数论简介、素数、算术基本定理

Author

李 辉

Published

April 2, 2026

数论简介

名人名言

  • (Johann Carl Friedrich Gauss 1777.4–1855.2)

  • Mathematics is the queen of sciences and number theory is the queen of mathematics. She often condescends to render service to astronomy and other natural sciences, but in all relations she is entitled to the first rank.

  • From “Gauss zum Gedächtniss”. Book by Wolfgang Sartorius von Waltershausen, 1856.

相关领域

  • Algebra

  • geometry

  • analysis

  • logic

  • topology

  • computer science

数的演化

  • 自然数\(N\) -> 取反

  • 整数\(Z\) -> 除

  • 有理数\(Q\) -> 实分析/Dedekind Cut

  • 实数\(R\) -> 负数开方

  • 复数\(C\)

Theorem 1 \(\sqrt{2}\) is an irrational number.

Proof. Assume that \(\sqrt{2}\) is a rational number. Then

\[ \sqrt{2}=a/b, \]

where \(a\) and \(b\) are coprime integers.

\[ a^2 / b^2 = 2,\,\,\,\,a^2 = 2 b^2 \]

So, \(a\) must be even.

Let \(a = 2k.\)

\[ 2 b^2 = (2k)^2,\,\,\,\,b^2 = 2k^2 \]

So, \(b\) must be even, which contradicts that a/b is irreducible. \(\;\square\)

从自然数开始

自然数的基本性质

  1. Successor Operation
  • \(s(n)=n+1\)
  1. PMI (Principle of Mathematical Induction)
  • \(p(1)\) is true and \(p(n)\Rightarrow p(n+1)\), then \(p(n)\) is true for all natural numbers.
  1. WOP (Well Ordering Principle)
  • Every nonempty subset of natural number has a smallest element.

Exercise 1 Prove (PMI): \(0 + 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}.\)

整除、带余除法和互素

Divisibility

\[ \begin{gathered} a|b \text{ if }b=ax \text{ for }a,b,x\in Z \text{ and }a\neq 0\\ \forall n\in N,\quad n|0\\ a|b,b|c \rightarrow a|c\\ a|b,a|c \rightarrow a|bx+cy\;\forall x,y\in Z \end{gathered} \]

Example 1 \[ 3|6,6|36 \rightarrow 3|36 \]

\[ 7|14,7|35 \rightarrow 7|(14\times 3+35\times 2=112) \]

Division with Remainder

Theorem 2 Given \(a,b\in Z\) with \(a>0\), \(\exists q,r\in Z\), such that \(b=aq+r\),\(0\le r<a\)

Proof. Let \(S=\{b+ka:k\in Z,b+ka\ge 0\}\),

\[ S\text{ is notempty:} \left\{ \begin{aligned} b>0 \text{ then } b+0a\in S \\ b<0 \text{ then adding a enough times to make it positive}\\ \end{aligned} \right. \]

Since \(S\) is nonempty, it has a smallest element \(r=b+ka\) for some \(k\) (WOP).

Setting \(q=-k\) results in \(r=b-qa\).

\(r\ge0\) because it is in \(S\), and \(r<a\) because if not, then \(b+(k-1)a\) would be smallest element in \(S\).\(\;\square\)

Example 2 \[ 311 = ? \times 13 + ?\,(a=13) \]

\[ -21 = ? \times 11 + ?\,(a=11) \]

Definition 1 GCD(Greatest Common Divisor)

If \(a\) and \(b\) are not both 0, then \(\gcd(a,b)\) or \((a,b)\) is the greatest common divisor of \(a\) and \(b\).

Example 3 \[ \gcd(24,38)=2 \]

Exercise 2 \[ \gcd(148, 111111)=? \]

Theorem 3 Let \(g=\gcd(a,b)\), then \(\exists v_0,y_0 \in Z\) such that \(g=ax_0+by_0\).

