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\documentclass[11pt]{article}
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\title{Prime Numbers and "Euclidean" Proofs}
\author{Nathaniel DeRousse}
\date{\textit{February 21, 2020}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}{Definition}[section]
\begin{document}
\maketitle
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\begin{abstract}
This paper will discuss prime numbers, beginning with Euclud's proof of the infinitude of primes, and then give an example of how this "Euclidean" method of proof can be extended to more specific properties of prime numbers such as the infinitude of prime numbers congruent to $3$ (mod $4$). Finally, the limits of the Euclidean method of proof when applied to arithmetic progressions of prime numbers is discussed.
\end{abstract}
\section{Introduction}
A prime number $p$ is a natural number that has exactly two factors, $1$ and itself. Every natural number can be formed by multiplying together prime numbers, and as such these "building blocks of number theory" are extremely important. \cite{Silverman} The study of prime numbers dates back to the time of Euclid, and it was Euclid himself who elegantly proved the infinitude of prime numbers.
\begin{theorem}[Infintely Many Primes Theorem \cite{Silverman}]
There are infinitely many prime numbers.
\end{theorem}
\begin{proof}
Suppose there is a list of primes $p_1,p_2,\dots,p_r$. Let $$A=p_1p_2 \cdots p_r+1.$$
If $A$ is prime, than the initial list is incomplete, and therefore there are infinitely many primes.\par
Suppose $A$ is not prime. Let $q$ be some prime dividing $A$, thus $$q|p_1p_2 \cdots p_r+1.$$
If $q$ were in our original list, then it would divide $A$, meaning it would have to divide $1$, which is impossible.\par
So $q$ is a prime number that is not in the original list, and hence there are infinitely many primes. \cite{Silverman}
\end{proof}
Interestingly, the argument that Euclid uses in this famous proof can be extended further into the study of prime numbers.
\section{Another Euclidean Proof}
Observe in the following list of prime numbers, $$3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 97, 101,\dots,$$ that none of the primes are odd, because $2$ is the only even prime. Furthermore, observe that each of these numbers is either congruent to $1$ (mod $4$) or $3$ (mod $4$). \cite{Silverman} These two categories can be thought of as either $1$ more that a multiple of $4$ or $1$ less than a multiple of $4$, respectively. Physically separating the two types of primes yields the following lists:\\
\begin{center}
\begin{tabular} { c c }
\hline
$p\equiv1 \text{ (mod} 4):$ & $5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101,\dots$ \\
\hline
$p\equiv3 \text{ (mod} 4):$ & $3, 7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83,\dots$ \\
\hline
\end{tabular}
\end{center}\par
Naturally, one might wonder whether each of these more specific list of prime numbers is infinite. As it turns out, the Euclidean argument can indeed be extended to each of these cases. Specifically, in the case of primes congruent to 3 modulo 4, the proof is a direct extension of Euclid's original argument.
\begin{theorem}[Primes $3$ (Mod $4$) Theorem \cite{Silverman}]
There are infinitely many primes that are congruent to 3 modulo 4.
\end{theorem}
\begin{proof}
Suppose there is a list of primes congruent to $3$ modulo $4$: $3,p_1,p_2,\dots,p_r$. Let $$A=4p_1p_2 \cdots p_r+3.$$
Let $A$ be written as the product of prime factors such that $$A=q_1q_2 \cdots q_s.$$
Note that because $A$ is defined to be congruent to to $3$ (mod $4$), none of its prime factors are $2$. Furthermore, all of its prime factors are congruent to either $3$ (mod $4$) or $1$ (mod $4$).\par
Now, because the product of numbers that are congruent to $1$ (mod $4$) is itself congruent to $1$ (mod $4$), one of the prime factors of $A$ must be congruent to $3$ (mod $4$) in order for $A$ to be congruent to $3$ (mod $4$).\par
Additionally, because none of the original list of primes $3,p_1,p_2,\dots,p_r$ divides $A$, then the prime factors of $A$ are not in our original list. Thus there exists a prime number congruent to $3$ (mod $4$) that is not in the original list.\par
Therefore, there are infinitely many primes congruent to $3$ modulo $4$. \cite{Silverman}
\end{proof}
The above proof demonstrates that the Euclidean argument can be extended to take on more complicated proofs involving prime numbers.
\section{Limits of the Euclidean Proof}
The study of prime numbers in arithmetic progressions was abstractly stated by Dirichlet in 1837. \cite{Silverman}
\begin{theorem}[Dirichlet's Theorem on Primes in Arithmetic Progressions \cite{Silverman}]
Let a and m be integers with gcd(a,m)=1. Then there are infinitely many primes that are congruent to a modulo m. That is, there are infinitely many prime numbers p satisfying $$p \equiv a(\text{mod }m).$$
\end{theorem}
The traditional proof offered by Dirichlet for this theorem is a complicated one, involving calculus with complex numbers. \cite{Silverman} It is natural to wonder if instead the simpler Euclidean argument can be applied to prove this abstract theorem. Unfortunately, this is not the case, as Murty and Thane have shown that a Euclidean argument can be constructed only under certain conditions. \cite{MurtyandThain}
\begin{theorem}[\cite{MurtyandThain}]
A "Euclidean proof" exists for the arithmetic progression l (mod k) if and only if $l^2 \equiv 1$ (mod k).
\end{theorem}
Thus, Euclid's argument, when applied to the arithmetic progression of prime numbers, has been exhausted. We will prove the "if" direction of Theorem 3.2 by following the outline of the argument given by Murty and Thane modulo some field theoretic results which are deemed outside the scope of this project.
\begin{thebibliography}{2}
\bibitem{Silverman} Joseph Silverman, \textit{A Friendly Introduction to Number Theory} p.83-140, Pearson, Boston, 4th edition, 2011.
\bibitem{MurtyandThain} M.R. Murty; N. Thain, \textit{Prime Numbers in Certain Arithmetic Progressions}, Funct. Approx. Comment. Math. 35 (2006), 249-259.
\bibitem{wolfram} Eric W. Weisstein. \textit{Cyclotomic Polynomial}. \\ URL: http://mathworld.wolfram.com/CyclotomicPolynomial.html.
\end{thebibliography}
\end{document}