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 $\equiv1\pmod{k}$ 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, $f'(b)\neq0\pmod{p^2}$, because $p\nmid D(f)$ implies that $f$ can have no double roots modulo $p$. Thus, it follows from $f(b)\equiv0\pmod{p^2}$ that $f(b+p)\neq0\pmod{p^2}$. 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 $c\equiv0\pmod{kQ}$. 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.
Suppose there is a list of primes $p_1,p_2,\dots,p_r$. Let $$A=p_1p_2\cdots p_r+1.$$
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
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 $$q\mid p_1p_2\cdots p_r+1.$$
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
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}
So $q$ is a prime number that is not in the original list, and hence there are infinitely many primes. \cite{Silverman}
@ -53,7 +53,7 @@
\begin{proof}
\begin{proof}
Suppose there is a list of primes congruent to $3$ modulo $4$: $3,p_1,p_2,\dots,p_r$. Let $$A=4p_1p_2\cdots p_r+3.$$
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 $$A=q_1q_2\cdots q_s.$$
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
Note that because $A$ is defined to be congruent to to $3\pmod{4}$, $2$ is not among its prime factors. Therefore, 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
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
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}
Therefore, there are infinitely many primes congruent to $3$ modulo $4$. \cite{Silverman}
@ -73,7 +73,7 @@
4^2\equiv1\pmod{5}.
4^2\equiv1\pmod{5}.
\end{eqnarray*}
\end{eqnarray*}
In this case, $1$ and $4$ are QRs modulo 5, and $2$ and $3$ are NRs modulo 5.\par
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
In order to simplify arithmetic involving quadratic residues, a special notation is often used. Assuming that $a$ is not a multiple of $p$, the \textit{Legendre symbol}\cite{Silverman} of $a$ modulo $p$ is
\begin{eqnarray*}
\begin{eqnarray*}
\left(\frac{a}{p}\right) =
\left(\frac{a}{p}\right) =
\left\{\begin{array}{rl}
\left\{\begin{array}{rl}
@ -121,15 +121,14 @@
If $f\in\mathbb{Z}[x]$ is non-constant, then f has infinitely many prime divisors.
If $f\in\mathbb{Z}[x]$ is non-constant, then f has infinitely many prime divisors.
\end{theorem}
\end{theorem}
\begin{proof}
\begin{proof}
To begin, notice that $f$ has at least one prime divisor, because $f(x)=\pm1$ has a finite number of solutions.
Now suppose $f(0)=c\neq0$ and that f has a finite number of prime divisors $p_1, p_2,\dots,p_k$. Let $$Q=p_1p_2\cdots p_k.$$\
To begin, notice that $f$ has at least one prime divisor, because $f(x)=\pm1$ has a number of solutions that is at most the degree of $f$. Additionally, $\{f(n)\,|\,n\in\mathbb{Z}\}$ is an infinite set. Now suppose $f(0)=c\neq0$ and that f has a finite number of prime divisors $p_1, p_2,\dots,p_k$. Let $$Q=p_1p_2\cdots p_k.$$\
Note that $f(x)$ has the form $c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x_1+c$ and
Note that $f(x)$ has the form $c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x_1+c$ and
where $g(x)\in\mathbb{Z}$ is of the form $1+c_1x+c_2x^2+\cdots$. \par
where $g(x)\in\mathbb{Z}[x]$ is of the form $1+b_1x+b_2x^2+\cdots+b_{n-1}x^{n-1}+b_nx^n$. \par
Now, $Q|c_i$ in for each $c_i$ in $g$. Additionally, $g\in\mathbb{Z}[x]$ must have at least one prime divisor, $p$, by the argument presented above. \par
Now, $Q|c_i$ in for each $c_i$ in $g$. Additionally, $g\in\mathbb{Z}[x]$ must have at least one prime divisor, $p$, by the argument presented above. \par
Finally, $p|g$ implies that $p|f$ and thus $p|Q$. However, this is impossible, because $p|Q$ also implies that $p|1$ (in order for $p$ to be a prime divisor of $g$), which is a contradiction. Therefore $f$ has infinitely many prime divisors.
