同余、中国剩余定理
同余
Definition 1 Let \(a,b,m\) be integers, with \(m\neq 0\). Say \(a\) is congruent to \(b\) modulo \(m\) (\(a\equiv b\,\mod\,m\)) if \(m|(a-b)\).
Example 1 \[ 3\equiv 27\,\mod\,12 \]
\[ -3\equiv 11\,\mod\,7 \]
Basic properties:
- Congruence compatible with usual arithmetic operations of addition and multiplication.
If \(a\equiv b\,(\mod\,m)\) and \(c\equiv d\,(\mod\,m)\),
\[ a+c \equiv b+d\,(\mod\,m) \]
\[ ac \equiv bd\,(\mod\,m) \]
Example 2 \[ 4+12 \equiv 26+1\,(\mod\,11), \,\,\,4\times 12 \equiv 26\times 1\,(\mod\,11) \]
Proof. \[ a=b+mk,\,\,c=d+ml \]
\[ a+c=b+d+m(k+l) \]
\[ ac=bd+bml+dmk+m^2kl=bd+m(bl+dk+mkl)\square \]
Likewise,
- if \(a\equiv b\,(\mod\,m)\), then \(a^k\equiv b^k\,(\mod\,m)\)
However, the follows are NOT TRUE:
If \(a\equiv b\,(\mod\,m)\) and \(c\equiv d\,(\mod\,m)\), then \(a^c\equiv b^d\,(\mod\,m)\).
If \(ax\equiv bx\,(\mod\,m)\), then \(a\equiv b\,(\mod\,m)\).
Example 3 \[ 4^{12} \neq 4^1\,(\mod\,11) \]
\[ 8^{2} \neq 3^{7}\,(\mod\,5) \]
\[ 5\times 2\equiv 2\times 2\,(\mod\,6) \]
欧拉定理
Definition 2 A complete residue system mod \(m\) is a collection of integer \(\{a_1,a_2,\cdots,a_m\}\) such that \(a_i\neq a_j\,\mod\,m\) if \(i\neq j\) and any integer \(n\) is congruent to some \(a_i\,\mod\,m\).
A Reduced residue system mod \(m\) is a collection of integer \(\{a_1,a_2,\cdots,a_m\}\) such that \(a_i\neq a_j\,\mod\,m\) if \(i\neq j\) and \(\gcd(a_i,m)=1\) for all \(i\), and any integer \(n\) coprime to \(m\) must be congruent to some \(a_i\,\mod\,m\).
Example 4 If m=9
Complete Residue System: \(\{1,2,3,4,5,6,7,8,9\}\)
Reduced Residue System: \(\{1,2,4,5,7,8\}\)
If m=10
Complete Residue System: \(\{1,2,3,4,5,6,7,8,9,10\}\)
Reduced Residue System: \(\{1,3,7,9\}\)
Definition 3 The number of elements in a reduced residue system mod \(m\) is called Euler’s Totient Function \(\phi(m)\).
Example 5 \[ \phi(9)=6 \]
\[ \phi(10)=4 \]
Theorem 1 Euler’s Theorem: If \(\gcd(a,m)=1,\) then \[ a^{\phi(m)}\equiv 1\,\mod\,m. \]
Example 6 \[ 3^{\phi(10)}=81\equiv 1\,\mod\,10. \]
Lemma 1 If \(\gcd(a,m)=1\) and \(r_1r_2\cdots r_k\) is a reduced residue system mod \(m\), \(k=\phi(m)\), then \(ar_1\,ar_2\cdots ar_k\) is also a reduced residue system mod \(m\).
Example 7 Reduced residue system mod 10: \[ s=\{1,3,7,9\} \]
\[ 7s=\{7,1,9,3\} \]
Proof. of the lemma.
We shall show that \(ar_i\) are all coprime to \(m\) and distinct mod \(m\). Since \(\gcd(r,m)\) and \(\gcd(a,m)=1\), so, \(\gcd(ar,m)=1\). Also, if \(ar_i\equiv ar_j\,\mod\,m\), then \(m|ar_i-ar_j=a(r_i-r_j)\).
If \(\gcd(a,m)=1\), then \(m|r_i-r_j\Rightarrow r_i\equiv r_j\,\mod\,m\), which won’t be true unless \(i=j\).\(\;\square\)
Proof. of Euler’ theorem.
