GCD of two polynomials in $ \mathbb{Z} [X]$ Let $(P,Q) \in ( \mathbb{Z} [X])^2$, such that $P$ and $Q$ don't have a common complex root, show that the sequence $\gcd(P(n),Q(n))_{n\ge0}$ is periodic. 
It seems to be a hard problem, please help.
 A: $P, Q$ not having a common complex roots means that in $\mathbb{Q}[x], \gcd(P,Q)= 1$. This means that there exist polynomials $A(x), B(x) \in \mathbb{Q}[x]$ such that $$P(x)A(x)+Q(x)B(x) = 1$$ by Bezout. Therefore, there exist polynomials $a(x), b(x) \in \mathbb{Z}[x]$ such that $$P(x)a(x)+Q(x)b(x) = M$$ (Just multiply $A(x), B(x)$ by an integer that kills all their fraction coefficients.) We will assume that $M$ is the smallest such integer.
Therefore, for all integers $n \ge 0$, $\gcd(P(n), Q(n))|M$. Now I will prove that $$\gcd(P(n+M), Q(n+M)) = \gcd(P(n), Q(n)).$$ Suppose $\gcd(P(n),Q(n)) = d$. Then again by Bezout, there exist integers $r,s$ such that $$P(n)r+Q(n)s = d.$$ By the identity $(a-b)|(f(a)-f(b))$, we have $M|(P(n+M)-P(n))$ and similar for $Q$. Therefore, $P(n) = P(n+M)+Mk_1$ and $Q(n) = Q(n+M)+Mk_2$ for integers $k_1, k_2$. Then $$(P(n+M)+Mk_1)r+(Q(n+M)+Mk_2)s = d$$ or, $$P(n+M)r+Q(n+M)s = d-rMk_1-sMk_2.$$ Taking this equation $\pmod {\gcd(P(n+M), Q(n+M))}$ gives $$\gcd(P(n+M), Q(n+M))|\gcd(P(n), Q(n)).$$ We can repeat the same argument for $\gcd(P(n+M), Q(n+M))$ in place of $\gcd(P(n), Q(n))$ to arrive at $$\gcd(P(n), Q(n))|\gcd(P(n+M), Q(n+M)).$$ Therefore, $$\gcd(P(n+M), Q(n+M)) = \gcd(P(n), Q(n))$$ so the sequence is periodic with period $M$ as desired. 