Finally, $p|g$ implies that $p|f$ and thus $p|Q$. However, this is impossible, because $p|Q$ also implies that $p|1$ (in order for $p$ to be a prime divisor of $g$), which is a contradiction. Therefore $f$ has infinitely many prime divisors.
\end{proof}
\end{proof}
@ -153,12 +152,13 @@
\begin{corollary}
\begin{corollary}
If $f(x)=ax+b$ and $g(x)$ is any polynomial, then $f$ and $g$ have infinitely many common prime divisors.
If $f(x)=ax+b$ and $g(x)$ is any polynomial, then $f$ and $g$ have infinitely many common prime divisors.
\end{corollary}
\end{corollary}
The effect of this corollary is that we can assume in what follows that $\deg f$, $\deg g\geq2$.
\begin{lemma}[\cite{Nagell}\cite{Loreaux}]
\begin{lemma}[\cite{Nagell}\cite{Loreaux}]
When proving whether or not two polynomials $f(x)$ and $g(x)$ with degrees $N$ and $M$ have infinitely many common prime divisors, one can assume that $N>M$.
When proving whether or not two polynomials $f(x)$ and $g(x)$ with degrees $N$ and $M$ have infinitely many common prime divisors, one can assume that $N>M$.
\end{lemma}
\end{lemma}
\begin{proof}
\begin{proof}
Assume that $f(x)$ and $g(x)$ both have degree $N$, and construct a new function $h(x)$ such that $h(x)=af(x+c)-bg(x)$, where $a,b,c\in\mathbb{Z}$. If $f(x)=r_Nx^N+r_{N-1}x^{N-1}+\cdots+r_1x+r_0$ and $g(x)=s_Nx^N+s_{N-1}x^{N-1}+\cdots+s_1x+s_0$, then $a$ and $b$ in $h$ can be chosen so that the leading term in $f$ and $g$ drop out ($ar_Nx^N=bs_Nx^N$), and thus guaranteeing that the degree of $h(x)$ is at most $N-1$. \par
Assume that $f(x)$ and $g(x)$ both have degree $N$, and construct a new function $h(x)$ such that $h(x)=af(x+c)-bg(x)$, where $a,b,c\in\mathbb{Z}$. If $f(x)=r_Nx^N+r_{N-1}x^{N-1}+\cdots+r_1x+r_0$ and $g(x)=s_Nx^N+s_{N-1}x^{N-1}+\cdots+s_1x+s_0$, then $a$ and $b$ in $h$ can be chosen so that the leading term in $f$ and $g$ drop out ($ar_Nx^N=bs_Nx^N$), and thus guaranteeing that the degree of $h(x)$ is at most $N-1$. \par
Now, choose a $c$ and $c_1$ ($c\neq c_1$) so that $$af(x+c)-bg(x)=k,$$$$af(x+c_1)-bg(x)=k_1,$$ where $k$ and $k_1$ are both constants. Subtracting these two quantities yields $$a(f(x+c)-f(x+c_1))=k-k_1,$$ which is also a constant. However, this is not possible, because when $N\geq2$, $f(x)$ is not linear, and thus the function given by$a(f(x+c)-f(x+c_1))$ cannot be a constant. \par
Now, choose a $c$ and $c_1$ ($c\neq c_1$) so that $$af(x+c)-bg(x)=k,$$$$af(x+c_1)-bg(x)=k_1,$$ where $k$ and $k_1$ are both constants. Subtracting these two quantities yields $$a(f(x+c)-f(x+c_1))=k-k_1,$$ which is also a constant. However, this is not possible, because when $N\geq2$, $f(x)$ is not linear, and thus the $N-1$ term of $a(f(x+c)-f(x+c_1))$ is not zero. So,$a(f(x+c)-f(x+c_1))$ cannot be a constant. \par
As a result, if it is proven that $h(x)$ and $f(x)$ have infinitely many prime divisors, then $bg(x)$ and $f(x)$ do as well, because any prime divisor of $h(x)$ and $f(x)$ is a prime divisor of $bg(x)$. And because the constant $b$ has only a finite number of prime divisors, there would also therefore be an infinite number of prime divisors shared by $f(x)$ and $g(x)$.
