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May 29, 2014 at 15:29 comment added Jeremy Rouse Oops. Certainly the number $C$ can be taken to be the resultant, but the resultant need not be minimal.
May 29, 2014 at 15:05 comment added Felipe Voloch @JeremyRouse No. See: mathoverflow.net/questions/17501/…
May 29, 2014 at 13:28 review Close votes
Jun 3, 2014 at 13:58
May 29, 2014 at 13:16 comment added Jeremy Rouse The minimal choice of $C$ is called the resultant of $p(x)$ and $q(x)$.
May 29, 2014 at 13:08 comment added Vesselin Dimitrov Sure: just use Bezout's identity. You have $A(x)p(x) + B(x)q(x) = C$ with $A,B \in \mathbb{Z}[x]$ and $C \in \mathbb{Z} \setminus \{0\}$, and then the GCD of any value has to divide $C$.
May 29, 2014 at 12:59 history asked joro CC BY-SA 3.0