分圆多项式、算术函数
欧拉函数的计算
Euler’s totient function
If the number \(m\) is prime, then \[\phi(m) = m-1.\]
Example 1 \[ \begin{aligned} 0\;&1\;&2\;&3\;&4\\ \end{aligned} \]
\[ \phi(5)=4. \]
If \(p\) is prime and \(k\ge 1\) then \[ \phi(p^k) = p^k -p^{k-1} =p^{k-1}(p-1) = p^k \left( 1 - \frac{1}{p} \right). \]
Proof. The multiples of \(p\) that are less than to \(p^k\) are \(0, p, 2p,\cdots, (p^{k-1}-1)p\), and there are \(p^{k-1}\) of them.
Therefore, the other \(p^k -p^{k - 1}\) numbers are all relatively prime to \(p^k\).\(\square\)
Example 2 \[ \begin{aligned} \underline{0}\;&1\;&2\\ \underline{3}\;&4\;&5\\ \underline{6}\;&7\;&8\\ \end{aligned} \]
\[ 9-3=6 \]
\[ \phi(3^2)=3^2-3^1=6. \]
If two numbers \(m\) and \(n\) are coprime, then \[\phi(mn) = \phi(m) \phi(n).\]
Example 3 \[ \begin{aligned} 0\;&1\;&2\;&3\;&4\\ 5\;&6\;&7\;&8\;&9\\ 10\;&11\;&12\;&13\;&14\\ \end{aligned} \]
\[ 5\times 3- 3 - 5 +1=(5-1)(3-1)=8. \]
\[ \phi(3\times 5)=\phi(3)\phi(5)=2\times 4=8. \]
The fundamental theorem of arithmetic states that if \(n > 1\) there is a unique expression for \(n\), \[ n = p_1^{k_1} \cdots p_r^{k_r}, \] where \(p_1 < p_2 < \cdots < p_r\) are prime numbers and each \(k_i \ge 1\). Then \[ \phi(n) =n \prod_{p\mid n} \left(1-\frac{1}{p}\right), \]
\[ \phi(n) = \phi(p_1^{k_1}) \phi(p_2^{k_2}) \cdots\phi(p_r^{k_r}) \]
\[ = p_1^{k_1} \left(1- \frac{1}{p_1} \right) p_2^{k_2} \left(1- \frac{1}{p_2} \right) \cdots p_r^{k_r} \left(1- \frac{1}{p_r} \right) \]
\[ = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} \left(1- \frac{1}{p_1} \right) \left(1- \frac{1}{p_2} \right) \cdots \left(1- \frac{1}{p_r} \right) \]
\[ =n \left(1- \frac{1}{p_1} \right)\left(1- \frac{1}{p_2} \right) \cdots\left(1- \frac{1}{p_r} \right). \]
Example 4 \[ \phi(225)=\phi\left(3^2 5^2\right)=225\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)=225\cdot\frac{2}{3}\cdot\frac{4}{5}=90. \]
Exercise 1 What is the last two digits of \(27^987654321\) ?
Property established by Gauss.
\[ \sum_{d\mid n}\phi(d)=n, \] where the sum is over all positive divisors \(d\) of \(n\).
