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\title{Prime Numbers and ``Euclidean'' Proofs}
\author{Nathaniel DeRousse}
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\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.
In doing so, he created a method of proof that could be extended to prove the infinitude of arithmetic progressions of prime numbers.

\section*{Establishing the ``Euclidean'' Method}
\begin{thmcustom}{Theorem 1 \cite{Silverman}}
        There are infinitely many prime numbers.
\end{thmcustom}
\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, then 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 $$q\mid p_1p_2\cdots p_r+1.$$
    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}
Euclid's general argument, assuming there are a finite number of prime numbers and then finding an additional prime not in the original list, is elegant and powerful. It also has the potential for further use in the study of prime numbers.

\section*{Extensions}
The Euclidean argument can be similarly used for different arithmetic progressions of prime numbers, and often with greater simplicity than would be possible through alternative methods. For example, in the case of $1\pmod{4}$ and $3\pmod{4}$ primes, which can be thought of as primes $1$ more and $1$ less than a multiple of $4$, respectively, the following two theorems can be established using the ``Euclidean'' method:
\begin{thmcustom}{Theorem 2 \cite{Silverman}}
    There are infinitely many primes that are congruent to 1 modulo 4.
\end{thmcustom}
\begin{thmcustom}{Theorem 3 \cite{Silverman}}
    There are infinitely many primes that are congruent to 3 modulo 4.
\end{thmcustom}

\section*{Limits}
The ``Euclidean'' method of poof, when applied to arithmetic progressions of prime numbers, does have its limits, however. In order to prove this, it will be necessary to employ an additional theorem and its corollary.
\begin{thmcustom}{Theorem 4 \cite{MurtyandThain}}
    Let $H$ be a subgroup of $(\mathbb{Z}/k\mathbb{Z})$*. Then there is an irreducible polynomial $f$ so that all the prime divisors of $f$, with a finite number of exceptions, belong to the residue class of $H$. Any such prime divisor belonging to any residue class of $H$ divides $f$.
\end{thmcustom}
\begin{thmcustom}{Corollary 1}
    If $\phi_k$ is the $k$-th cyclotomic polynomial, then all of the prime divisors of $\phi_k$ are $\equiv 1\pmod{k}$ or divide $k$. 
\end{thmcustom}
The limits of the Euclidean argument can be abstractly established in the proof of the following theorem:
\begin{thmcustom}{Theorem 5 \cite{MurtyandThain}}
    If $l^2\equiv1\pmod{k}$ then there are infinitely many primes $\equiv l\pmod{k}$, provided there is at least one.
\end{thmcustom}
\begin{proof}
    Let $l$ be an integer. If $l\equiv1\pmod{k}$, apply Corollary 1. Thus we are left with the case in which $l$ is not congruent to $1\pmod{k}$. \par
    We will now consider two groups: the group $(\mathbb{Z}/k\mathbb{Z})$*, which is the integers modulo $k$, and one of its subgroups $H=\{1,l\}$. Now, letting $\zeta$ be a primitive $k$-th root of unity, construct the term $h(\zeta)=(u-\zeta)(u-\zeta^l)$, with $u$ to be chosen momentarily. If $m_1,m_2,\dots,m_s$ are the coset representatives of $H$ in $(\mathbb{Z}/k\mathbb{Z})$*, then we want to choose $u$ so that $h(\zeta^{m_i})$ is distinct for $i=1,2,\dots,s$. The case in which they are not distinct is only possible for a finite number of choices of $u$. Then, according to Theorem 3 with subgroup $H$ and $\eta=h(\zeta)$, there exists a polynomial $$f(x)^2=\prod_{(a,k)=1}(x-(u-\zeta^a)(u-\zeta^{la})),$$ whose prime divisors, except for a finite number, are congruent to $1$ or $l\pmod{k}$. Additionally, $f(0)=\phi_k(u)$, where $\phi_k(u)$ is the $k$-th cyclotomic polynomial. If we then let $u$ be a non-zero multiple of $k$, then $f(0)=\phi_k(u)\equiv1\pmod{k}$. Thus every prime divisor $p$ of $f(0)$ is such that $p\equiv1\pmod{k}$, because Corollary 1 guarantees that every prime divisor of $\phi_k$ is either congruent to $1\pmod{k}$ or is a multiple of $k$. \par
    We will now consider a prime $p\equiv l\pmod{k}$ that does not divide the discriminant of $f$. By Theorem 3, $p\mid f(b)$ for some $b\in\mathbb{Z}$. If $p^2\mid f(b)$, then $f(b+p)=f(b)+pf'(b)\equiv pf'(b)\pmod{p^2}$. However, $f'(b)\neq 0\pmod{p^2}$, because $p\nmid D(f)$ implies that $f$ can have no double roots modulo $p$. Thus, it follows from $f(b)\equiv 0\pmod{p^2}$ that $f(b+p)\neq 0\pmod{p^2}$. In this case, replace $b$ with $b+p$ so that the choice of $b$ is always such that $p^2\nmid f(b)$. \par
    Finally, we proceed with a Euclidean argument: assume there are a finite number of primes congruent to $l$ modulo $k$, $p,p_1,\dots,p_m$. Additionally, let $q_1,q_2,\dots,q_t$ be the prime divisors of $D(f)$ and $$Q=p_2p_3\cdots p_mq_1q_2\cdots q_t.$$ Note that $p^2$ and $kQ$ are relatively prime, and thus by the Chinese Remainder Theorem there exists a $c$ such that $c\equiv b\pmod{p^2}$ and $c\equiv 0\pmod{kQ}$. This also implies that $f(c)\equiv f(b)\pmod{p^2}$ and $f(c)\equiv f(0)\pmod{kQ}$. As already mentioned, all but a finite number of the prime divisors of $f$ are congruent to either $1$ or $l$ modulo $k$. This remaining finite set of primes must either divide $k$ or the discriminant of $f$. Because $f(0)$ is only divisible by primes that are congruent to $1$ modulo $k$, so is $f(c)$. Also $p^2\nmid f(c)$, which implies that $f(c)\equiv l\pmod{k}$. But $f(c)\equiv f(0) \equiv1 \pmod{k}$, which is a contradiction. Therefore, there must be infinitely many prime numbers congruent to $l$ modulo $k$.
\end{proof}

