Does $c(f) = \gcd(\{ f(n) | n \in \mathbb{Z} \})$? Consider $\sum_{i = 0}^n a_i x^i \in \mathbb{Z}[x]$. Recall that the content of a polynomial is the gcd of its coefficients. I'm wondering whether the content is equal to $\gcd ( \{ \sum_{i = 0}^n a_i k^i | k \in \mathbb{Z}  \} )$.
Context: I thought of the question when trying to think of a proof of Gauss's lemma, in which it is shown that c(fg) = c(f)c(g) for any two polynomials f and g. Of course the question above is true for degree 1 polynomials and constant polynomials, and so it seemed natural that it would hold for all degrees (and so I never thought to look for counterexamples in higher degrees as described below).
 A: Let $f(X)\in\mathbb{Q}[X]$ be integer-valued (i.e., $f(k)\in\mathbb{Z}$ for all $k\in\mathbb{Z}$).  If $f(X)$ is of degree $n$, then $f(X)=\sum_{i=0}^n\,A_i\,\binom{X}{i}$ for some $A_0,A_1,\ldots,A_n\in\mathbb{Z}$.  If the modified content $\tilde{C}(f)$ of $f$ is defined to be the gcd of all the $A_i$'s, then $\tilde{C}(f)$ is precisely the gcd of all $f(k)$ for $k\in\mathbb{Z}$.
In Andre Nicolas's first example, $X^2+X=0\binom{X}{0}+2\binom{X}{1}+2\binom{X}{2}$ so that $\gcd(0,2,2)=2$ is the gcd of all $k^2+k$ with $k\in\mathbb{Z}$.  For the second one, $X^3-X=0\binom{X}{0}+0\binom{X}{1}+6\binom{x}{2}+6\binom{X}{3}$ so that $\gcd(0,0,6,6)=6$ is the gcd of all $k^3-k$ with $k\in\mathbb{Z}$.
I believe that this is also true.  Let $S$ be an infinite subset of $\mathbb{Z}$ such that, for every prime $p\in\mathbb{N}$, integer $r>0$, and $j\in\left\{0,1,2,\ldots,p^r-1\right\}$, there exists $s\in S$ such that $s\equiv j\pmod{p^r}$.  Then, $\tilde{C}(f)$ is also the gcd of all $f(k)$ for $k\in S$.

  Why do I need the ridiculous requirement on $S$?  First, it is clear that $S$ should be infinite for the gcd to be established for every integer-valued polynomial in $\mathbb{Q}[X]$.  Then, it follows that, for any $j\in\mathbb{Z}$, $\gcd(S-j)=1$, otherwise $f(X)=X-j$ is a counterexample.  This implies that $\gcd(S-S)=1$.  However, this is still not enough (originally, I thought $\gcd(S-S)=1$ would be sufficient as mentioned in some of the comments below).  If there exist a prime $p\in\mathbb{N}$, an integer $r>0$, and $j\in\left\{0,1,2,\ldots,p^r-1\right\}$ such that $s\not\equiv j\pmod{p^r}$ for all $s\in S$, then we may assume that $j=p^r-1$ and consider $f(X)=\binom{X}{p^r-1}$.  In this case, $\tilde{C}(f)=1$ but it is easily seen by Lucas's Theorem that $p$ divides $f(k)$ for all $k\in S$.  Hence, $s\equiv j\pmod{p^r}$ must have a solution $s\in S$ in order to satisfy the condition that $\tilde{C}(f)$ is the gcd of $f(k)$ for $k\in S$. 

