解析数论引论第1章算术基本定理

1.1 Introduction

The principle of induction 归纳法则
意思应该是：证明了1成立，如果n成立意味着n+1成立，那么对于所有大于等于1的数都成立
Well-ordering principle 良序原理
在学习本书过程中作为公理用

1.3 Greatest common divisor

1.4 Prime numbers

定理1.9书中证明不清楚，查看了wiki百科证明如下
px+ay=1
xpb+yab=b
p|ab
所以p|(xpb+yab)=b

1.6 The series of reciprocals of the primes

这一章最难懂的证明就是下面这步：
为什么说不等式右边项包含全部左边项
对于任意11+nQ\frac 1{1+nQ}，将其表示为素数因子的乘积，假设因子个数为t(高于1次算多个)，则其必然位于右边项t的式子中，且次数大于等于1。

为什么∑rn=111+nQ\sum_{n=1}^r\frac1{1+nQ}发散？
因为11+nQ<1n\frac1{1+nQ}，而调和级数发散

Exercise

1=ax+by, a=cp, b=dq, so 1=cpx+dqy
1=ax+by, 1=ax+cz, so 1=(ax+by)(ax+cz)=x(aax+by+cz)+bcyz
(a, b)=1, so a and b have no commom prime factor, so do ana^n and bkb^k
1=ax+by, let m=a+b, n=a-b, then 2=(m+n)x+(m-n)y=m(x+y)+n(x-y)
if 2|x+y and 2|x-y then (a+b, a-b)=1 else (a+b, a-b)=2
as exercise 4, but more complex:
1=ax+by, let m=a+b, n=a2−ab+b2a^2-ab+b^2, then a/b=3m±12n−3m2√6\frac{3m\pm\sqrt{12n-3m^2}}6
so (12n−3m2)(x−y)2=(6−3m(x+y))2(12n-3m^2)(x-y)^2=(6-3m(x+y))^2
so 3=n(x−y)2+m(3(x+y)−m(x2+xy+y2))=ni+mk3=n(x-y)^2+m(3(x+y)-m(x^2+xy+y^2))=ni+mk
so the result is 1 or 3 depend on whether 3|(i,k)
1=ax+by, cd=a+b then 1=(cd-b)x+by=dcx+b(y-x) so (d,b)=1
(a/b)+(c/d)=n, so ad+bc=nbd, so (a-bn)d+bc=0, because (a,b)=1, b|d, in the same way d|b so |d|=|b|
if n is squarefree, choose the max integer c such that c2|nc^2|n, then a=c, b=n/c2b=n/c^2, if n is not squarefree, then a=1, b=n
(a) 4|36 and 9|36 and 4<9 but 2∤32\nmid3
(b)b2b^2 is the largest, so a2|b2a^2|b^2, so a|b
m=ax+by,n=cx+dy, so (a,b)|m and (a,b)|n. so (a,b)|(m,n), the prove of (m,n)|(a,b) is similar
the lowest digit must be odd:1 3 5 7, 5|14+4or34+4or74+41^4+4or3^4+4or7^4+4, only leave numbers of n=10k+5, how to prove this?
n4+4=(n2+2n+2)(n2−2n+2))n^4+4=(n^2+2n+2)(n^2-2n+2)) from newsmth.net
(a) true let n=1 we get a|b, even n>1, we have m =
(b/a)n(b/a)^n, m is not integer unless b/a is integer, so a|b
(b) false 44|10104^4|10^{10}, however 4∤104\nmid10
(c) true the difference with (a) is only factor 2, let a=2kaa=2^{k_a} b=2kbb=2^{k_b}, we get nka<nkb+1nk_a for all n, so ka<kbk_a
(a) am=nbma^m=nb^m, so bm|amb^m|a^m, so b=1 because 12(a)
(b) let c=n1/m=p/qc = n^{1/m}=p/q so (p/q)m=n(p/q)^m=n, so q=1 and pm=np^m=n, contradict with the assumption of n is not the mth power of n
assume a=x^m and m

Chapter2 Arithmetical Functions and Dirichlet Multiplication

定理2.2的证明中说所有A(d)的并是S，因为S中任意数与n的gcd肯定是n的因子，某个d；而且对于某个确定的数a，gcd(a,n)是固定的，因此a只属于某个A(d)，而且不同的d，A(d)没有交集
A(d)与0<q≤n/d(q,n/d)=10\lt q\le n/d (q, n/d)=1的对应关系是，如果(q,n/d)=x>1(q, n/d)=x>1，那么(dq,n)=xd>d