Example 5 \[ \phi(20)+\phi(10)+\phi(5)+\phi(4)+\phi(2)+\phi(1)=20 \]
\[ 8+4+4+2+1+1=20 \]
\[ \begin{aligned} \frac{ 1}{20},\,\frac{ 2}{20},\,\frac{ 3}{20},\,\frac{ 4}{20},\, \frac{ 5}{20},\,\frac{ 6}{20},\,\frac{ 7}{20},\,\frac{ 8}{20},\, \frac{ 9}{20},\,\frac{10}{20},\\ \frac{11}{20},\,\frac{12}{20},\, \frac{13}{20},\,\frac{14}{20},\,\frac{15}{20},\,\frac{16}{20},\, \frac{17}{20},\,\frac{18}{20},\,\frac{19}{20},\,\frac{20}{20}\\ \end{aligned} \]
Put them into lowest terms:
\[ \begin{aligned} \frac{ 1}{20},\,\frac{ 1}{10},\,\frac{ 3}{20},\,\frac{ 1}{ 5},\, \frac{ 1}{ 4},\,\frac{ 3}{10},\,\frac{ 7}{20},\,\frac{ 2}{ 5},\, \frac{ 9}{20},\,\frac{ 1}{ 2},\\ \frac{11}{20},\,\frac{ 3}{ 5},\, \frac{13}{20},\,\frac{ 7}{10},\,\frac{ 3}{ 4},\,\frac{ 4}{ 5},\, \frac{17}{20},\,\frac{ 9}{10},\,\frac{19}{20},\,\frac{ 1}{ 1} \end{aligned} \]
\[ \phi(20) \rightarrow \left(\frac{ 1}{20},\,\frac{ 3}{20},\,\frac{ 7}{20},\,\frac{ 9}{20},\,\frac{ 11}{20},\,\frac{13}{20},\,\frac{17}{20},\,\frac{19}{20}\right) \]
\[ \phi(10) \rightarrow \left(\frac{ 1}{10},\,\frac{ 3}{10},\,\frac{ 7}{10},\,\frac{ 9}{10}\right), \,\, \phi(5)\rightarrow \left(\frac{ 1}{5}\,\,\frac{ 2}{5},\,\frac{ 3}{5},\,\frac{ 4}{5}\right) \]
\[ \phi(4) \rightarrow \left(\frac{1}{4},\,\frac{3}{4} \right),\,\,\phi(2) \rightarrow \left(\frac{1}{2} \right),\,\, \phi(1) \rightarrow \left(\frac{1}{1} \right) \]
分圆多项式
Definition 1 The \(n\)-th Cyclotomic Polynomials, for any positive integer n, is the monic polynomial which is a divisor of \(x^n-1\) and is not a divisor of \(x^k-1\) for any \(k < n\).
Its roots are the \(n\)-th primitive roots of unity \(e^{2i\pi\frac{k}{n}},\) where \(k\) runs over the integers lower than \(n\) and coprime to \(n\).
In other words, \(n\)-th Cyclotomic Polynomials is equal to
\[ \Phi_n(x) = \prod_\stackrel{1\le k\le n}{\gcd(k,n)=1}\left(x-e^{2i\pi\frac{k}{n}}\right) \]
Example 6 Start with \(x^6-1=(x^3-1)(x^3+1)\).
Throw out \((x^3-1)\) due to \(3|6\).
Then \(x^3+1=(x+1)(x^2-x+1)\).
Throw out \((x+1)\) due to \(x+1|x^2-1\) and \(2|6\).
So, \(\Phi_6(x)=x^2-x+1.\)
Example 7 \[ \Phi_1(x) = x - 1 \]
\[ \Phi_2(x) = x + 1 \]
\[ \Phi_3(x) = x^2 + x + 1 \]
\[ \Phi_4(x) = x^2 + 1 \]
\[ \Phi_5(x) = x^4 + x^3 + x^2 + x +1 \]
\[ \Phi_6(x) = x^2 - x + 1 \]
\[ \Phi_7(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 \]
\[ \Phi_8(x) = x^4 + 1 \]
Exercise 2 \[ \Phi_9(x) = ? \]
Theorem 1 \[ x^n - 1 = \prod_{d\mid n}\Phi_d(x), \] which means that each \(n\)-th root of unity is a primitive \(d\)-th root of unity for a unique \(d\) dividing \(n\).