Proof. Let \(S=\{ax+by:x,y\in Z,ax+by>0\}\), and assume \(a,b\) not both 0. Assume \(a\neq 0\),

\[ S\text{ is notempty}: \left\{ \begin{gathered} a>0 \Rightarrow a\in S \\ a<0 \Rightarrow -a\in S \end{gathered} \right. \]

Since \(S\) is notempty, it has a smallest element \(g=ax+by\). Next, we shall prove \(g|a\) and \(g\) is largest common.

Prove \(g|a\) (by contradiction):

\[ \begin{gathered} a=gq+r,\;0<r<g\\ r=a-gq=a-q(ax+by)=a(1-qx)-b(qy)\\ \Rightarrow r\in S \end{gathered} \] However, \(r<g\), so \(g\) isn’t the smallest.

Prove \(g\) is largest common:

If \(d|a\) and \(d|b\), then \(d|ax+by=g\). Since \(g|a\),\(g|b\), and \(g\) is largest common divisor, then \(g\) is \(\gcd(a,b)\). \(\;\square\)

Exercise 3 \[ \gcd(24,34)=2 \]

\[ 2=24x+34y,\,x,y\in Z \]

\[ x=?, \,y=? \]

Definition 2 Coprime

If \(\gcd(a,b)=1\), then \(a\) and \(b\) are coprime.

Example 4  

  • 10 and 9 are coprime

  • 10 and 12 are not coprime

Corollary 1 If \(\gcd(a,m)=1\) and \(\gcd(b,m)=1\), then \(\gcd(ab,m)=1\).

Proof. \[ \begin{gathered} 1=ax+my,ax=1-my\\ 1=bx\prime+my\prime,bx\prime=1-my\prime\\ abxx\prime=(1-my)(1-my\prime)\\ =1-my-my\prime+m^2yy\prime\\ =1+m(-y-y\prime+myy\prime)\\ 1=ab(xx\prime)+m(y+y\prime-myy\prime)\square \end{gathered} \]

Example 5 \(\gcd(5, 24)=1\), \(\gcd(7,24)=1\). So, \(\gcd(35,24)=1\).

Corollary 2 If \(c|ab\) and \(\gcd(c,a)=1\), then \(c|b\).

Proof. \[ \begin{gathered} \gcd(a,c)=1 \Rightarrow 1=ax+cy \Rightarrow b=abx+bcy\\ c|ab,c|bc \Rightarrow c|(abx+bcy)=b\square \end{gathered} \]

Example 6 \[ 35|1050, \,\gcd(35,6)=1 \]

\[ 35|175 \]

Euclidean GCD Algorithm (辗转相除法): Given \(a,b\in Z\), not both 0, one can find \(\gcd(a,b)\) as follows.

  1. If \(a,b<0\), replace with negative.

  2. If \(a>b\), switch \(a\) and \(b\).

  3. If \(a=0\), return \(b\).

  4. Since \(a>0\), write \(b=aq+r\) with \(0\le r<a\). Replace \(\gcd(a,b)\) with \(\gcd(r,a)\) and go to step 3.

Proof. Step 1 and 2 do not affect the GCD. So only need to prove \(\gcd(a,b)=\gcd(r,a)\) where \(b=aq+r\). Let \(d=\gcd(r,a)\) and \(e=\gcd(a,b)\),

\[ \begin{aligned} d=\gcd(r,a)\\ \Rightarrow d|a,d|r \\ \Rightarrow d|aq+r=b \\ \Rightarrow d|\gcd(a,b)=e \end{aligned} \]

\[ \begin{aligned} e=\gcd(a,b)\\ \Rightarrow e|a,e|b \\ \Rightarrow e|b-aq=r \\ \Rightarrow e|\gcd(r,a)=d \end{aligned} \]

Since \(d\) and \(e\) are positive and divide each other, and thus being equal. \(\;\square\)

Exercise 4 again

\[ \gcd(148, 111111)=? \]

Definition 3 A prime number is an integer \(p>1\) such that it cannot be written as \(p=ab\) where \(a,b>1\).

