Add research_paper.tex

research_paper.tex

@@ -0,0 +1,146 @@
+\documentclass[11pt]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{graphicx,amssymb,amsmath,amsthm,hyperref,setspace}
+\usepackage[a4paper, total={6in, 8in}, margin=1in]{geometry}
+
+\title{Prime Numbers and "Euclidean" Proofs}
+\author{Nathaniel DeRousse}
+\date{\textit{November 9, 2018}}
+
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{definition}{Definition}[section]
+
+\begin{document}
+    \maketitle
+    \doublespacing
+    
+    \begin{abstract}
+        This paper will discuss prime numbers, beginning with Euclud's proof of the infinitude of primes, and then demonstrate how this "Euclidean" method of proof is a powerful tool for further investigation of the properties of prime numbers commonly encountered in number theory. These properties include the infinitude of prime numbers congruent to $3 \pmod{4}$ and $1 \pmod{4}$, as well as more complex cases that highlight the limits of the Euclidean method of proof.
+    \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{More Examples of Euclidean Proofs}
+    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 $1 \pmod{4}$ or $3 \pmod{4}$. \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
+            $p\equiv1 \pmod{4}:$ & $5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101,\dots$ \\
+            \hline
+            $p\equiv3 \pmod{4}:$ & $3, 7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83,\dots$ \\
+            \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.
+    \subsection{Primes congruent to $3$ Modulo $4$}
+    \begin{theorem}[Primes $3 \pmod{4}$ 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 $3 \pmod{4}$, none of its prime factors are $2$. Furthermore, all of its prime factors are congruent to either $3 \pmod{4}$ or $1 \pmod{4}$.\par
+        Now, because the product of numbers that are congruent to $1 \pmod{4}$ is itself congruent to $1 \pmod{4}$, one of the prime factors of $A$ must be congruent to $3 \pmod{4}$ in order for $A$ to be congruent to $3 \pmod{4}$.\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 $3 \pmod{4}$ 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 is a relatively direct extension of Euclid's original argument. In contrast, the next proof utilizes another important concept in number theory while also demonstrating the flexibility of Euclid's argument.
+    \subsection{Primes congruent to $1$ Modulo $4$}
+    Before continuing, some familiarity with the concept of quadratic reciprocity is needed.
+    \begin{definition}[Quadratic Residue \cite{Silverman}]
+        A nonzero number that is congruent to a square modulo $p$ is called a \textbf{quadratic residue (QR) modulo $p$}. A number that is not congruent to a square modulo $p$ is called a \textbf{(quadratic) nonresidue (NR) modulo $p$}. 
+    \end{definition}
+    For example, under integers modulo $5$,
+    \begin{eqnarray*}
+        0^2\equiv0 \pmod{5}\\
+        1^2\equiv1 \pmod{5}\\
+        2^2\equiv4 \pmod{5}\\
+        3^2\equiv4 \pmod{5}\\
+        4^2\equiv1 \pmod{5}.
+    \end{eqnarray*}
+    In this case, $1$ and $4$ are QRs modulo 5, and $2$ and $3$ are NRs modulo 5.\par
+    In order to simplify arithmetic involving quadratic residues, a special notation is often used. The \textit{Legendre symbol} \cite{Silverman} of $a$ modulo $p$ is
+    \begin{eqnarray*}
+        \left(\frac{a}{p}\right) =
+        \left\{\begin{array}{rl}
+            1 & \text{if } a \text{ is a quadratic residue modulo } p,\\
+            -1 & \text{if } a \text{ is a nonresidue modulo } p.
+        \end{array}\right.
+    \end{eqnarray*}\par
+    Finally, one more necessary theorem:
+    \begin{theorem}[Quadratic Reciprocity \cite{Silverman}]
+        Let $p$ be an odd prime. Then
+        \begin{center}
+            $\left(\frac{-1}{p}\right)=1$ if $p\equiv1\pmod{4}$, and\\
+            $\left(\frac{-1}{p}\right)=-1$ if $p\equiv3\pmod{4}.$
+        \end{center}
+    \end{theorem}
+    With the Theorem 2.2 in hand, it is now possible to prove the infinitude of $1$ modulo $4$ primes using the Euclidean argument.
+    \begin{theorem}[Primes $1 \pmod{4}$ Theorem \cite{Silverman}]
+        There are infinitely many primes that are congruent to 1 modulo 4.
+    \end{theorem}
+    \begin{proof}
+        Suppose there is a list of primes congruent to $1$ modulo $4$:  $p_1,p_2,\dots,p_r$. Let $$A=(2p_1p_2 \cdots p_r)^2+1.$$
+        Let $A$ be written as the product of prime factors such that $$A=q_1q_2 \cdots q_s.$$
+        Since none of the original list of primes $p_1,p_2,\dots,p_r$ divides $A$, the prime factors of $A$ are not in our original list.\par
+        Because $A$ is odd by definition, all of its prime factors $q_1,q_2,\dots,q_r$ are also odd.\par
+        Now, since the prime factor $q_i$ divides $A$, $$(2p_1p_2 \cdots p_r)^2+1=A\equiv0\pmod{q_i}.$$
+        Let $x=(2p_1p_2 \cdots p_r)^2$ and rearrange the congruence so that $$x^2\equiv-1\pmod{q_i}\text{ or }\left(\frac{-1}{q_i}\right)=1.$$
+        By Theorem 2.2, $q_i$ is therefore congruent to $1 \pmod{4}$).\par
+        Thus, the original list of primes is incomplete and there are infinitely many primes congruent to $1$ modulo $4$ \cite{Silverman}.
