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 $a \neq0\pmod{p}$. 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 $N \geq2$, 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 $N \geq2$, $f(x+c)-f(x+c_1)$ 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.
