in calendars and crypto.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #同余
米勒-拉宾素性测试是什么？
What is the Miller–Rabin primality test?
它是一种概率性素性测试：对候选奇数 n，选随机“见证”a 进行检验；若通过若干轮则 n 极可能是素数。RSA 等大数素性判断主要靠它快速筛选。
The Miller–Rabin test is a probabilistic primality test using random 'witness' bases; passing many rounds makes n overwhelmingly likely prime, used to generate RSA primes.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #素性测试
费马大定理说的是什么，谁证明了它？
What does Fermat's Last Theorem say, and who proved it?
它断言当整数n大于2时，方程x^n+y^n=z^n没有正整数解。怀尔斯在1994年用椭圆曲线与模形式最终证明。
It states that for integer n>2, x^n+y^n=z^n has no positive integer solutions. Andrew Wiles proved it in 1994 using elliptic curves and modular forms.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#费马大定理 #椭圆曲线 #怀尔斯
什么是模运算（同余）？
What is modular arithmetic (congruence)?
模运算是“除以某个数看余数”的运算。例如 11 ≡ 2 (mod 3) 表示11和2除以3余数相同，都是2。
Modular arithmetic works with remainders; e.g. 11 ≡ 2 (mod 3) means 11 and 2 have the same remainder when divided by 3.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #同余
数字签名是怎样防止伪造和篡改的？
How does a digital signature prevent forgery and tampering?
发送方先用哈希函数把消息压缩成固定长度的“指纹”，再用自己的私钥加密这段指纹得到签名。验证方用公钥解密，并和自己算出的哈希比对：消息任何改动都会改变指纹，使签名失效。因为只有持有私钥的人能生成有效签名，所以身份可确认，签名者也无法抵赖。
The sender hashes the message into a fixed-length fingerprint, then encrypts that fingerprint with their private key to form the signature. The verifier decrypts with the public key and compares against a freshly computed hash: any alteration changes the fingerprint and breaks the signature. Since only the private-key holder can produce a valid signature, authenticity is assured and repudiation is impossible.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数字签名 #哈希 #公钥密码
威尔逊定理给出了怎样的素数判别法？
What primality test does Wilson's theorem provide?
它说正整数p是素数当且仅当(p−1)! ≡ −1 (mod p)。虽然理论优美，但阶乘计算太慢，实际并不用于大数素性检测。
It states p>1 is prime iff (p-1)! ≡ -1 (mod p). Elegant in theory, but factorial computation is too slow for practical primality testing.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#威尔逊定理 #阶乘 #素数判别
什么是孪生素数猜想？
What is the twin prime conjecture?
孪生素数猜想认为存在无穷多对相差2的质数，如(3,5)、(11,13)、(17,19)。2013年张益唐证明了存在无穷多对间隔小于7000万的质数，但“相差2”仍未证明。
The twin prime conjecture says there are infinitely many prime pairs differing by 2, e.g. (3,5),(11,13),(17,19). In 2013 Yitang Zhang proved there are infinitely many prime pairs with a gap under 70,000,000, but the gap-2 case remains open.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-03
来源 · Source
张益唐《Bounded Gaps Between Primes》，Annals of Mathematics 179(3):1121–1176, 2014
Y. Zhang, 'Bounded Gaps Between Primes', Annals of Mathematics 179(3):1121–1176, 2014
独立核验依据 · Independent verification: 核验依据 1 ARCHIVE 标识 arxiv-1409.8361：The "bounded gaps between pri
#数论 #猜想
算术基本定理保证什么，为什么它重要？
What does the Fundamental Theorem of Arithmetic guarantee, and why is it important?
它保证每个大于1的整数都能唯一分解为素数乘积（不计顺序）。这是整除、最大公约数和整个数论大厦的基石。
It guarantees every integer greater than 1 factors uniquely into primes (up to order). It is the foundation of divisibility, gcd, and much of number theory.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#质因数分解 #唯一性 #算术基本定理
钟表上现在 9 点，再过 5 小时是几点？这和数学里的“模运算”有什么关系？
It is 9 o'clock now; what time will it be 5 hours later, and how does that connect to modular arithmetic?
