From f1317040168ac2f26e33a71b9be85366de4ef92b Mon Sep 17 00:00:00 2001 From: nderousse Date: Mon, 2 Mar 2020 09:57:34 -0600 Subject: [PATCH] Research Paper Update Added lemma's 0, 1, and 3 as a build-up towards Murty & Thane's Theorem 2. --- research_paper.tex | 38 +++++++++++++++++++++++++++++++++++++- 1 file changed, 37 insertions(+), 1 deletion(-) diff --git a/research_paper.tex b/research_paper.tex index 3c69bd6..8ddd258 100644 --- a/research_paper.tex +++ b/research_paper.tex @@ -9,6 +9,8 @@ \newtheorem{theorem}{Theorem}[section] \newtheorem{definition}{Definition}[section] +\newtheorem{corollary}{Corollary}[theorem] +\newtheorem{lemma}[theorem]{Lemma} \begin{document} \maketitle @@ -154,7 +156,41 @@ where $g(x) \in \mathbb{Z}$ is of the form $1+c_1x+c_2x^2+ \cdots$. \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. - + \end{proof} + + \section{Murty and Thane Theroem 3} + \begin{lemma} + If $ax+b \in \mathbb{Z} [x]$ and $p$ is a prime such that $p \nmid a$, then $p$ is a prime divisor of $ax+b$. Hence, all but finitely may primes are divisors of $ax+b$. + \end{lemma} + \begin{proof} + Let $ax+b$ be such that $a \neq 0 \pmod{p}$. Then there is a $c$ such that $ac \equiv -b \pmod{p}$. Now, let $x=p+c$. Then + \begin{align*} + ax+b&=a(p+c)+b \\ + &=ap+ac+b \\ + &\equiv 0-b+b \pmod{p} \\ + &\equiv 0 \pmod{p}. + \end{align*} + Thus $p$ is a prime divisor of $ax+b$. The only primes for which this is not the case are the finite set that make up the prime factorization of $a$, and therefore there are only a finite number of primes that are not divisors of $ax+b$. + \end{proof} + This above lemma naturally leads to the following result: + \begin{corollary} + If $f(x)=ax+b$ and $g(x)$ is any polynomial, then $f$ and $g$ have infinitely many common prime divisors. + \end{corollary} + \begin{lemma} + If $f(x)$ and $g(x)$ are any polynomials that both have degree $N \geq 2$, then they have infinitely many common prime divisors. + \end{lemma} + \begin{proof} + Using $f(x)$ and $g(x)$, 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 $N-1$. Additionally, $c$ can be chosen so that $h$ is not 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 $c \neq c_1$ and $N \geq 2$, $f(x+c)-f(x+c_1)$ cannot be constant. + \end{proof} + \begin{lemma} + If $g_1(x)$ from ? has degree $M>1$, then there are only a finite number of values of c for which $g_1(x)$ is constant relative to x. + \end{lemma} + \begin{proof} + Let $c$ be a value such that $g_1(x)=k(c)$, which is a polynomial only depending on $c$. Furthermore, let $\eta$ be a root of $g(x)=0$. Then the expression $f(x+c)=g(x)h(x)+g_1(x)$ can be written as $$f(\eta+c)=k(c).$$ + Now, if $\eta'$ is the number conjugate of $\eta$, then we also have $$f(\eta'+c)=k(c).$$ + Finally, we can can construct $$f(\eta+c)-f(\eta'+c)=0,$$ + which, given $f(x)$ and $\eta$, can only have a finite number of solutions in c. Therefore, there are infinitely many values of $c$ for which $g_1(x)$ is not constant. \end{proof} \begin{thebibliography}{2}