SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Without loss of generality one may assume that $f(1)\in \mathbb{Z}$. Since for all positive $k$, we have $\frac{f(k+1)}{f(1)}$ is integer, then we conclude that on all positive integer points our polynomial gets integer values. Assume that $\mathrm{deg}(f)=d$ then, according to Lagrange interpolation formula we get $$f(x)=\sum _{i=1}^{d+1}f(i)\frac{(x-1)\dots (x-(i-1))(x-(i+1))\dots (x-(d+1))}{(i-1)\dots 1(-1)\dots (i-(d+1))},$$ and all numbers are rational, so $p(x)$ is a polynomial with rational coefficients. By multiplying by a constant we can get $f(x)$ with integer coefficients. If $f(0)=0$ then we are done. Assume that $f(0)\ne 0$. Let's fix positive integer $k$ and denote $f(n)=f(n+1)f(n+2)\dots f(n+k)$ and $M=f(0)$. According to the problem condition, we have $\frac{f(n)}{M}\in \mathbb{Z}$ for all positive integer $n$. Since $f(x)$ is polynomial with integer coefficients, then $$f(-1)\equiv f(2|M|-1)\equiv 0(\mathrm{mod}M),$$ which means $$\frac{f(-1)}{M}=\frac{f(0)}{f(k)}\in \mathbb{Z},$$ for all positive integers $k$. It means $f(0)=0$ or $f(x)$ is divisible by $x$. $\square$