Example 7  

  • 11 is a prime number

  • 111 is not a prime number

Theorem 4 (Fundamental Theorem of Arithmetic). Every positive integer can be written as a product of primes (possibly with repetion) and any such expression is unique up to a permutation of the prime factors.

Example 8 \[ 72=2^3\times 3^2 \]

\[ 999999=3^3\times 7\times 11\times 13\times 37 \]

Proof. of Existence (by contradiction):

Let \(S\) be the set of numbers which cannot be written as a product of primes. Assume \(S\) is not empty, it has a smallest element \(n\) by WOP. \(n=1\) is not possible by definition, so \(n>1\). \(n\) cannot be prime, since if so it’d be a product with one term, and so wouldn’t be in \(S\).

Hence, \(n=ab\) with \(a,b>1\).

Also, \(a,b<n\) so they cannot be in \(S\) by minimality of \(n\), and so \(a\) and \(b\) are the product of primes. \(n\) is the product of the two, and so is also a product of primes, and so cannot be in \(S\), and hence \(S\) is empty.

Lemma 1 If \(p\) is prime and \(p|ab\), then \(p|a\) or \(p|b\).

Proof. Assume \(p\nmid a\), and let \(g=\gcd(p,a)\), since \(p\) is prime, \(g=1\) or \(p\), and \(g\) cannot be \(p\) because \(g|a\) and \(p\nmid a\), so \(g=1\). So, \(p|b\).\(\square\)

Corollary 3 If \(p|a_1a_2\cdots a_n\), then \(p|a_i\) for some \(i\).

Proof. If \(n=1\) then the corollary is true. Suppose it holds for \(n=k\). Let \(n=k+1\),

\[ p|\overbrace{a_1a_2\cdots a_k}^{A}\overbrace{a_{k+1}}^{B} \]

\[ p|AB \Rightarrow \left\{ \begin{aligned} p|A \; = p|a_1a_2\cdots a_k\,\,\Rightarrow p|a_i \text{ for some }i\\ p|B \; \Rightarrow p|a_{k+1}\square \end{aligned} \right. \]

Proof. of Uniqueness.

Suppose \(n=p_1p_2\cdots p_r=q_1q_2\cdots q_s\), \(p_1|n=q_1q_2\cdots q_s\), so \(p_1|q_i\) for some \(i\). Since \(p_1\) and \(q_i\) are prime, \(p_1=q_i\).

Canceling one by one, one can obtain \(r=s\) and \(p_1p_2\cdots p_r\) is permutation of \(q_1q_2\cdots q_s\). \(\square\)

Theorem 5 There are infinitely many primes.

Proof. Suppose there are finitely many primes \(p_1,p_2,\cdots,p_n\), with \(n\ge 1\). consider \(N=p_1p_2\cdots p_n+1\), and so by the Fundamental Theorem of Arithemtic there must be a prime \(q\) dividing \(N\). Using Euclidean gcd algorithm, \((p_i,p_1p_2\cdots p_n+1)=(p_i,1)=1\), and so \(p_i\nmid N\). So, \(q\neq p_i\) for any \(i\), and \(q\) is a new prime. \(\square\)

Proof. Another proof by Euler\(^{\star\star\star}\)

\[ 1+\left(\frac{1}{p}\right)+\left(\frac{1}{p}\right)^2+\left(\frac{1}{p}\right)^3+\ldots=\frac{1}{1-1/p}, \]

\[ \prod_{p}\frac{1}{1-1/p}=\prod_{p}(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\ldots). \] After expanding \(\Sigma\), we can pick out any combination of terms to obtain

\[ \prod_{p}\frac{1}{1-1/p}= (\cdots\frac{1}{p_1^{e_1}}\cdots)(\cdots\frac{1}{p_2^{e_2}}\cdots)\cdots(\cdots\frac{1}{p_m^{e_m}}\cdots)\cdots=\sum_{n=1}^{\infty}\frac{1}{n} \square \]

与素数相关的著名猜想

  • Goldbach Conjecture

  • Twin Prime Conjecture

  • Mersenne Prime Conjecture