9+5=14，但钟面没有 14 点，于是要减去 12 变成 2 点。数学里我们写 14≡2 (mod 12)，意思是 14 和 2 除以 12 的余数相同。这种“只看余数、周期循环”的运算叫模运算，日历计算、ISBN 校验码和现代加密都建立在它上面。
9+5=14, but a clock face has no 14 o'clock, so we subtract 12 and land on 2. Mathematically we write 14≡2 (mod 12), meaning 14 and 2 leave the same remainder upon division by 12. Arithmetic that keeps only remainders in a cyclic pattern is modular arithmetic, and it underlies calendar math, ISBN check digits, and modern encryption.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-10-03
来源 · Source
《什么是数学》(柯朗)（数论章：同余）
What Is Mathematics? (Courant), chapter on number theory (congruence)
#同余 #模运算 #余数
什么是连分数？为什么 √2 可以写成无限的连分数？
What is a continued fraction, and why can √2 be written as an infinite one?
连分数是把一个数写成“整数加分数、分数里又套分数”的嵌套形式。比如 √2 = 1 + 1/(2 + 1/(2 + ...))，记作 [1;2,2,2,...]，因为 √2 是无理数，它的连分数无限且循环。截断连分数得到的 1、3/2、7/5、17/12 等分数交替落在 √2 两侧，误差随项数迅速变小，是它的“最佳有理逼近”。
A continued fraction writes a number as a nested “integer plus fraction” form. For instance √2 = 1 + 1/(2 + 1/(2 + ...)), written [1;2,2,2,...]; since √2 is irrational, the expansion is infinite and periodic. Truncating gives fractions such as 1, 3/2, 7/5, 17/12 that alternate around √2 with errors shrinking quickly—they are its best rational approximations.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-10-03
来源 · Source
《什么是数学》(柯朗)（连分数章）
What Is Mathematics? (Courant), chapter on continued fractions
#连分数 #无理数逼近 #最佳逼近
什么是模运算？
What is modular arithmetic?
模运算（“时钟算术”）只关心两个整数相除的余数，记作 a ≡ b (mod n)。它在密码学、计算机科学和校验码设计中被广泛使用。
Modular arithmetic, or clock arithmetic, cares only about the remainder when two integers are divided, written a ≡ b (mod n). It is widely used in cryptography, computer science, and checksum design.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#模运算 #余数 #数论
计算机怎样快速判断一个很大的数是不是素数？
How do computers quickly decide whether a very large number is prime?
对巨大数字逐个试除完全不可行。实用上先用 Miller–Rabin 这类随机算法：它非常快，还能可靠地“捉出”合数，反复测试可以把误判概率压到任意低。2002 年 AKS 算法给出了第一个确定性的多项式时间检验，从理论上彻底回答了“素数性能否快速判定”这个问题。
Trial division is hopeless for huge numbers. In practice randomized tests such as Miller–Rabin run fast and reliably catch composites; repeating the test pushes the error probability arbitrarily low. In 2002 the AKS algorithm provided the first deterministic polynomial-time test, settling in theory whether primality can be decided quickly.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#素数检验 #Miller-Rabin #AKS算法
什么是质因数分解？
What is prime factorization?
质因数分解是把一个合数写成质数乘积，例如 12 = 2×2×3。算术基本定理保证这种分解在不计顺序时唯一。
Prime factorization writes a composite as a product of primes, e.g. 12=2×2×3; the Fundamental Theorem of Arithmetic guarantees uniqueness.
难度 进阶 · 适合 小学高段(9-12)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #分解
什么是卡迈克尔数，为什么它让素性检测变难？
What is a Carmichael number, and why does it complicate primality testing?
卡迈克尔数是合数却满足对所有与之互质的a有a^(n−1)≡1 (mod n)，能通过费马测试却不是素数。它被称为伪素数，需更强的检测来排除。
A Carmichael number is composite yet satisfies a^(n-1)≡1 (mod n) for all coprime a, passing Fermat's test without being prime. Called a pseudoprime, it needs stronger tests to catch.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-10-04
来源 · Source
K. H. Rosen《Elementary Number Theory and Its Applications》（第6版），Pearson，2011，§6.2
K. H. Rosen, Elementary Number Theory and Its Applications, 6th ed., Pearson, 2011, §6.2
独立核验依据 · Independent verification: 核验依据 1 ISBN 9780321717757 (2011)
#卡迈克尔数 #伪素数 #素性测试
中国剩余定理能解决什么类型的问题？
What kind of problems does the Chinese Remainder Theorem solve?
