Let $P(x)$ and $Q(x)$ be two polynomials with integer coefficients. If $\frac{P(n)}{Q(n)}$ is an integer for every integer $n$, prove that there exists a polynomial $S(x)$ with rational coefficients such that $P(x)=Q(x)S(x)$.
I feel that this problem isn't that hard, but I don't see a rigorous approach that proves this. Any help would be appreciated.
Thanks!!!
Let's assume the leading coefficient is positive.
First,$deg (P(x)) \geq deg(Q(x))$,for if $deg (P(x)) <deg(Q(x))$,$\frac{P(n)}{Q(n)}$ tends to zero when $n$ tends to zero,for large $n$ ,it must less than $1$,then for $m $ large than $n$,it is not integer.
Second,use $P(x)=f(x)Q(x)+r(x), 0\leq deg(r(x))< deg (Q(x))$, and get $$\frac{P(x)}{Q(x)}=f(x)+\frac{r(x)}{Q(x)}$$ use the reason like "First",for large $n$ ,$\frac{r(x)}{Q(x)}$ must less than $1$,but $f(x)$is rational coefficient,it must be integer for some large $n$, then they plus each other are not integer ,so $f(x)$ is $0$ or $r(x)$ is $0$,both means exist $S(x)$ makes $P(x)=S(x)Q(x)$