+    \end{proof}
+    This proof demonstrates that the Euclidean argument, simply by including other results from across number theory, can be expanded to tackle more complicated proofs.
+    
+    \section{Euclidean Proofs and Cyclotomic Polynomials}
+    The above "Euclidean" proofs generally follow the argument that Euclid made in his original proof of the infinitude of prime numbers. Through adapting the argument to the cases of $3 \pmod{4}$ and $1 \pmod{4}$ primes, the utility and flexibility of that argument has been demonstrated. However, "Euclidean" proofs do not stop there. Before generalizing further, it is necessary to restructure the proof for the infinitude of prime numbers congruent to $1 \pmod{4}$ in terms of polynomials.
+    \begin{proof}
+        Suppose there is a list of primes congruent to $1 \pmod{4}$:  $p_1,p_2,\dots,p_r$. Let $$f(x)=4x^2+1.$$
+        Then $$f(p_1,p_2, \cdots p_r)=4(p_1,p_2, \cdots p_r)^2+1$$
+        If $q$ is a prime divisor of $4(p_1,p_2, \cdots p_r)^2+1$, then $\left(\frac{-1}{q}\right)=1$ using the same logic as the original proof. Hence, again by Theorem 2.2, $q$ is congruent to $1 \pmod{4}$ and our original list is incomplete. \cite{MurtyandThain}
+    \end{proof}
+    A similar change can be made to the proof of the infinitude of $3 \pmod{4}$ primes. In fact, thinking of the Euclidean proof in terms of polynomials allows for even greater extension of Euclid's argument. Such an extension hinges around the concept of cyclotomic polynomials.
+    \begin{definition}[Cyclotomic Polynomials \cite{wolfram}]
+        For any positive integer n, the \textbf{nth cyclotomic polynomial, $\phi_n(x)$,} is the unique irreducible polynomial with integer coefficients that is a divisor of $x^n-1$ and is not a divisor of $x^k-1$ for any $k<n$.
+    \end{definition}
+    For example, the $3$rd cyclotomic polynomial, $\phi_3(x)$, is given by $$\phi_3(x)=x^2+x+1,$$
+    where $\phi_3(x)$ is irreducible, has integer coefficients, and is a divisor of $x^3-1$ but not a divisor of $x^k-1$ for any $k<n$. \par Cyclotomic polynomials have many useful properties, but one in particular will help to connect it to the Euclidean proof method. Such a property involves the prime divisors of cyclotomic polynomials.
+    \begin{definition}[Prime Divisors \cite{MurtyandThain}]
+        A \textbf{prime divisor} p of a polynomial f with integer coefficients is any prime number such that $p | f(n)$ for some integer n.
+    \end{definition}
+    \begin{theorem}[\cite{MurtyandThain}]
+        The prime divisors of the nth cyclotomic polynomial consist of the prime divisors of n and prime numbers $p \equiv 1 \pmod{k}$.
+    \end{theorem}
+    Murty and Thain demonstrate that this property of cyclotomic polynomials can be used to show that the "Euclidean polynomial" $f(x)=4x^2+1$ used in the above proof has an infinite number of prime divisors congruent to $1 \pmod{4}$, which shows that there are infinitely many primes in the $1 \pmod{4}$ arithmetic progression. This concept of "Euclidean polynomials" can then be extended to examine primes in arithmetic progressions of the form $l \pmod{k}$. The details of this connection are beyond the scope of this paper, but are available in Murty and Thane's work. \cite{MurtyandThain}
+    
+    \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 $$p \equiv a \pmod{m}.$$
+    \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 (see Section 3) only under certain conditions. \cite{MurtyandThain}
+    \begin{theorem}[\cite{MurtyandThain}]
+        A "Euclidean proof" exists for the arithmetic progression l $\pmod{k}$ if and only if $l^2 \equiv 1 \pmod{k}$.
+    \end{theorem}
+    Thus, Euclid's argument, when applied to the arithmetic progression of prime numbers, has been exhausted.
+    
+    \section{Conclusions}
+    Euclid, when proving that there is an infinite number of prime numbers, formulated an argument whose utility remained untapped for many years. Throughout this paper, that utility has been demonstrated by proving the infinitude of different categories of prime numbers. Though the argument does have its limits, it provides both elegant and relatively simple proofs when compared to alternative methods, and at the same time demonstrates that proof techniques can be have many useful applications beyond their original use.
+    
+    \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}