Choose a reduced residue system \(r_1r_2\cdots r_k\,\mod\,m\) with \(k=\phi(m)\). By lemma, \(ar_1\,ar_2\cdots ar_k\) is also a reduced residue system. These two must be permutation of each other mod \(m\). So,
\[ r_1r_2\cdots r_k\equiv ar_1\,ar_2\cdots ar_k\,\mod\,m \]
\[ r_1r_2\cdots r_k\equiv a^{\phi(m)}r_1\,r_2\cdots r_k\,\mod\,m \]
\[ a^{\phi(m)}\equiv 1\,\mod\,m, \] because
\[ \gcd(r_1r_2\cdots r_k,m)=1. \]
Corollary 1 Fermat’s little Theorem: If \(p\) is a prime and \(a\) is an integer, then
\[ a^p\equiv a(\mod\,p) \]
Proof. \[ \phi(p)=p-1\square \]
Example 8 \[ 3^5\equiv 3\,\mod\,5 \]
\[ 2^{11}\equiv 2\,\mod\,11 \]
Exercise 1 What is the last digit of \(27^{123456789}?\)
模逆元
Definition 4 If \(\gcd(a,m)=1\), then there is a unique integer \(b\mod\,m\) such that \(ab\equiv 1 \mod\,m\). The \(b\) is denoted as \(\frac{1}{a}\) or \(a^{-1}\mod\,m\).
Example 9 \[ \frac{1}{5} \mod\,7 = 5^{-1} \mod\,7 =3. \]
Proof. of Existence
Since \(\gcd(a,m)=1\), \(ax+my=1\) for some integers \(x,y\), so \(ax\equiv 1 \mod\, m\).
Set \(b=x\).
of Uniqueness
If \(ab_1\equiv 1\,\mod\,m\) and \(ab_2\equiv 1\,\mod\,m\), then \[ ab_1\equiv ab_2\,\mod\,m \Rightarrow m|a(b_1-b_2). \] Since \(\gcd(m,a)=1\), \[ m|b_1-b_2\Rightarrow b_1\equiv b_2\,\mod\,m.\square \]
Theorem 2 Wilson’s Theorem: If \(p\) is a prime then \((p-1)!\equiv -1\,\mod\,p\)
Example 10 \[ 4!=24\equiv -1\,\mod\,5 \]
Lemma 2 The congruence \(x^2\equiv 1\,\mod\,p\) has only the solutions \(x\equiv \pm 1\,\mod\,p\).
Proof. \[ x^2\equiv 1\,\mod\,p \]
\[ \Rightarrow p|(x^2-1) \]
\[ \Rightarrow p|(x+1)(x-1) \]
\[ \Rightarrow p|x\pm 1 \]
\[ \Rightarrow x\equiv \pm 1\,\mod\,p\square \]
Proof. of Wilson’s Theorem
Assume that \(p\) is odd, note that \[ x^2\equiv 1\,\mod\,p\Rightarrow \gcd(x,p)=1 \] \(x\) has inverse and \(x\equiv x^{-1}\,\mod\,p\). \(\{1,2,\cdots, p-1\}\) is a reduced residue system mod \(p\).
Pair up elements \(a\) with inverse \(a^{-1} \mod\,p\). Only sigletons will be 1 and -1.
\[ (p-1)!\equiv (a_1a_1^{-1})(a_2a_2^{-1})\cdots(a_ka_k^{-1})(1)(-1)\,\mod\,p \equiv -1\,\mod\,p\square \]
同余方程(组)
Definition 5 A congruence equation is of the form \(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0\equiv 0\,\mod\,m\) where \(\{a_n,a_{n-1},\cdots, a_0\}\) are integers.
Solution of the congruence equation are integers or residue classes mod \(m\) that satisfy the equation.
Example 11 \(x^2\equiv -1\,\mod\,13\). Answer is \(\{5,8\}\).
\(x^2\equiv 1\,\mod\,15\). Answer is \(\{\pm 1,\pm 4\,\mod\,15\}\).
Definition 6 Linear Congruence Equation A congruence equation of degree 1 (\(ax\equiv b\,\mod\,m\))
Theorem 3 Let \(g=\gcd(a,m)\), then there is a solution to \(ax\equiv b\,\mod\,m\) if and only if \(g|b\). If it has solutions, then it has exactly \(g\) solutions mod \(m\).
Example 12 \(4x\equiv 5 \,\mod\, 10\) has no solution, because \(g=\gcd(4,10)\nmid 5\).
\(4x\equiv 6 \,\mod\, 10\) has solution \(x=4\).