它给出一组两两互素的模数下同时满足多个同余式x≡a_i (mod m_i)的解，且解在乘积模下唯一。可用于日历换算、密码学与并行计算。
It finds a number x satisfying several congruences x≡a_i (mod m_i) with pairwise coprime moduli, unique modulo their product. It applies to calendars, cryptography, and parallel computation.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#中国剩余定理 #同余 #模运算
逆向相加能否让每个数都变成回文数？196为什么特殊？
Does repeatedly reversing and adding make every number a palindrome? Why is 196 special?
对多数数，把数字倒序相加反复迭代会得到回文数，但196在亿万次迭代后仍未出现回文，被称为利克瑞尔数候选，其是否终成回文至今未知。
For most numbers, reversing and adding repeatedly yields a palindrome, but 196 has resisted billions of iterations. It is the leading candidate Lychrel number, and whether it ever becomes a palindrome is unknown.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#回文数 #利克瑞尔 #迭代
分圆多项式 Φₙ(x) 有什么作用，它和正 n 边形尺规作图有什么关系？
What is the role of the nth cyclotomic polynomial Φₙ(x), and how does it connect to straightedge-and-compass constructibility of regular n-gons?
分圆多项式 Φₙ(x) 是其本原 n 次单位根的最小整系数多项式，其次数等于欧拉函数 φ(n)。高斯证明：正 n 边形可用尺规作出的充要条件是 n 为 2 的幂与若干互异费马素数之积，而分圆域 Q(ζₙ) 的次数 φ(n) 的 2 幂性正是这一判据的代数核心。
The nth cyclotomic polynomial Φₙ(x) is the minimal integer polynomial whose roots are the primitive nth roots of unity, and its degree is Euler's totient φ(n). Gauss showed a regular n-gon is constructible with straightedge and compass exactly when n is a power of two times distinct Fermat primes, and the 2-power nature of the cyclotomic field degree φ(n) is the algebraic heart of that criterion.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-04
来源 · Source
Apostol, T. M., 《解析数论导引》, Springer
Apostol, T. M., 'Introduction to Analytic Number Theory', Springer
独立核验依据 · Independent verification: 核验依据 1 ISBN 9780387901633 (1998)
#分圆多项式 #单位根 #尺规作图 #高斯
盈数、亏数和完全数如何区分？
How do abundant, deficient and perfect numbers differ?
把一个数的所有真因数（不含自身）相加：和等于自身为完全数（如6），大于自身为盈数（如12），小于自身为亏数（如8）。
Sum the proper divisors: equal to the number = perfect (6), greater = abundant (12), less = deficient (8).
难度 进阶 · 适合 初中(12-15)
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #分类
什么是模逆元？
What is a modular multiplicative inverse?
在模 m 下，a 的逆元是满足 a·x ≡ 1 (mod m) 的 x，仅当 a 与 m 互质时存在。它是 RSA 解密运算的基础。
The inverse of a mod m is x with a·x≡1 (mod m), existing iff gcd(a,m)=1; it underlies RSA decryption.
难度 进阶 · 适合 初中(12-15)
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #模运算
考拉兹（冰雹）猜想是什么？
What is the Collatz (hailstone) conjecture?
对任意正整数，若是偶数就除以2，是奇数就乘3加1，反复迭代，猜想最终都会落到1。它已被验证到极大数，却仍未被证明。
For any positive integer, if even divide by 2, if odd multiply by 3 and add 1, repeating. The conjecture says every sequence reaches 1. Verified to huge bounds but still unproven.
难度 进阶 · 适合 初中(12-15)
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#考拉兹 #冰雹序列 #未解猜想
为什么已知最大的素数几乎都是梅森素数？
Why are nearly all known largest primes Mersenne primes?
形如 2ᵖ−1 的素数叫梅森素数。若 2ⁿ−1 是素数则 n 必为素数；更重要的是，卢卡斯-莱默检验能以极快的速度判定这类数的素性，比通用检验快得多，特别适合大规模搜索。GIMPS 项目 2024 年 10 月发现的第 52 个梅森素数 2^136279841−1，创下了已知最大素数的纪录。
A Mersenne prime is a prime of the form 2ᵖ−1. If 2ⁿ−1 is prime then n itself must be prime; crucially, the Lucas–Lehmer test decides primality of these numbers far faster than general tests, making large-scale searching feasible. The 52nd Mersenne prime, 2^136279841−1, found by GIMPS in October 2024, set the record for the largest known prime.
