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I asked this question on MSE 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?

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  • $\begingroup$ If you're looking for an integer-valued polynomial such that $y_i=P(x_i)$ for each $i$, you should certainly assume the $x_i$ are distinct(!) $\endgroup$ – Anthony Quas Aug 18 '14 at 6:36
  • $\begingroup$ Yes, of course, I edited the question, thanks! $\endgroup$ – Alessandro Codenotti Aug 18 '14 at 6:53
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    $\begingroup$ Please include a link to the m.se question, and include a link there to this question. $\endgroup$ – Gerry Myerson Aug 18 '14 at 12:44
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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 $\left( - \frac{1}{2}, \frac{1}{3} \right)$.

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