\section*{An Example}
Theorem 5 has established the cases in which the conditions exist for the creation of a ``Euclidean'' proof for the infinitude of arithmetic progressions of prime numbers. While it is unfortunate that this argument cannot be universally applied, its utility still remains unquestioned. As a demonstration of this fact, Theorem 2 can be proven using the methods of the Theorem 5. The same basic method can be applied to any progression that is guaranteed to work by Theorem 5.
\begin{proof}
    We are considering the subgroup $H=\{1,3\}$ of $(\mathbb{Z}/4\mathbb{Z})$*. First, note that $3^2=9\equiv1\pmod{4}$, which is why we can use the methods of Theorem 5 to prove this result. Let $\zeta=i$ and construct the polynomial
    \begin{align*}
        f(x)&=x-(u-\zeta)(u-\zeta^3) \\
        f(x)&=x-(u-i)(u-i^3) \\
        f(x)&=x-(u^2-iu+iu+1) \\
        f(x)&=x-u^2-1.
    \end{align*}
    Since $1$ and $2$ are the coset representatives of $H$, we must choose a $u$ such that $f(\zeta^1)=f(i)$ and $f(\zeta^2)=f(-1)$ are distinct. This is clearly the case for $u=4$, so we now have $f(x)=x-17$. \par
    Now, notice that $2$ is not a prime divisor of the translation $f(4x+20)=4x+3$, and so all of its prime divisors are either $1\pmod{4}$ or $3\pmod{4}$. Additionally, $f(4x+20)\equiv3\pmod{4}$, so not all of its prime divisors are $1\pmod{4}$. Suppose that $f(4x+20)$ has a finite number of prime divisors congruent to $3\pmod{4}$ and let $Q$ be their product. Then consider $f(4Q+20)=4Q+3$, which is clearly $3\pmod{4}$. But, $Q$ and $4Q+3$ are coprime, so it has no prime divisor that is $3\pmod{4}$. This is a contradiction, and we have thus proven the infinitude of $3\pmod{4}$ primes.
\end{proof}
    
\section*{Conclusions}
Euclid, when proving that there is an infinite number of prime numbers, formulated an argument whose utility remained untapped for many years. Murty and Thain established that these proofs can only be constructed for a progression $l\pmod{k}$ if $l^2\equiv1\pmod{k}$ \cite{MurtyandThain}. However, despite this limit, the Euclidean technique 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}{11}
    \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.
\end{thebibliography}

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