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Suppose you have two polynomials $P$ and $Q$ with integer coefficients. Let their GCD (more specifically, the smallest integer multiple of the GCD in $\mathbb Q$ such that all the coefficients are integral) be $D$.

If, for a particular value $x_0$, we have that $P(x_0)$ and $Q(x_0)$ are both integers, does it follow that $D(x_0)$ is an integer? Does $D(x_0)$ divide $P(x_0)$ and $Q(x_0)$?

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    $\begingroup$ @Felipe, whatever is meant by gcd, I don't think the OP is assuming $D$ is monic. Consider $P=Q=2x$. Then you'd better take $D=2x$ as well, otherwise $x_0=1/2$ would be a cheap counterexample. $\endgroup$
    – Barry Cipra
    Apr 18, 2013 at 22:24
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    $\begingroup$ However it's defined (monic or not), surely you must have $P/D$ and $Q/D$ polynomials with integer coefficients. What do you take for $2x$ and $2x^2+x$? Either it's a counterexample, or you aren't requiring the quotients to be in $\mathbb{Z}[x]$. $\endgroup$
    – Zack Wolske
    Apr 18, 2013 at 22:43
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    $\begingroup$ GCD should be considered in $\mathbb{Z}[x]$ as Zack pointed out. $\endgroup$
    – Sungjin Kim
    Apr 19, 2013 at 0:40
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    $\begingroup$ $\mathbb{Z}[x]$ is not PID, but it is a UFD. So, it will be clear if we think GCD as in UFD. $\endgroup$
    – Sungjin Kim
    Apr 19, 2013 at 2:09
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    $\begingroup$ Then Zach already gave a counterexample... $\endgroup$
    – Felipe Voloch
    Apr 19, 2013 at 3:21

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