Finding a common factor of two coprime polynomials I have two coprime univariate integer polynomials. Although they have no common factor as polynomials, they may have common factors at some (integer) values. How can I find such a value?
For example, suppose the polynomials are $x^3-x^2+3x-1$ and $x^3+2$. They are coprime in $\mathbb{Z}[x]$, but $\gcd(27^3-27^2+3\cdot27-1, 27^3+2)=\gcd(19034,19685)=31.$
 A: Here is an explanation.
Let $f=x^3-x^2+3x-1$ and $g=x^3+2$.
Even though $\gcd(f,g)=1$ in $\mathbb{Z}[x]$, we cannot write $1=uf+vg$ with $u,v \in \mathbb{Z}[x]$ (because $\mathbb{Z}[x]$ is not a PID). But we can, if we allow $u,v \in \mathbb{Q}[x]$. Indeed, WA tells us that
$$
1 = \dfrac1{31} (-6 x^2-7 x-3)f(x)+\dfrac1{31}(6 x^2+x+14)g(x)
$$
Therefore
$$
31 = (-6 x^2-7 x-3)f(x)+(6 x^2+x+14)g(x)
$$
for all $x \in \mathbb Z$.
This only proves that $\gcd(f(x),g(x))=1$ or $31$ (because $31$ is prime), but it points to $31$.
So we try to solve $f(x)\equiv g(x) \equiv 0 \bmod 31$ and find that $x=27$ is a solution.
Therefore, $\gcd(f(x),g(x))=31$ for all $x=27+31k$.
For all other values, $\gcd(f(x),g(x))=1$.
A: We know that : 
"two polynomials in $\mathbb{F}[x]$ have a common root (possibly in some field extension of $\mathbb{F}$)" $\Leftrightarrow$ "the resultant of these two polynomials is zero", which implies that for some prime $p$: 
"there is a positive integer $n$ such that $\gcd\big(f(n), g(n)\big)=p$" $\Rightarrow$ "the polynomials $f,g$ have a common root $\mod p$" $\Rightarrow$ "their resultant is zero $\mod p$" $\Leftrightarrow$ "their resultant is a multiple of $p$" 
(Here, we view the polynomials $f(x)=x^3-x^2+3x-1$, $g(x)=x^3+2$  as polynomials of $\mathbb{Z}_p[x]$) 
Thus, it suffices to investigate the prime divisors of the resultant: if $r$ is a divisor of the resultant $Res\big(f,g\big)=k$, then any solution of the simultaneous polynomial equations:
$$f(x) \equiv 0 \mod r  \\
g(x) \equiv 0 \mod r $$
provides a value of $x=n$ for which: $\gcd\big(f(n),g(n)\big)\geq r>1$.
So, using Mathematica:
In[1]:= Resultant[x^3 - x^2 + 3 x - 1, x^3 + 2, x]
Out[1]= 31
and thus:
$$
\bigg\{
\begin{array}{c}
  n^3 - n^2 + 3 n - 1\equiv 0\mod 31 \\
  % \\
  n^3+2\equiv 0\mod 31
\end{array}\bigg\} \Rightarrow n\equiv27\mod 31
$$
Therefore, as user lhf already mentioned in his answer: $\gcd(f(n),g(n))=31$ for all values $n=27+31k$, where $k\geq 0$. For all other values of $n$, $\gcd(f(n),g(n))=1$.
A: Easier: by a $\rm\color{#c00}{tw}\color{#0a0}{ist}$ on  Euclidean algorithm [cf. here & here] and a few minutes mental arithmetic
$$\begin{align} 
(x^3\!-\!x^2\!+\!3x\!-\!1,\,x^3\!+\!2) &= (-x^2\!+\!3x\!-\!3,\,x^3\!+\!2)\\
&= (-x^2\!+\!3x\!-\!3,\,6x\!-\!7)\\
&= (\color{#c00}6^2(-\color{#c00}x^2\!+\!3\color{#c00}x\!-\!3),\ \color{#0a0}{6x\!-\!7})\\
&=(-31,\,6x\!-\!7)\ \ {\rm by}\ \ \color{#c00}{6x}\equiv 7\!\!\!\pmod{\!\color{#0a0}{6x-7}}\\
&=(31,\, x\!+\!4)\ \ {\rm by}\ \bmod 31\!:\ x\equiv \small \frac{7}6 \equiv \frac{35}{30}\equiv \frac{4}{-1}
\end{align}\qquad\quad$$
So they are noncoprime for integer $x\iff x\equiv -4\equiv 27\pmod{\!31}$