In fact, it has \(g=2\) solutions. The other solution is \(x=9.\)
Proof. Suppose \(g\nmid b\).
Suppose \(x_0\) is a solution \(\Rightarrow ax_0=b+mk\) for some integer \(k\).
Since \(g|a,g|m\), \(g\) divides \(ax_0-mk=b\), which is a contradict.
So \(g|b\).
\(g=ax_0+my_0\) for integer \(x_0\),\(y_0\).
Let \(b=b^\prime g\), multiply by \(b^\prime\) to obtain
\[ b=b^\prime g=b^\prime(ax_0+my_0)=a(b^\prime x_0)+m(b^\prime y_0) \]
\[ \Rightarrow a(b^\prime x_0)\equiv b\,(\mod\,m) \] So, \(x=b^\prime x_0\) is a solution.
Prove that there are exactly \(g\) solutions.
Suppose there is one solution \(x_1\).
Then \[ ax\equiv b\equiv ax_1\,(\mod\,m) \]
\[ a(x-x_1)\equiv 0\,(\mod\,m) \]
\(a(x-x_1)=mk\) for some integer \(k\)
\[ g=\gcd(a,m)\Rightarrow a=a^\prime g,m=m^\prime g \] So, \[ \gcd(a^\prime,m^\prime)=1. \]
Then \[ a^\prime g(x-x_1)=m^\prime gk \] \(\Rightarrow a^\prime (x-x_1)=m^\prime k\) for some \(k\).
So, \[ m^\prime|x-x_1, \]
Then \[ x\equiv x_1\,\mod\,m^\prime. \]
So, all solutions are \[ x_1,x_1+m^\prime,x_1+2m^\prime,\cdots,x_1+(g-1)m^\prime.\square \]
欧几里德扩展算法
Extended Euclidean Algorithm is used to obtain \[ a^{-1}\,\mod\,n \] when \(\gcd(a,n)=1\).
Example 13 \[ 41=1\times 23+18 \]
\[ 23=1\times 18+5 \]
\[ 18=3\times 5+3 \]
\[ 5=1\times 3+2 \]
\[ 3=1\times 2+1 \]
\[ 2=2\times 1 \]
\[ 1=3-1\times 2=3-(5-1\times 3)=-1\times 5+2\times 3 \]
\[ =-1\times 5+2\times (18-3\times 5)=2\times 18-7\times 5 \]
\[ =2\times 18-7\times (23-1\times 18)=-7\times 23+9\times 18 \]
\[ =-7\times 23+9\times (41-1\times 23)=9\times 41-16\times 23, \] So, \[ 23^{-1}\,\mod\,41=-16\,or\,25. \]
Exercise 2 \[ 13^{-1}\,\mod\,43 \]
Exercise 3 \[ 999^{-1}\,\mod\,2021 \]
中国剩余定理
Example 14 \[ \left\{ \begin{aligned} x\equiv a_1&(\mod\,m_1)\\ x\equiv a_2&(\mod\,m_2)\\ \cdots &\cdots \\ x\equiv a_k&(\mod\,m_k)\\ \end{aligned} \right. \]
Example 15 \[ \left\{ \begin{aligned} x\equiv 5&(\mod\,7)\\ x\equiv 3&(\mod\,11)\\ x\equiv 10&(\mod\,13)\\ \end{aligned} \right. \]
Solution (Chinese Remainder Theorem)
\[ M_i=\frac{\prod m_i}{m_i} \]
\[ y_i=M_i^{-1}\,\mod\,m_i \]
\[ x=\sum a_iM_iy_i \,\mod\,(\prod m_i) \]
| a | m | M | y | |
|---|---|---|---|---|
| 1 | 5 | 7 | 143 | 5 |
| 2 | 3 | 11 | 91 | 4 |
| 3 | 10 | 13 | 77 | 12 |
So, the result (\(\Sigma aMy\)) is 894 mod 1001.
\[ \left\{ \begin{aligned} 894\equiv 5&(\mod\,7)\\ 894\equiv 3&(\mod\,11)\\ 894\equiv 10&(\mod\,13)\\ \end{aligned} \right. \]
Exercise 4 \[ \left\{ \begin{aligned} x\equiv 1&(\mod\,7)\\ x\equiv 3&(\mod\,11)\\ x\equiv 5&(\mod\,13)\\ x\equiv 7&(\mod\,19)\\ \end{aligned} \right. \]