% The beginning of every LaTeX document is the a preamble which gives some basic instructions to the LaTeX compiler.
% Lines beginning with % are ignored by the compiler.
% I recommend using article format or the American Mathematical Society format (amsart):
%\documentclass{amsart}
\documentclass{article}
% Now we tell the LaTeX compiler that we want access to math symbols and theorem environments.
\usepackage{amsmath, amssymb, amsthm}
% Define some theorem environments.
\newtheorem{proposition}{Proposition}
\theoremstyle{definition}
\newtheorem{definition}{Definition}
% Define some convenient symbols
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\RR}{\mathbb{R}}
% Tell the compiler your name and the problem number
\author{Your Name Here}
\title{Problem something of chapter something}
% We're done with the preamble: begin the actual document.
\begin{document}
% Have the compiler make a title
\maketitle
If you want to write something you can just type like this.
\LaTeX\@ will automatically indent and space as appropriate to the document.
If you want a new paragraph, you'll need to skip a line.
You can put each sentence on its own line and put as many spaces in as you want: \LaTeX\@ ignores most whitespace.
% that \@ puts a space after \LaTeX
\begin{definition} \label{not-really-a-definition}
% Labels make it easy to refer back to something.
A definition could go here if you wanted.
\end{definition}
Right now Definition \ref{not-really-a-definition} isn't really a definition.
But we can still talk about it using the ref command.
\begin{proposition}
% Math goes between dollar signs or \[ in here \]
The equation $x^2 - 4y - 2 = 0$ has no integer solutions.
\end{proposition}
In symbols the proposition is \[
\forall x, y \in \ZZ, x^2-4y-2 \neq 0 .
\]
\begin{proof}
Suppose (by way of contradiction) the equation $x^2 - 4y- 2 = 0$ has an integer solution.
Let $a, b \in \ZZ$ be that solution.
Hence $a^2 - 4b - 2 = 0$.
Consequently $a^2 = 4b + 2 = 2(2b + 1)$.
Thus $a^2$ is even.
It then follows from earlier work that $a$ is even.
By definition (of even) there is an integer $c$ such that $a = 2c$.
Hence $a^2 = (2c)^2 =4c^2 = 2(2b + 1)$.
Dividing by $2$ shows that $2c = 2b+1$.
This is a problem because $2c$ is even but $2b+1$ is odd.
We have reached a contradiction; our starting assumption must have been false.
Therefore the equation $x^2 - 4y - 2 = 0$ has no integer solutions.
\end{proof}
% Everything that begins also has to end, including this document
\end{document}