Example 8 \[ x^6-1=\Phi_1(x)\Phi_2(x)\Phi_3(x)\Phi_6(x)=(x-1)(x+1)(x^2 + x + 1)(x^2 - x + 1) \]
Proof. \[ x^n-1 = \prod_{1\le k\le n} \left(x-e^{2i\pi\frac{k}{n}}\right) \]
If \(\gcd(k,n)=d\) then \(x-e^{2i\pi\frac{k}{n}}=x-e^{2i\pi\frac{k^\prime}{n^\prime}},\) where \(k^\prime=k/d\), \(n^\prime=n/d\), and \(\gcd(k^\prime,n^\prime)=1\). \(x-e^{2i\pi\frac{k^\prime}{n^\prime}}\) is one of the factors of \(\Phi_{n^\prime}(x)\), for every \({n^\prime}\) dividing \(n\), exactly once. So,
\[ x^n-1 = \prod_{n^\prime | n} \Phi_{n^\prime}(x)\square \]
Theorem 2 The degree of \(\Phi_n(x)\), or in other words the number of \(n\)-th primitive roots of unity, is \(\phi(n)\), where \(\phi\) is Euler’s totient function.
Theorem 3 \(\Phi_n(x)\) has integer coefficients.
Proof. (By PMI) \(\Phi_{1}(x)=x-1\). Suppose the claim is true for \(k<m\). Then
\[ x^m-1=\prod_{d | m} \Phi_{d}(x) = \left(\prod_\stackrel{d|m}{d<m} \Phi_{d}(x)\right)\cdot\Phi_{m}(x), \]
The first part is monic with integer coefficients. So, \(\Phi_{m}(x)\) also has integer coefficients. \(\square\)
Theorem 4 For \(n\ge 2\), \(\Phi_n(x)\) is reciprocal. Namely, \[ \Phi_n\left(\frac{1}{x}\right)\cdot x^{\phi(n)}= \Phi_n(x). \]
Example 9 \[ \Phi_6(x) = x^2 - x + 1 \] \[ \left[{\left(\frac{1}{x}\right)}^2 - {\left(\frac{1}{x}\right)} + 1\right]\cdot x^2=x^2 - x + 1 \]
Proof. (By BMI) It is true for \(n=2\) because \(\Phi_2(x)=x+1\) and \[ \Phi_2\left(\frac{1}{x}\right)\cdot x^{1}= \Phi_2(x). \]
Suppose it is true for \(n<m\).
\[ {\left(\frac{1}{x}\right)}^m-1=\prod_{d | m}\Phi_{d}\left(\frac{1}{x}\right) =\left(\prod_\stackrel{d|m}{1<d<m} \Phi_{d}{\left(\frac{1}{x}\right)}\right)\cdot\Phi_{m}{\left(\frac{1}{x}\right)}\cdot{\left(\frac{1}{x}-1\right)} \]
Multiply by \(x^m=x^{\sum_{d|m}{\phi(d)}}=\prod_{d|m}{x^{\phi(d)}}\) on both sides.
\[ 1-x^m=\left(\prod_\stackrel{d|m}{1<d<m} \Phi_{d}{\left(\frac{1}{x}\right)}x^{\phi(d)}\right)\cdot\Phi_{m}{\left(\frac{1}{x}\right)}x^{\phi(m)}\cdot{\left(\frac{1}{x}-1\right)x^1} \]
\[ =\left(\prod_\stackrel{d|m}{1<d<m} \Phi_{d}(x)\right)\cdot\Phi_{m}{\left(\frac{1}{x}\right)}x^{\phi(m)}{(-\Phi_{1}(x))} \]
\[ \prod_{d | m}\Phi_{d}(x)=\Phi_{1}(x)\left(\prod_\stackrel{d|m}{1<d<m} \Phi_{d}(x)\right)\cdot\Phi_{m}{\left(\frac{1}{x}\right)}x^{\phi(m)} \]
Cancelling the common facts, we obtain: \(\Phi_{m}(x)=\Phi_{m}{\left(\frac{1}{x}\right)}x^{\phi(m)}.\square\)
Example 10 If \(n\) is a prime number then
\[ x^n-1=(x-1)(x^{n-1}+\cdots+x^2+x+1). \] \[ ~\Phi_n(x) = 1+x+x^2+\cdots+x^{n-1}. \] If \(n=2p\) where \(p\) is an odd prime number then \[ x^{2p}-1=(x^{p}-1)(x+1)(x^{p-1}-\cdots+x^2-x+1). \]
\[ ~\Phi_{2p}(x) = x^{p-1}-\cdots+x^2-x+1. \]
Exercise 3 \[ \Phi_{24}(x) = \,? \]
Lemma 1 \(^{\star\star\star}\). Let \(p\nmid n\) (\(p\) is a prime) and \(m|n\) be a proper divisor of \(n\) (\(m\ne n\)). Then \(\Phi_n(x)\) and \(x^m − 1\) cannot have a common root mod \(p\).
