## 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.

### Introduction to Number Theory by Apostol #

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.

1. 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.
2. 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 $$(a,bc)=1$$.
3. 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 $$n>k$$,