难度 进阶 · 适合 大众
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#梅森素数 #卢卡斯-莱默检验 #GIMPS
欧拉函数φ(n)表示什么，如何计算？
What does Euler's totient function φ(n) count, and how is it computed?
φ(n)是1到n中与n互质的整数个数。若n的素数分解为各p_i^k_i之积，则φ(n)=n∏(1−1/p_i)。它在RSA等公钥密码中至关重要。
φ(n) counts integers from 1 to n coprime to n. For n=∏p_i^k_i, φ(n)=n∏(1−1/p_i). It is central to public-key cryptography such as RSA.
难度 进阶 · 适合 初中(12-15)
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#欧拉函数 #互质 #数论
什么情况下一个数在模运算下存在“倒数”？
When does a number have an "inverse" under modular arithmetic?
a 在模 m 下的逆元是满足 ax≡1 (mod m) 的数 x。它存在当且仅当 a 与 m 互素，即最大公约数为 1。例如模 7 下 3 的逆元是 5，因为 3×5=15≡1 (mod 7)。模逆元是解同余方程、设计密码系统的常用工具，求法用扩展欧几里得算法。
The inverse of a modulo m is a number x with ax≡1 (mod m). It exists if and only if a and m are coprime, i.e., their greatest common divisor is 1. For example, modulo 7 the inverse of 3 is 5, since 3×5=15≡1 (mod 7). Modular inverses are essential for solving congruences and building cryptographic systems, computed with the extended Euclidean algorithm.
难度 进阶 · 适合 大众
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#模逆元 #同余 #互素
格林–陶定理（Green–Tao theorem）证明了素数在什么意义上是“有结构”的？
What does the Green–Tao theorem reveal about the structure of the prime numbers?
格林与陶哲轩在 2008 年证明：素数序列中包含任意长的等差数列，也就是说无论你想要多长的等间隔素数串都存在。该结果把加性组合学（特别是 Szemerédi 定理）与素数分布联系起来，说明素数并非完全随机散布。
Green and Tao proved in 2008 that the primes contain arithmetic progressions of arbitrary length, so no matter how long an evenly spaced run of primes you ask for, one exists. The proof transplants Szemerédi's theorem on dense sets into the primes via additive combinatorics, showing the primes are not merely random noise.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-03
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Green, B. 与 Tao, T., 《素数含任意长等差数列》, Annals of Mathematics 167(2), 2008
Green, B. & Tao, T., 'The primes contain arbitrarily long arithmetic progressions', Annals of Mathematics 167(2), 2008
独立核验依据 · Independent verification: 核验依据 1 ARCHIVE 标识 arxiv-math0404188：The primes contain arbitrar
#素数 #等差数列 #加性组合学
费马小定理在模运算和公钥密码中有什么作用？
What role does Fermat's little theorem play in modular arithmetic and public-key cryptography?
费马小定理指出若p为素数且a不被p整除，则a的p-1次方模p等于一。它提供了快速检验幂的同余关系，并是大整数分解相关密码体制的数学基础之一。哈代与赖特以及尼文等人的数论教材都把它作为核心结论。
Fermat's little theorem states that if p is prime and a is not divisible by p, then a to the power p-1 is congruent to 1 modulo p. It gives a quick test for power congruences and underlies parts of public-key cryptography based on factoring. Hardy & Wright and Niven et al. treat it as a core result.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-04
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Hardy 与 Wright《An Introduction to the Theory of Numbers》；Niven、Zuckerman 与 Montgomery《An Introduction to the Theory of Numbers》
Hardy & Wright, An Introduction to the Theory of Numbers; Niven, Zuckerman & Montgomery, An Introduction to the Theory of Numbers
独立核验依据 · Independent verification: 核验依据 1 ISBN 9783540643326 (1938)
#费马小定理 #模运算 #公钥密码
欧几里得算法如何求最大公约数？
How does the Euclidean algorithm compute the GCD?