As a result, if it is proven that $h(x)$ and $f(x)$ have infinitely many prime divisors, then $bg(x)$ and $f(x)$ do as well, because any prime divisor of $h(x)$ and $f(x)$ is a prime divisor of $bg(x)$. And because the constant $b$ has only a finite number of prime divisors, there would also therefore be an infinite number of prime divisors shared by $f(x)$ and $g(x)$.
\end{proof}
\end{proof}
\begin{lemma}[\cite{Mohamed}]
\begin{lemma}[\cite{Mohamed}]
@ -200,18 +200,18 @@
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 $f$.
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{theorem}
\end{theorem}
As an example, consider the subgroup $H=\{1,3\}$ in $(\mathbb{Z}/k\mathbb{Z})$* and the irreducible polynomial $f(x)=4x+3$. Any prime $p$ that divides $f$ is either $1\pmod{4}$ or $3\pmod{4}$ (with only a finite number of exceptions). In addition, if any prime is $1\pmod{4}$ or $3\pmod{4}$, it is guaranteed to divide $f$. \par
As an example, consider the subgroup $H=\{1,3\}$ in $(\mathbb{Z}/k\mathbb{Z})$* and the irreducible polynomial $f(x)=4x+3$. Any prime $p$ that divides $f$ is either $1\pmod{4}$ or $3\pmod{4}$ (with only a finite number of exceptions). In addition, if any prime is $1\pmod{4}$ or $3\pmod{4}$, it is guaranteed to divide $f$. \par
There is also one specific case of of this theorem that will be needed in the final proof:
There is also one specific case of of this theorem that will be needed in the final proof. Recall that the roots of the $k$-th cyclotomic polynomial $\phi_k$ are exactly the primitive $k$-th roots of unity. We then have the following:
\begin{corollary}
\begin{corollary}
If $\phi_k$ is the $k$-th cyclotomic polynomial, then all of the prime divisors of $\phi_k$ are $\equiv1\pmod{k}$ or divide $k$.
If $\phi_k$ is the $k$-th cyclotomic polynomial, then all of the prime divisors of $\phi_k$ are $\equiv1\pmod{k}$ or divide $k$.
\end{corollary}
\end{corollary}
An immediate consequence of Corollary 4.1.1 is that there are infinitely many prime divisors of $\phi_k$ that are congruent to $1\pmod{k}$. This is because there are only a finite number of primes that divide $k$, and we have already proven that $\phi_k$ must have infinitely many prime divisors. Combining this logic with that of Theorem 3.6, because any two polynomials have an infinite number of common prime divisors, \textit{any} polynomial with integer coefficients must have infinitely many prime divisors that are congruent to $1\pmod{k}$. With this remarkable result in hand, we finally have the tools necessary to prove to define the limits of ``Euclidean'' proofs:
An immediate consequence of Corollary 4.1.1 is that there are infinitely many prime divisors of $\phi_k$ that are congruent to $1\pmod{k}$. This is because there are only a finite number of primes that divide $k$, and we have already proven that $\phi_k$ must have infinitely many prime divisors. Combining this logic with that of Theorem 3.6, because any two polynomials have an infinite number of common prime divisors, \textit{any} polynomial with integer coefficients must have infinitely many prime divisors that are congruent to $1\pmod{k}$. With this remarkable result in hand, we finally have the tools necessary to prove to define the limits of ``Euclidean'' proofs. The proof that follows is very close to the one given by Murty and Thain, as some of the techniques that are used are very sophisticated.
\begin{theorem}[\cite{MurtyandThain}]
\begin{theorem}[\cite{MurtyandThain}]
If $l^2\equiv1\pmod{k}$ then there are infinitely many primes $\equiv l\pmod{k}$, provided there is at least one.
If $l^2\equiv1\pmod{k}$ then there are infinitely many primes $\equiv l\pmod{k}$, provided there is at least one.
