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
定理1.2用归纳法证明其实原理还是a-b b和a b有共同的最大公约数 定理1.4(b)证明 d = (a,(b,c))，则d|a,d|(b,c)，那么d|a,d|b,d|c 那么d|(a,b)，那么d|((a,b),c) 同理，如果e=((a,b),c)，则e|(a,(b,c)) 所以d=e
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，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发散？ 因为11+nQ<1n，而调和级数发散
Exercise
1=ax+by, a=cp, b=dq, so 1=cpx+dqy1=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 an and bk1=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)=2as exercise 4, but more complex: 1=ax+by, let m=a+b, n=a2−ab+b2, then a/b=3m±12n−3m2√6 so (12n−3m2)(x−y)2=(6−3m(x+y))2 so 3=n(x−y)2+m(3(x+y)−m(x2+xy+y2))=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|n, then a=c, b=n/c2, if n is not squarefree, then a=1, b=n(a) 4|36 and 9|36 and 4<9 but 2∤3 (b)b2 is the largest, so a2|b2, so a|bm=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 similarthe lowest digit must be odd:1 3 5 7, 5|14+4or34+4or74+4, only leave numbers of n=10k+5, how to prove this? n4+4=(n2+2n+2)(n2−2n+2)) from newsmth.net(a) true let n=1 we get a|b, even n>1, we have m = (b/a)n, m is not integer unless b/a is integer, so a|b (b) false 44|1010, however 4∤10 (c) true the difference with (a) is only factor 2, let a=2ka b=2kb, we get nka<nkb+1 for all n, so ka<kb(a) am=nbm, so bm|am, so b=1 because 12(a) (b) let c=n1/m=p/q so (p/q)m=n, so q=1 and pm=n, contradict with the assumption of n is not the mth power of nassume a=x^m and mChapter2 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)=1的对应关系是，如果(q,n/d)=x>1，那么(dq,n)=xd>d
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