反复用较大数除以较小数并取余数，直到余数为0，最后的除数即最大公约数。例如 GCD(48,18): 48=2×18+12, 18=1×12+6, 12=2×6+0，故 GCD=6。
Repeatedly replace (a,b) with (b, a mod b) until remainder 0; the last nonzero remainder is the GCD. For 48 and 18: 48 = 2×18 + 12, 18 = 1×12 + 6, 12 = 2×6 + 0, so the GCD is 6.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #算法
欧拉函数 φ(n) 是什么？
What is Euler's totient function φ(n)?
φ(n) 表示不超过 n 且与 n 互质的正整数个数。例如 φ(9)=6，φ(p)=p−1 当 p 为质数。它在欧拉定理 a^φ(n) ≡ 1 (mod n) 中起作用。
φ(n) counts integers up to n coprime to n; e.g. φ(9)=6. It appears in Euler's theorem a^φ(n)≡1 (mod n).
难度 进阶 · 适合 高中(15-18)
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #欧拉函数
质数会越来越少吗？在很大的范围里质数到底有多稀？
Do primes become rarer as numbers grow, and just how sparse are they in large ranges?
会的，但永远不会断。素数定理说，不超过 x 的质数个数 π(x) 约等于 x/ln(x)：质数在 1 到 100 里有 25 个，而在 100 万以内只占约 7.8%。比例越来越小，但质数依然无穷无尽——两个事实同时为真，这正是数论的奇妙之处。
Yes, they thin out, but they never stop. The prime number theorem says the count π(x) of primes up to x is approximately x/ln(x): there are 25 primes below 100, but within a million they make up only about 7.8%. The share shrinks while the primes still go on forever — both facts hold at once, which is part of number theory's charm.
难度 进阶 · 适合 大众
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#素数定理 #质数分布 #密度
为什么1不是质数？
Why is 1 not considered a prime number?
质数定义要求恰好有两个不同的正因数，而1只有一个正因数（它自己）。若把1算作质数，每个数的质因数分解就不再唯一。
A prime must have exactly two distinct positive divisors; 1 has only one, and including it would break the uniqueness of prime factorization.
难度 进阶 · 适合 初中(12-15)
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #质数定义
怎样用奇偶位差判断一个数是11的倍数？
How does the alternating-digit-sum rule test divisibility by 11?
把一个数奇数位数字之和减去偶数位数字之和，若差是11的倍数（含0），原数就能被11整除。它来自10≡−1 (mod 11)的性质。
Subtract the sum of digits in even positions from the sum in odd positions; if the difference is a multiple of 11 (including 0), the number is divisible by 11. It follows from 10≡−1 (mod 11).
难度 进阶 · 适合 大众
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#整除 #11倍数 #模运算
模运算为什么常被比作钟表算术？
Why is modular arithmetic often compared to clock arithmetic?
模n把数按除以n的余数分类，超过n就绕回，正如12点之后回到1点。它把无限整数压缩成有限个剩余类，便于处理周期与整除。
Modulo n groups numbers by remainder after division by n, wrapping around like a clock past 12. It compresses the integers into finitely many residue classes, handy for cycles and divisibility.
难度 进阶 · 适合 大众
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#模运算 #同余 #钟表算术
什么是完全数？目前已知的完全数有什么规律？
What are perfect numbers, and what pattern do known ones follow?
完全数是指恰好等于其真因子（不含自身）之和的数，例如 6 = 1+2+3，28 = 1+2+4+7+14。欧几里得-欧拉定理证明：一个偶数是完全数，当且仅当它形如 2^(p−1)(2^p − 1)，其中 2^p − 1 是梅森素数。至今尚未发现任何奇完全数，其是否存在仍是未解之谜。
A perfect number equals the sum of its proper divisors (excluding itself), e.g., 6 = 1+2+3 and 28 = 1+2+4+7+14. The Euclid–Euler theorem shows an even number is perfect iff it has the form 2^(p−1)(2^p − 1) where 2^p − 1 is a Mersenne prime. No odd perfect number has ever been found, and whether one exists remains unknown.
难度 进阶 · 适合 初中(12-15)
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#完全数 #真因子 #梅森素数 #欧几里得-欧拉
卡迈克尔数（伪素数）是什么？
What are Carmichael numbers (pseudoprimes)?
