" wrote:
we will show that the problem holds for $ k\geq 4$.
first of all note that $ F(k+1)-F(0)$ is a multiple of $ (k+1)$,on the other hand we know that $ \left|F(k+1)-F(0)\right|$ doesn't exceed $ k$,so we must have $ F(k+1)-F(0)=0$ i.e. $ F(k+1)=F(0)$ so there exists a polynomial $ G(x)$ with integer coefficients such that:
$ F(x)-F(0)=x(x-k-1)\cdot G(x)$
hence:
$ k\geq\left|F(c)-F(0)\right|=c(k+1-c)\cdot\left|G(c)\right|\textrm{ for } c=1,2,\ldots,k$ (1)
also note that the inequality $ c(k+1-c)>k$ holds whenever $ 2\leq c\leq k-1$ because:
$ c(k+1-c)>k\iff (c-1)(k-c)>0\iff 2\leq c\leq k-1$.
now note that if $ k\geq 3$,then the set $ \{2,\ldots,k-1\}$ would not be empty,so according to (1) we have $ \left|G(c)\right|<1$ which yields to the fact that for $ 2\leq c\leq k-1$,$ G(c)=0$.so there exists a polynomial $ H(x)$ with integer coefficients such that:
$ F(x)-F(0)=x(x-2)(x-3)\cdots (x-k+1)(x-k-1)\cdot H(x)$ (2)
now in order to complete our proof,it's sufficient to show that $ H(1)=H(k)=0$,which is easy enough,because for $ c=1,c=k$,the equation (2) is equivalent to:
$ k\geq\left|F(c)-F(0)\right|=(k-2)!\cdot k\cdot\left|H(c)\right|$.
now if $ k\geq 4$,then $ (k-2)!>1$ which yields to $ H(c)=0$.
now for $ k=1,2,3$ let:
$ k=1: F(x)=x(2-x)$
$ k=2: F(x)=x(3-x)$
$ k=3: F(x)=x(4-x)(x-2)^2$
so the problem holds iff $ k\geq 4$.
QED