Suppose $f(x)$ and $g(x)$ are two integer polynomials with no common complex roots. For every integer $n,$ let $a_n = \gcd(f(n),g(n))$. Show that the sequence $\{a_n\}$ is periodic.
Note that $f$ and $g$ are coprime polynomials, since if they had a nontrivial factor in common, they would also have a complex root in common. So by the Euclidean algorithm, there exist rational polynomials $F$ and $G$ so that $f(x)F(x) + g(x)G(x) = 1.$ Multiplying all the coefficients of $F$ and $G$ by their least common denominator, we get that there is some positive integer $A$ so that $f(x)J(x) + g(x)K(x) = A$ where $J(x)=AF(x), K(x)=AG(x).$ Then I think one can show $a_{n+A} = a_n$ for all n. This clearly holds if $A = 1$ so we may assume otherwise. Also the gcd of $f(n)$ and $g(n)$ divides $A$ for every n and since $f$ and $g$ are integer polynomials, $f(n) \equiv f(m)\mod A$ for any $n\equiv m$ mod A. Hence $\gcd(f(n+A), g(n+A)) = \gcd(f(n)+Aj, g(n) + Ak)$ for some integers j and k. So $a_n | a_{n+A}.$ Thus we just need to show $a_{n+A}$ divides $a_n$.
Best Answer
Using your notation, it is clear that $gcd(f(m),g(m))$ will divide $A$ for any $m$. And again we can write $f(n+A)=f(n)+j_n A$ for some $j_n$ and $g(n+A)=g(n)+k_n A$.
So let $d$ be a common divisor of $f(n+A)$ and $g(n+A)$. Then $d$ will divide $\alpha (f(n)+j_n A)+\beta(g(n)+k_n A)$ for any choice of $\alpha,\beta\in\mathbb{Z}$. But since $d$ divides $A$, this implies $d$ divides $\alpha f(n)+\beta g(n)$, so $d|gcd(f(n),g(n))$ and therefore $gcd(f(n+A),g(n+A))|gcd(f(n),g(n))$. This will hold for all integers $n$.
Note that we could have played the same game by taking $-A$ instead of $A$ and note that $f(n)-f(n-A)=j'_n A$ for some integer choice $j'_n$ just by expanding the binomials, so we can get in the same way that $gcd(f(n),g(n))|gcd(f(n+A),g(n+A))$ by plugging in $n-A$ for $n$.