Proof. (By contradiction) Suppose \(a\) is a common root mod \(p\). Then \(a^m\equiv 1 \,\mod\, p\) forces \(\gcd(a,p) = 1\). Next,
\[ x^n-1=\Phi_n(x)\left(\prod_\stackrel{d|n}{d<n}\Phi_{d}(x)\right) \] \(x^m-1=\prod_{d|m}\Phi_d(x)\) has all factors in the last product.
So \(x^n-1\) should have a double root at \(a\), one for \(\Phi_n(x)\), the other for \(x^m-1\) or \(x^m-1=\prod_{d|m}\Phi_d(x)\).
Thus \[ x^n-1\equiv(x-a)^2f(x)\,\mod\,p \] for some \(f(x)\).
Then \(na^{n-1}\equiv 0 \,\mod\,p\).
However, \(p\nmid n\) and \(p\nmid a\), make a contradiction.
Theorem 5 \(^{\star\star\star}\) Let \(n\) be a positive integer. There are infinitely many primes congruent to 1 mod \(n\).
Proof. (By contradiction) Suppose not, let \(\{p_1,p_2,\cdots,p_N\}\) be all the primes congruent to 1 mod \(n\).
Choose some large number \(l\) and let \[ M = \Phi_n (lnp_1\cdots p_N). \] Since \(\Phi_n(x)\) is monic, if \(l\) is large enough, \(M\) will be \(> 1\) and so divisible by some prime \(p\).
First, \(p\) cannot equal \(p_i\) for any \(i\) , since \(\Phi_n(x)\) has constant term 1, and so \(p_i\) divides every term except the last of \(\Phi_n(lnp_1\cdots p_n)\Rightarrow\) it doesn’t divide \(M\).
For the same reason, \(p \nmid n\).
In fact, \(\gcd(p,a) = 1\) where \(a = lnp_1\cdots p_N\).
Now \(\Phi_n(a)\equiv 0\,\mod\,p\) by definition.
By the lemma, we cannot have \[ a^m\equiv 1\,\mod\,p \] for any \(m|n\), \(m < n\).
So the order of \(a\) \((\mod\, p)\) is exactly \(n\), which means that \(n|p − 1\), \[\Rightarrow p\equiv 1\,\mod\,n.\] So, \(p\) is another prime \(\equiv 1 \,\mod\, n\). Contradiction. \(\square\)
算术函数
Definition 2 An arithmetic function is a function f: \(\mathcal{N} \rightarrow \mathcal{C}\)
Example 11 Define \(\nu_{p_i}(n)\) as the exponent of the highest power of the prime \(p_i\) that divides \(n\).
That is to say, \(a_i=\nu_{p_i}(n)\), otherwise it is zero. Then \[ n=\prod_{i}^k p_i^{a_i}=\prod_{i}^k p_i^{\nu_{p_i}(n)}. \]
In terms of the above definition, functions \(\omega\) and \(\Omega\) are defined by
\[ \omega(n) = k, \]
\[ \Omega(n) = a_1 + a_2 + \cdots + a_k. \]
Definition 3 An arithmetic function \(f\) is
additive: if \(f(mn) = f(m) + f(n)\) for all coprime natural numbers \(m\) and \(n\)
multiplicative: if \(f(mn) = f(m)f(n)\) for all coprime natural numbers \(m\) and \(n\)
not coprime \(\Leftrightarrow\) completely \(\cdots\)
Some functions
\(\sigma_k(n)\) is the sum of the \(k\)-th powers of the positive divisors of \(n\), including 1 and \(n\), where \(k\) is a complex number.
