Polynomial $p(x)$ with real coefficients, which is different from the constant, has the following property:
for any naturals $n$ and $k$ the $\frac{p(n+1)p(n+2)...p(n+k)}{p(1)p(2)...p(k)}$ is an integer.
Prove that this polynomial is divisible by $x$.
wlog we can assume that $P(1)\in \mathbb{Z}$.
Claim 1 – $p(n)\in \mathbb{Z}, \forall n\in \mathbb{N}$.
Sketch of proof. From substitution $k=1$ we can conclude that $\dfrac{p(n+1)}{p(1)}\in \mathbb{Z}, \forall n \implies p(n)\in \mathbb{Z}, \forall n\in \mathbb{N}$.
Claim 2 – $p\in \mathbb{Q}[x]$.
Sketch of proof. Use Lagrange interpolation formula + $p(n)\in \mathbb{Z}$.
Claim 3 – $p(0)=0$.
Sketch of proof. Prove that $p(0)\ \vdots\ p(k), \forall k\in \mathbb{N}$. Fix $k$. Consider $P(n)=p(n+1)\dots p(n+k)$.
So, by polynomial remainder theorem we have that $p(x)\ \vdots\ (x-0)$.