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Let $f(x)\in\mathbb{Q}[x]$ be a polynomial of degree $n$ such that $f(\mathbb{Z})\subseteq \mathbb{Z}$.

I want to show that $f$ has the following form $$f(x)=\sum_{j=0}^{j=n} a_{j}\binom{x}{j}$$ with $a_{j}\in \mathbb{Z}$

Attempt:

Base case $n=0$, is clear

Induction Hypothesis: Assume the result for degree$=n-1$

Consider the polynomial $\Delta f(x)=f(x+1)-f(x)$. One can observe that it has degree $n-1$, so $\Delta f$ has the above mentioned form.

Can I deduce something from here ?

Any other approach ?

Kindly correct the tags if necessary, I am studying Hilbert Polynomial and Hilbert Series.

If I am correct, The above problem characterize all numerical polynomial.

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    Related: http://math.stackexchange.com/q/410148 "This is a special case of a classical result of Polya and Ostrowski $(1920).$ Namely, $\,f(x) \in \mathbb Q[x]$ is an integer-valued polynomial, i.e. $f(\mathbb Z)\subset \mathbb Z\:,\:$ iff $f(x)$ is an integral linear combination of binomial coefficients ${x\choose k},\:$ for $\: k\le \deg f$. For a proof see e.g. Polya And Szego, *Problems and theorems in analysis*, vol II, [Problem 85 p. 129](http://books.google.com/books?id=pCa2CmAt8pwC&pg=PA129) and its [solution on p. 320.](http://books.google.com/books?id=pCa2CmAt8pwC&pg=PA320)" – 2017-02-05
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    Hint: by induction, you know what is $f(x+1)-f(x)$. So it is easy to find what is $f(n)$ for integers $n$. – 2017-02-05

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