\(\sigma_1(n)\), the sum of the (positive) divisors of \(n\), is usually denoted by \(\sigma(n)\).
Since a positive number to the zero power is one, \(\sigma_0(n)\) is therefore the number of (positive) divisors of \(n\), denoted by \(d(n)\).
Definition 4 (Möbius function)***
\[ \mu(n)=\begin{cases} (-1)^{\omega(n)}=(-1)^{\Omega(n)} &\text{if }\; \omega(n) = \Omega(n)\\ 0&\text{if }\;\omega(n) \ne \Omega(n).\end{cases} \]
This implies that \(\mu(1) = 1\).
(Because \(\Omega(1)= \omega(1) = 0\).)
Definition 5 \[ f(n)= \begin{cases} 1&\text{if }n=1\\ 0&\text{otherwise. } \end{cases} \] is completely multiplicative. It’s sometimes called \(\mathcal{I}\).
Example 12
\(f(n) = n^k\) for some fixed \(k\in N\) is also completely multiplicative.
\(\phi(n)\) is multiplicative.
Additive functions
\(\Omega(n)\), \(\omega(n)\), and \(ν_p(n)\).
Example 13 \(2^{\omega(n)}\) is multiplicative
Neither multiplicative nor additive
\(\pi(n)\), the prime counting function, is the number of primes not exceeding \(n\).
Definition 6 A perfect number \(n\) is one for which \(\sigma(n) = 2n\).
Example 14 6, 28, 496
An open conjecture: Every perfect number is even ?
Exercise 4 Please write a program to find a perfect number other than 6, 28, 496.
卷积***
\[ c(n) = \sum_{ij = n} a(i)b(j) = \sum_{i\mid n}a(i)b\left(\frac{n}{i}\right) , \] This function \(c(n)\) is called the Dirichlet convolution of \(a\) and \(b\), and is denoted by \(a*b\). Similar to: \[ (f * g )(t)\ \ \, \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau)\, g(t - \tau)\, d\tau = \int_{-\infty}^\infty f(t-\tau)\, g(\tau)\, d\tau. \]
Example 15 \(f {\star} \mathcal{I} = \mathcal{I} {\star} f = f\) for every \(f\).
Theorem 6 If \(f\) and \(g\) are multiplicative then \(f {\star} g\) is multiplicative.
Proof. Suppose \(m\) and \(n\) are coprime. Then any divisor of \(mn\) is of the form \(d_1,d_2\) , where \(d_1|m\) and \(d_2|n\), uniquely.
So we have
\[ (f*g)(mn)=\sum_{d|mn}f(d)g(\frac{mn}{d}) =\sum_{d_1|m}\sum_{d_2|n}f(d_1d_2)g \left(\frac{m}{d_1}\cdot\frac{n}{d_2}\right) \]
\[ =\sum_{d_1|m}\sum_{d_2|n}f(d_1)f(d_2)g \left(\frac{m}{d_1}\right)g\left(\frac{n}{d_2}\right) =\left(\sum_{d_1|m}f(d_1)g\left(\frac{m}{d_1}\right)\right) \left(\sum_{d_2|n}f(d_2)g\left(\frac{n}{d_2}\right)\right) \]
\[ = (f {\star} g)(m)(f{\star} g)(n)\square \]
Let \(U(n)=1\) for all \(n\).
Then for any arithmetic function f, we have
\[ (f*U)(n)=\sum_{d|n}f(d)U\left(\frac{n}{d}\right)=\sum_{d|n}f(d) \]
This is called \(F(n)\).
Proof. of “\(\sigma_k(n)\) is multiplitive”.
For the function \(r_k(n)=n^k\), we have
\[ (r_k*U)(n)=\sum_{d|n}d^k=\sigma_k(n), \] which is therefore multiplicative. \(\square\)
Other important properties
convolution is commutative
\[ f*g = g*f \]
convolution is associative
\[ f*(g*h) = (f*g)*h \]
The proof is not provided.