卡迈克尔数满足对所有与其互质的 a 都有 a^(n−1) ≡ 1 (mod n)，却并非质数，因此能骗过费马素性测试。最小的是 561。
A Carmichael number n satisfies a^(n−1)≡1 (mod n) for all coprime a yet is composite, fooling Fermat tests; smallest is 561.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #伪素数
什么是超越数？
What is a transcendental number?
超越数是不能作为任何整系数多项式方程根的实数或复数，如 π 和 e。与之相对的是代数数。林德曼证明 π 是超越数，从而肯定“化圆为方”不可能用尺规完成。
A transcendental number is not a root of any nonzero polynomial with integer coefficients, like π and e. Lindemann proved π transcendental, settling that squaring the circle by ruler and compass is impossible.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #超越数 #代数数
什么是孪生素数，张益唐的工作有什么意义？
What are twin primes, and why was Yitang Zhang's work important?
孪生素数是相差2的素数对（如11和13）。张益唐2013年证明存在无穷多对素数其差小于7000万，开启了有界间隙研究的热潮。
Twin primes are prime pairs differing by 2 (e.g., 11 and 13). Zhang Yitang's 2013 result showed infinitely many prime pairs within 70 million, launching the bounded-gaps research boom.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-11-03
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张益唐《Bounded Gaps Between Primes》，Annals of Mathematics 179(3):1121–1176, 2014
Y. Zhang, 'Bounded Gaps Between Primes', Annals of Mathematics 179(3):1121–1176, 2014
独立核验依据 · Independent verification: 核验依据 1 ARCHIVE 标识 arxiv-1409.8361：The "bounded gaps between pri
#孪生素数 #素数间隙 #张益唐
卡迈克尔函数 λ(n) 与欧拉定理有什么联系，它为何在 RSA 密码里更精确？
How does Carmichael's function λ(n) relate to Euler's theorem, and why is it more precise than φ(n) in RSA?
卡迈克尔函数 λ(n) 给出使 a^λ(n) ≡ 1 (mod n) 对所有与 n 互素的 a 成立的最小指数，它总是整除欧拉函数 φ(n)。在 RSA 中，用 λ(n) 计算解密指数能得到更小、同样正确的密钥，因为它刻画的是群 (Z/nZ)× 的精确指数（最大阶）。
Carmichael's function λ(n) is the smallest exponent for which a^λ(n) ≡ 1 (mod n) for every a coprime to n, and it always divides Euler's φ(n). In RSA, computing the private exponent modulo λ(n) yields a smaller yet valid key because λ(n) is exactly the exponent (maximum order) of the multiplicative group modulo n.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-04
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Menezes、van Oorschot、Vanstone《Handbook of Applied Cryptography》（CRC Press）
Menezes, van Oorschot, Vanstone, Handbook of Applied Cryptography, CRC Press
独立核验依据 · Independent verification: 核验依据 1 ISBN 978-1-4398-2191-6 (CRC Press)
#卡迈克尔函数 #欧拉定理 #RSA #群指数
为什么欧几里得算法（辗转相除法）一定能求出两个正整数的最大公约数？
Why does the Euclidean algorithm always succeed in finding the greatest common divisor of two positive integers?
核心是恒等式 gcd(a,b)=gcd(b, a mod b)：设 a=bq+r，凡同时整除 a 与 b 的数必整除 r，反之亦然，所以新旧两对数的公因数集合完全相同，最大公约数保持不变。由于每步余数严格变小且非负，算法必在有限步内遇到余数 0，此时最后一个非零除数就是所求的 gcd，终止性与正确性由此同时得到保证。
The engine is the identity gcd(a,b)=gcd(b, a mod b): writing a=bq+r, any common divisor of a and b divides r, and conversely, so the old and new pairs share exactly the same set of common divisors and hence the same gcd. Because each remainder is strictly smaller and nonnegative, the algorithm reaches a zero remainder in finitely many steps; the last nonzero divisor is the answer. Termination and correctness follow together.
