I asked this question on [MSE](https://math.stackexchange.com/questions/732854/integer-valued-polynomial-through-some-known-points) about 5 months ago, but, even after offering a bounty, I didn't receive any answer, I hope this question isn't too easy for MO.


If we have a set of points $(x\_i,y\_i)$ with rational coordinates and distinct $x$-coordinates, such that for every (if any) $x\_i\in\Bbb{N}$ the corresponding $y\_i$ is also an integer, is it always possible to build an integer valued polynomial with rational coefficents passing through those points?



Integer valued polynomial through some points with rational coordinates
No. The integer-valued polynomials have a basis (over $\mathbb{Z}$) given by the Newton polynomials 


$$\displaystyle {x \choose n} = \frac{x (x - 1)\dots(x - (n-1))}{n!}$$


and as such if 


$$\displaystyle f(x) = \sum a\_n {x \choose n}, a\_n \in \mathbb{Z}$$ 


is an integer-valued polynomial, then, for example, $f \left( - \frac{1}{2} \right)$ must have denominator a power of $2$, since


$$\displaystyle {-\frac{1}{2} \choose n} = \frac{(-1)(-3)\dots(-2n+1)}{2^n n!} = (-1)^n \frac{ {2n \choose n} }{2^n}$$


and hence no such polynomial $f$ can pass through a point like $\left( - \frac{1}{2}, \frac{1}{3} \right)$. 