\end{theorem}
\end{theorem}
\begin{proof}
\begin{proof}
Let $l$ be an integer. If $l\equiv1\pmod{k}$, apply Corollary 4.1.1. Thus we are left with the case in which $l$ is not congruent to $1\pmod{k}$. \par
Let $l$ be an integer. If $l\equiv1\pmod{k}$, apply Corollary 4.1.1. 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 polynomial$h(\zeta)=(u-\zeta)(u-\zeta^l)$. 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 4.1 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 4.1.1 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 5, $p\mid f(b)$ 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, $f'(b)\neq0\pmod{p^2}$, because $p\nmid D(f)$ implies that $f$ can have no double roots modulo $p$. Thus, it follows from $f(b)\equiv0\pmod{p^2}$ that $f(b+p)\neq0\pmod{p^2}$. In this case, replace $b$ with $b+p$ so that the choice of $b$ is always such that $p^2\nmid f(b)$. \par
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 4.1 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 4.1.1 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 4.1, $p\mid f(b)$ 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, $f'(b)\neq0\pmod{p^2}$, because $p\nmid D(f)$ implies that $f$ can have no double roots modulo $p$. Thus, it follows from $f(b)\equiv0\pmod{p^2}$ that $f(b+p)\neq0\pmod{p^2}$. 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 $c\equiv0\pmod{kQ}$. 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 $k$.
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 $c\equiv0\pmod{kQ}$. 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}
\end{proof}
This theorem 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, it still provides a relatively simple and elegant method of proof for many progressions. As a final demonstration of this fact, we will return to the proof of Theorem 2.1 ($3\pmod{4}$ primes) and restructure it to use the methods of the above proof. The same basic method can be applied to any progression that is guaranteed to work by Theorem 4.2.
This theorem 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, it still provides a relatively simple and elegant method of proof for many progressions. As a final demonstration of this fact, we will return to the proof of Theorem 2.1 ($3\pmod{4}$ primes) and restructure it to use the methods of the above proof. The same basic method can be applied to any progression that is guaranteed to work by Theorem 4.2.
@ -221,13 +221,13 @@
f(x)&=x-(u-\zeta)(u-\zeta^3) \\
f(x)&=x-(u-\zeta)(u-\zeta^3) \\
f(x)&=x-(u-i)(u-i^3) \\
f(x)&=x-(u-i)(u-i^3) \\
f(x)&=x-(u^2-iu+iu+1) \\
f(x)&=x-(u^2-iu+iu+1) \\
f(x)&=x-u^2+1.
f(x)&=x-u^2-1.
\end{align*}
\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
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.
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}
\end{proof}
\section{Conclusions}
\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, including the abstract case. 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.
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, including the abstract case. Murty and Thain established that these proofs can only be constructed for a progression $l\pmod{k}$ if $l^2\equiv1\pmod{k}$. 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}{2}
\begin{thebibliography}{2}
\bibitem{Silverman} Joseph Silverman, \textit{A Friendly Introduction to Number Theory} p.83-140, Pearson, Boston, 4th edition, 2011.
\bibitem{Silverman} Joseph Silverman, \textit{A Friendly Introduction to Number Theory} p.83-140, Pearson, Boston, 4th edition, 2011.
@ -238,20 +238,3 @@
\bibitem{Mohamed} Mohamed Ayad; Omar Kihel; Jesse Larone, \textit{When Does a Given Polynomial with Integer Coefficients Divide Another?}, The American Mathematical Monthly (2016), 123:4, 376-381, DOI: 10.4169/amer.math.monthly.123.4.376
\bibitem{Mohamed} Mohamed Ayad; Omar Kihel; Jesse Larone, \textit{When Does a Given Polynomial with Integer Coefficients Divide Another?}, The American Mathematical Monthly (2016), 123:4, 376-381, DOI: 10.4169/amer.math.monthly.123.4.376
\end{thebibliography}
\end{thebibliography}
\end{document}
\end{document}
%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.
% 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{theorem}[\cite{MurtyandThain}]
% The prime divisors of the nth cyclotomic polynomial consist of the prime divisors of n and prime numbers $p\equiv1\pmod{k}$.
%\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}