难度 进阶 · 适合 初中(12-15)
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G. H. Hardy 与 E. M. Wright，《An Introduction to the Theory of Numbers》（数论导引），第6版，Oxford University Press，2008，第1章
G. H. Hardy & E. M. Wright, An Introduction to the Theory of Numbers, 6th ed., Oxford University Press, 2008, Chapter I
独立核验依据 · Independent verification: 核验依据 1 ISBN 9783540643326 (1938) · 核验依据 2 ISBN 9783540643326 (2008)
#数论 #欧几里得算法 #最大公约数
合数 341 = 11 × 31 为什么能骗过费马小定理检验，使 2^340 ≡ 1 (mod 341) 依然成立？这提醒我们做素性检测时要注意什么？
Why can the composite 341 = 11 × 31 fool the Fermat test, so that 2^340 ≡ 1 (mod 341) still holds — and what does this warn us about primality testing?
费马小定理只保证素数满足 2^(p-1) ≡ 1，逆命题不真。对 341：模 11 时 2 的阶为 10；模 31 时由 2^5 ≡ 1 知阶为 5；340 是两者的倍数，两模皆余 1，中国剩余定理合并得 2^340 ≡ 1 (mod 341)。这就是伪素数：单底检验误报，实用须换底数或用 Miller-Rabin。
Fermat's little theorem guarantees the congruence only for primes; its converse fails. For 341, modulo 11 the element 2 has order 10 and 340 is a multiple of 10; modulo 31, since 2^5 = 32 ≡ 1, its order is only 5 and 340 is likewise a multiple of 5. Both congruences hold, so the Chinese remainder theorem merges them into 2^340 ≡ 1 (mod 341). Such composites are pseudoprimes to base 2: a single-base Fermat test gives false positives, so practical testing varies the bases or turns to Miller-Rabin.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-10-04
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K. H. Rosen《Elementary Number Theory and Its Applications》（第6版），Pearson，2011，§6.2
K. H. Rosen, Elementary Number Theory and Its Applications, 6th ed., Pearson, 2011, §6.2
独立核验依据 · Independent verification: 核验依据 1 ISBN 9780321717757 (2011)
#数论 #费马小定理 #伪素数 #中国剩余定理
两千多年前就证明了质数无穷多,为什么相邻质数之谜至今没解开?
Euclid proved primes infinite over two millennia ago — why does the twin-prime mystery remain open?
欧几里得的论证简洁有力:假设质数有限,把全部相乘再加一,新数除以任何已知质数都余一,于是必有未知质数,矛盾。而孪生质数(相差2的对,如11与13)问的是这种成对结构是否无穷无尽——这要求控制质数之间的精确间隔,难度完全不在一个层级。2013年张益唐首次证明存在无穷多对间隔小于七千万的质数,随后全球协作把界压到246,但离2仍隔着深壑;它与黎曼猜想同列解析数论的圣杯,根源在于乘法结构整齐而加法间隔混沌。
Euclid's proof is elegantly forceful: multiply all assumed-finite primes and add one; the result leaves remainder one modulo every known prime, forcing an unknown prime — contradiction. Twin primes (pairs like 11 and 13) ask instead whether such pairs recur forever, which demands controlling the exact spacing between primes — a different universe of difficulty. In 2013 Yitang Zhang proved infinitely many prime pairs closer than seventy million apart; the bound has since fallen to 246, but the gap to 2 remains a chasm, tied to the same multiplicative-versus-additive tension as the Riemann hypothesis.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-11-03
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Zhang,《Bounded gaps between primes》,Annals of Mathematics 第179卷,2014
Y. Zhang, Bounded gaps between primes, Annals of Mathematics 179, 2014
独立核验依据 · Independent verification: 核验依据 1 ARCHIVE 标识 arxiv-1402.4849：Bounded gaps between primes（扫
#质数 #孪生质数 #张益唐
哥德巴赫猜想是什么，目前进展如何？
What is the Goldbach conjecture, and what is its status?
它猜想每个大于2的偶数都能写成两个素数之和。已对极大范围被计算机验证，但仍未被证明，是著名未解难题。
It conjectures every even number greater than 2 is the sum of two primes. Verified computationally for enormous ranges, but still unproven — a famous open problem.
难度 入门 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#哥德巴赫 #素数 #未解猜想
什么是最大公约数？
What is the greatest common divisor?
最大公约数是两个或多个整数共有约数中最大的一个。例如12和18的公约数有1、2、3、6，最大的是6。
The greatest common divisor (GCD) is the largest positive integer dividing all given numbers; e.g. GCD(12,18)=6.
