## Analytic number theory #

Analytic number theory uses tools and ideas from real and complex analysis to approach and solve problems in the field of number theory.

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*Introduction to Number Theory* by Apostol
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Found an online reading group for Apostol’s *Introduction to Number Theory* on Hackernews in March 2021. Have been looking for something mathematical and interesting think about as my default instead of “doomscrolling”. This is a topic I’ve always wanted to know more about so I thought I might give it a go. Specifically, will give a few of the early chapters a go to get a feel for the material and get back into learning some mathematics.

#### Notes on how to read this book #

- Focus on the exercises.
- “Carry” an exercise around with me until I’ve figured it out.
- Read through the preceding chapter for ideas.
- If I get really stuck, look through solutions for ideas.
- Write up my own solutions in org-mode.

#### Chapter 1 - The Fundamental Theorem of Arithmetic #

In these exercises, lower case latin letters \(a,b,c,\ldots,x,y,z\) represent integers. Prove each of the statements in Exercises 1 through 6.

- If \((a,b)=1\) and if \(c\mid a\) and \(d\mid b\), then \((c,d)=1\). In words, if two numbers are coprime then all their factors are also coprime. This is straightforward: we must have \(\alpha a + \beta b = 1\) for some \(\alpha, beta\) and also \(a=\lambda c\) and \(b=\mu d\) for some \(\lambda, \mu\). Hence \((\alpha \lambda)c + (\beta\mu)d =1\) and so \(c\) and \(d\) are coprime.
- If \((a,b)=(a,c)=1\) then \((a, bc)=1\). In words, if \(a\) is coprime to both \(b\) and \(c\) then it is comprime to their product. This is straighforward too: we must have \(\alpha_1 a + \beta b = 1\) and \(\alpha_2 a + \gamma c = 1\) and so multiplying the two equations together we see \[ \alpha_1\alpha_2 a^2 + \alpha_1\beta ab + \alpha_2\gamma ac + \beta\gamma bc = 1 \] i.e. \[ a(\alpha_1\alpha_2 a + \alpha_1\beta b+ \alpha_2\gamma c) + (\beta\gamma)bc= 1 \] which is the definition of \((a,bc)=1\).
- If \((a,b)=1\) then \((a^n, b^k)=1\) for all \(n\geq 1\), \(k\geq 1\). In words, if \(a\) is coprime to \(b\) then all their powers are also coprime. Indeed, suppose \(\alpha a+\beta b = 1\). Then if \(n>k\),