难度 入门 · 适合 小学高段(9-12)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #公约数
为什么素数（质数）有无限多个？
Why are there infinitely many prime numbers?
欧几里得在《几何原本》第九卷中用反证法证明了素数无穷多：假设只有有限个素数，把它们全部相乘再加 1，所得的新数要么本身是新的素数，要么含有不在原列表中的素因子，从而与假设矛盾。这个简洁的证明至今仍是数学中最优美的论证之一。
Euclid proved in Book IX of the Elements by contradiction that primes are infinite: if you assume finitely many primes, multiply them all and add 1, the result is either itself a new prime or has a prime factor not in the original list, contradicting the assumption. This elegant argument remains one of the most beautiful proofs in mathematics.
难度 入门 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
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Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#素数 #欧几里得 #反证法 #数论
什么是质数？
What is a prime number?
质数（素数）是大于1且只能被1和自身整除的自然数，例如2、3、5、7。
A prime number is a natural number greater than 1 whose only positive divisors are 1 and itself, e.g. 2, 3, 5, 7.
难度 入门 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #质数
什么是质数（素数）？
What is a prime number?
质数（素数）是大于 1、且除了 1 和它本身以外没有其他正因数的自然数，如 2、3、5、7。2 是唯一的偶质数，其余质数都是奇数。
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself, such as 2, 3, 5, and 7. Two is the only even prime; all other primes are odd.
难度 入门 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#质数 #素数 #因数
什么是完全平方数？
What is a perfect square?
完全平方数是某个整数的平方，例如 1、4、9、16、25。判断一个数是否为完全平方数，可看它的平方根是否为整数。
A perfect square is an integer square (1,4,9,16,25…); a number is one iff its square root is an integer.
难度 入门 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #平方数
什么是亲和数？
What are amicable numbers?
亲和数是一对正整数，其中每个数等于另一个数的真因数（不含自身）之和。最小的一对是 (220, 284)：220 的真因数之和为 284，284 的则为 220。
Amicable numbers are a pair where each equals the sum of the other's proper divisors; the smallest pair is (220, 284).
难度 入门 · 适合 小学低段(6-8)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #亲和数
一个完全平方数的末位数字可能有哪些？
Which digits can appear as the last digit of a perfect square?
只可能是0、1、4、5、6、9，绝不可能是2、3、7、8。这由模10的平方剩余直接决定，可用来快速排除非平方数。
Only 0, 1, 4, 5, 6, 9 are possible; never 2, 3, 7, 8. This follows from quadratic residues mod 10 and helps quickly rule out non-squares.
难度 入门 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#平方数 #末位 #整除
贝祖等式是什么？
What is Bézout's identity?
贝祖等式说：对任意整数 a,b，存在整数 x,y 使 a·x + b·y = gcd(a,b)。它既说明最大公约数是 a,b 的线性组合，也是欧几里得算法的理论依据。
Bézout's identity: for integers a,b there exist x,y with a·x+b·y=gcd(a,b); it shows the gcd is a linear combination and underlies the Euclidean algorithm.
难度 入门 · 适合 小学高段(9-12)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #贝祖等式
判断一个数能否被3整除有什么简便方法？
What is an easy rule to check divisibility by 3?
把各位数字相加，若和是3的倍数，原数就能被3整除。例如 123：1+2+3=6，是3的倍数，所以123能被3整除。
Sum the digits; if that sum is divisible by 3, so is the number. For 123, 1+2+3=6, so 123 is divisible by 3.
难度 入门 · 适合 小学低段(6-8)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #整除
偶数和奇数有什么区别？
What is the difference between even and odd numbers?
能被2整除的整数是偶数（如2、4、6），除以2余1的整数是奇数（如1、3、5）。
Even integers are divisible by 2 (2,4,6); odd integers leave remainder 1 when divided by 2 (1,3,5).
难度 入门 · 适合 小学低段(6-8)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Courant & Robbins,《What Is Mathematics?》Oxford University Press,1996 第二版;中译《什么是数学》
R. Courant and H. Robbins, What Is Mathematics?, Oxford University Press, 1996; Chinese ed.
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL811195W：What is mathematics?（按标题
#数论 #奇偶